Analytical Chemistry ¥° ('95 1st term, 2nd Exam. 95.5.11)

1. What volume of 0.0634M K₃Fe(CN)₆ solution contains 232.4mg of K⁺ ? (10 points)
   K:39.0983,  Fe:55.847,  C:12.011,  N:14.00674

2. Exactly 50.00mL of a solution of HClO₄ was added to a solution containing 0.5276g of primary-standard grade Na₂CO₃. The solution was boiled to remove CO₂, and the excess HClO₄ was back-titrated with 9.54mL of a NaOH solution. In a separate experiment, 30.00mL of NaOH neutralized 32.63mL of HClO₄. Calculate the molarity of HClO₄ and that of the NaOH. (15 points)
Na:22.98977,  C:12.011,  O:15.9994

3. List the following compounds in order of increasing molar solubility in water.
   Al(OH)₃,  Hg₂Br₂,  FeS,  Ce(IO₃)₄                               (10 points)
   For Al(OH)₃,  K_sp = 2.0 x 10^-32      For Hg₂Br₂,   K_sp = 5.8 x 10^-23
   For FeS,      K_sp = 6.0 x 10^-18      For Ce(IO₃)₄, K_sp = 4.7 x 10^-17

4. Calculate the Hg²⁺ and SCN^- concentration in a solution prepared by mixing 20.0mL of
   0.0300M Hg²⁺ with 30.0mL of 0.0500M SCN^-. (10 points)
   The equation of this complex-formation is
            Hg²⁺  +  2SCN^-  =  Hg(SCN)₂      ¥â₂ = 2.0 x 10¹⁷

5. Use activities to calculate the solubility of Ce(OH)₃ in the solution that results upon mixing 40.0mL of 0.0200M CeCl₃ with 60.00mL of 0.0200M Ba(OH)₂. (15 points)
   For Ce(OH)₃, Ksp = 2.0 x 10^-20,   ¥á(Ce³⁺) = 9.0¡Ê,  ¥á(OH^-) = 3.5¡Ê

6. What is the molar solubility of calcium carbonate, CaCO₃, in water ? (15 points)
   For CaCO₃, K_sp = 4.8 x 10^-9,  For H₂CO₃, K₁ = 4.45 x 10^-7,  K₂ = 4.7 x 10^-11

7. Using 1.0 x 10^-6M as criterion for quantitative removal, determine whether it is feasible to use IO₃^- to separate In³⁺ and Tl⁺ in a solution that is initially 0.12M in In³⁺ and 0.065M in Tl⁺. (10 points)
   For In(IO₃)₃, K_sp = 3.3 x 10^-11,     For TlIO₃, K_sp = 3.1 x 10^-6

8. Find that undissociated silver chloride in aqueous solution, AgCl(aq), is about 80% of the total dissolved silver at 0.0030M KCl solution. (15 points)
   For AgCl, K_sp = 1.82 x 10^-10            For AgCl(aq), K_d = 3.9 x 10^-4
   For AgCl₂^-,  K_f2 = 2.0 x 10^-5           For AgCl₃^2-,  K_f3 = 1.0

Analytical Chemistry ¥° ('95 1st term, 1st Exam. 95.3.30)

1. Describe each of the steps in a typical quantitative analysis. (10 points)

2. What are the sources of determinate error ?  And explain these sources. (8 points)

3. Calculate the result and estimate its absolute standard deviation for the following calculation. (7 points)
       y = [74(¡¾3) - 36(¡¾1)]/[123(¡¾2) + 65(¡¾4)] = ? (¡¾?)

4. Two sets of data were obtained by replicate measurements.
     set A :  43.23, 42.76, 43.54, 40.78, 42.94, 43.41, 42.62
     set B :  41.52, 40.98, 41.33, 42.67, 41.98, 42.36, 40.56
  (1) For set A, calculate the (a) mean,(2 points) (b) median,(2 points) (c) spread,(2points) (d) standard deviation,(3 points) (e) coefficient of variation(2 points).
  (2) Calculate the 95% confidence interval for data of set A and explain the meaning having this value. (7 points)
  (3) The fourth value in data of set A is likely to be anomalous. Should it be retained or rejected ?  Determine by using (a) T_n test( 95% confidence level) (5 points), and (b) Q test ( 90% confidence level ) (5 points)
  (4) Calculate the pooled standard deviation for data of set A and set B. (5 points)
  (5) Is it different the mean of set A from that of set B at 95% confidence level ? (7 points)

           Table.1 Value of t for various levels of probability

            Table.2 Critical Values for Rejection Quotient Q

            Table.3 Critical Values for Rejection Quotient T_n

5. 1.3256g sulfate compound sample was dried and found to contain 6.50% water. The sample yields 1.0362g of barium sulfate. Calculate the percentage of sulfur in dry sample.(Ba:137.34, S:32.064, O:15.9994, H:1.008) (7 points)

6. A mixture containing only FeCl₃ and AlCl₃ weighs 5.95g. The chlorides are converted to the hydrous oxides and ignited to Fe₂O₃ and Al₂O₃. The oxide mixture weighs 2.62g. Calculate the percent Fe and Al in the original mixture. (Fe:55.847, Al:26.9815, Cl:35.453, O:15.9994) (10 points)

7. A following calibration curve data for the colorimetric determination of phosphorus in urine is prepared by reacting standard solution of phosphate with molybdenum(¥µ) and reducing the phosphomolybdic acid complex to produce the characteristic blue color.
      ppm, P     1.00     2.00     3.00     4.00      urine sample
         A       0.205    0.410    0.615    0.820        0.625
  (a) Determine the linear equation by using the method of least square, and calculate the uncertainties in slope and intercept of this line. (12 points)
  (b) Calculate the phosphorus concentration and its uncertainty in urine sample.
      Absorbance 0.625 is the mean of five measurements. (6 points)

 ¨ç S_xx = ¥Ò(x_i - x)² =  ¥Òx_i² - (¥Òx_i)²/N            ¨è   S_yy = ¥Ò (y_i - y)² = ¥Ò y_i² - ( ¥Òy_i)²/N
 ¨é S_xy =  ¥Ò(x_i - x)(y_i - y) = ¥Ò x_iy_i - ¥Ò x_i¥Òy_i/N       ¨ê   x  =  ¥Ò x_i/N      ¨ë   y  =  ¥Ò y_i/N
 ¨ì slope    m = S_xy/S_xx      ¨í  intercept   b = y - mx     ¨î s_y = [(S_yy - m²S_xx)/(N-2)]^1/2  ¨ï s_m  = s_y/(S_xx)^1/2      ¨ð  s_b = s_y[1/{N-{(¥Òx_i)²/¥Òx_i²}}]^1/2
 ¨ñ s_c = (s_y/m)[(1/M) + (1/N) + {(y_c - y)²/m²S_xx}]^1/2

Analytical Chemistry ¥° ('94  1st term, 4th Exam. 94.6.21)

1. For a triprotic system , derive the following equation. (12 points)
   ¥á₂ = [HA^2-]/F = {[H⁺]K₁K₂}/{[H⁺]³ +[H⁺]²K₁ + [H⁺]K₁K₂ + K₁K₂K₃}
2. Describe the following terms. (12 points)
   (a) chelate effect, (b) displacement titration  (c) Jones reductor

3. Find the pH of 0.01M ammonium bicarbonate. (10 points)
   pK_b(NH₃) = 4.745,  for H₂CO₃, pK₁ = 6.352, pK₂ = 10.329

4. Expect the potential range over which the redox indicator color will change, when a saturated calomel electrode is used as the reference electrode. (12 points)

5. A 1.000mL sample of unknown containing Co²⁺ and Ni²⁺ was treated with 25.00mL of 0.03872M EDTA. Back titration with 0.02127M Zn²⁺ at pH 5 required 23.54mL to reach the xylenol orange end point. A 2.000mL sample unknown was passed through an ion exchange column that retards Co²⁺ more than Ni²⁺. The Ni²⁺ that passed through the column was treated with 25.00mL of 0.03872M EDTA and required 25.63mL of 0.02127M Zn²⁺ for back  titration. The Co²⁺ emerged from the column later. It, too, was treated with 25.00mL of 0.03872M EDTA. How many milliliters of 0.02127M Zn²⁺ will be required for back titration. (10 points)      

6. A 50.0mL solution containing Ni²⁺ and Zn²⁺ was treated with 25.0mL of 0.0383M EDTA to bind all the metal. The excess unreacted EDTA required 10.4mL of 0.0112M Mg²⁺ for  complete reaction. An excess reagent 2,3-dimercapto-1-propanol was then added to displace the EDTA from zinc. Another 28.3mL of Mg²⁺ was required for reaction with the  liberated EDTA. Calculate the molarity of Ni²⁺ and Zn²⁺ in the original solution. (10 points)

7. Sulfide ion was determined by indirect titration with EDTA. To a solution containing 25.00mL of 0.04312M Cu(ClO₄)₂ plus 15mL of 1M acetate buffer (pH 4.5) was added 25.00mL of unknown sulfide solution with vigorous stirring. The CuS precipitate was filtered and washed with hot water. Then ammonia was added to the filtrate (which contains excess Cu²⁺) until the blue color of Cu(NH₃)₄²⁺ was observed. Titration with 0.03835M EDTA required 12.05mL to reach the murexide end point. Calculate the molarity of sulfide in the unknown. (10 points)    

8. Calculate the potential (versus S.C.E. anode) at each point listed for the titration of 25.00mL of 0.0200M Cr²⁺ with 0.0100M Fe³⁺ : (a) 5.00mL, (b) 25.00mL, (c) 50.00mL, (d) 100.00mL. Esce = 0.241V, E¡Æ(Cr³⁺/Cr²⁺) = - 0.42V, E¡Æ(Fe³⁺/Fe²⁺) = 0.771V (12 points)

9. Nitrite(NO₂^-) can be determined by oxidation with excess Ce⁴⁺, followed by back titration of the unreacted Ce⁴⁺. A 5.166g sample of solid contaning only NaNO₂ and NaNO₃ was dissolved in 500.0mL. A 25.00mL sample of this solution was treated with 50.00mL of 0.1188M Ce⁴⁺ was back-titrated with 30.12mL of 0.04132M ferrous ammonium sulfate. Na:22.9898  N:14.0067  O:15.9994

2Ce⁴⁺+ NO₂^- + H₂O ¡æ 2Ce³⁺+ NO₃^- + 2H⁺       Ce⁴⁺+ Fe²⁺ ¡æ Ce³⁺+ Fe³⁺

Calculate the weight percent of NaNO₂ in the solid. (12 points)

Analytical Chemistry ¥° ('94  1st term, 3rd Exam. 94.6.2)

   Na:22.990   Cl:35.453    Br:79.904    K:39.098   O:15.999    H:1.008      C:12.011

1. Describe the following terms. ( 12 points )
   (a) Blank titration
   (b) Standardization
   (c) Back titration

2. Explain the Volhard titration, one of three argentometric titrations. (8 points)

3. 0.2458g sample which contained only NaCl and KBr was dissolved in water. When this  sample was titrated with 0.05027M , it was required 46.78mL for complete titration of both halides. Calculate the weight percent of Cl in the solid sample. (12 points)

4. Find that the pH of amphiprotic compound HL is (pK₁ + pK₂)/2. (10 points)

5. Calculate how many milliliters of 0.100M HCl should be added to how many grams of sodium acetate dihydrate (CH₃COONa 2H₂O) at 5¡É to prepare 250.0mL of 0.100M  buffer, pH 5.00.  At 5¡É, pK_a for acetic acid is 4.770. (10 points)

6. Explain the preparation of a buffer in real experiments. (13 points)

7. Describe the Gran plot. (10 points)

8. Consider the titration of 50.0mL of 0.0500M malonic acid (, K_a1 = 1.40 x 10^-3, K_a2 = 2.01 x 10^-6 ) with 0.100M NaOH. Calculate the pH at each point listed.  
   V_b = 0.0, 7.0, 12.5, 18.3, 25.0, 31.7, 37.5, 43.8, 50.0 and 56.3mL. (20 points)

9. The base B is too weak to titrate in aqueous solution. Which solvent, pyridine or acetic acid, would be more suitable for titration of B with HCl ? Why ? (5 points)

Analytical Chemistry ¥° ('94  1st term, 2nd Exam. 94.5.2)

1. Explain the following terms
   (1) disproportionation (5 points)
   (2) common ion effect (5 points)
   (3) peptization (5 points)

2. If AgCl(K_sp = 1.8 x 10^-10), AgBr(K_sp = 5.0 x 10^-13), AgI(K_sp = 8.3 x 10^-17), and  Ag₂CrO₄(K_sp = 1.1 x 10^-11) precipitate are in aqueous solution, what order will the solubility (in molarity) be increased ? (10 points)  

3. Explain the principle that HCl foundation is made. (10 points)

4. Is the solubility of a salt in aqueous solution increased or decreased by the addition of ions ?   Why ? (10 points)

5. Find the concentration of Ba²⁺ in a 0.0333M Mg(IO₃)₂ solution saturated with
   Ba(IO₃)₂ (K_sp = 1.57 x 10^-9).    (Ba²⁺) = 500pm,  (IO₃^-) = 450pm   (12 points)

6. Calculate the molarity of Ag⁺ in a saturated aqueous solution of Ag₃PO₄ at pH 5.0.
   K_sp(Ag₃PO₄) = 2.8 x 10^-18, for H₃PO₄,  K_a1 = 7.1 x 10^-3    K_a2 = 6.3 x 10^-8  
   K_a3 = 7.1 x 10^-13 (13 points)

7. State the way how larger particle is produced. (12 points)

8. Why is it less desirable to wash AgCl precipitate with aqueous NaNO₃ than with HNO₃ solution ? (5 points)

9. Twenty dietary iron tablets with a total mass of 22.131g were ground and mixed thoroughly. Then 2.998g of the powder was dissolved in HNO3 and heated to convert all the iron to Fe³⁺. Addition of NH₃ caused quantitative precipitation of Fe(OH)₃, which was ignited to give 0.246g of Fe₂O₃. What is the average mass of FeSO₄ 7H₂O in each tablet ? (13 points)  Fe = 55.847,  S = 32.066,  O = 15.999,  H = 1.008

Analytical Chemistry ¥° ('94  1st term, 1st Exam. 94.3.24)

1. A 43%(wt/wt) aqueous solution of HBr(MW= 80.912g) has a density of 1.40g/mL.
  (a) What is the formal concentration(M) of the solution? (7 points)
  (b) What is the mass of solution that contains 41.2g of HBr? (3 points)
  (c) What volume(mL) of solution contains 254mmole of HBr? (3 points)
  (d) How much this HBr solution is needed to prepare 0.250L of 0.150M HBr? (4 points)

2. Explain the principle that an electronic balance is operated. (10 points)

3. What do the symbols "TD" and "TC" mean on volumetric glassware ? (5 points)

4. Find the absolute uncertainty and percent relative uncertainty for each calculation. Express the answers with a reasonable number of significant figure.
  (a) { 42.3(¡¾ 1.0) + 90.2( ¡¾ 0.2) } ¡À  21.1(¡¾ 0.2) =     (5 points)
  (b) 0.002364( ¡¾ 0.000003) ¡À  0.02500(¡¾ 0.00005) =    (5 points)

5. Two methods were used to measure the specific activity of an enzyme.
     Method 1     143      147      156      148      145    
     Method 2     148      159      156      162      157
  (1) For method 2, calculate the (a) mean (2 points) (b) median (2 points)  (c) spread (2 points) (d) standard deviation. (3 points)
  (2) Calculate the 95% confidence interval for method 2, and explain the meaning of this confidence interval. (5 points)
  (3) The third value in method 1 is anomalous. Determine whether that value is rejected or retained by Q test( 90% confidence ) (5 points)
  (4) Calculate the pooled standard deviation, based upon data in method 1 and method 2 (5 points)
  (5) Is there significant difference between the means of method 1 and method 2 at the 95% confidence level ?  Why ? (5 points)

6. The following are relative peak area for chromatogram of standard solutions of benzene.
   Concentration of benzene(mM)   0.500   1.50   2.50    3.50     4.50    5.50      
        Relative peak area        3.76    9.16   15.03   20.42    25.33   31.97
  (1) Derive a least squares expression assuming the variables bear a linear relationship to one another. (12 points)
  (2) Plot the least-squares line as well as the experimental point. (3 points)
  (3) Calculate the standard deviation of intercept as well as the standard deviation of the slope. (5 points)
  (4) An unknown sample containing benzene yielded relative peak area of 27.50. Calculate the concentration and standard deviation of benzene in this sample. (9 points)    

            Table.1 Value of t for various levels of probability

            Table.2 Critical Values for Rejection Quotient Q

                                                                 
    m = |  ¥Òx_iy_i  ¥Òx_i |¡À   D           D = | ¥Ò(x_i²)  ¥Òx_i  |
           | ¥Òy_i      n   |                         | ¥Òx_i        n   |
                             
    b = | ¥Ò(x_i²)  ¥Òx_iy_i  | ¡À  D         ¥ò_y ¡Ö  s_y = [¥Ò(d_i-d)2/(degree of freedom)]^1/2
          | ¥Ò x_i      ¥Òy_i    |

     ¥ò_m² = ¥ò_y²n/D         ¥ò_b² = ¥ò_y²¥Ò(x_i²)/D
 

Analytical Chemistry ¥° ('93 1st term, 4th Exam. 93.6.18)

1. Explain the reasons why anhydrous acetic acid is usually chosen as a solvent for neutralization titration of very weak bases.(10 points)

2. Chromel is an alloy composed of Ni, Fe, and Cr. A 1.2825g sample was dissolved and  diluted to 500.0mL. When a 50.0mL aliquot of 0.05260M EDTA was mixed with an equal volume of the diluted sample, all three ions were  chelated, and a 6.23mL back-titration with 0.06241M Cu²⁺ was required. The Cr in a second 50.0mL aliquot was masked through the addition of hexamethylenediamine ; titration of the Fe and Ni required 38.32mL of 0.05260M EDTA. Fe and Cr were masked with pyrophosphate in a third 50.0mL aliquot, and Ni was titrated  with 26.32mL of 0.05260M EDTA solution. Calculate the percentage of Ni, Cr, and Fe in the alloy.(10 points) (Cr = 51.996, Fe = 55.847, Ni = 58.70)

3. The 25.00mL of 0.0200M Ni²⁺ solution maintained at pH 10.00 with NH₃/NH₄⁺ buffer was titrated with 0.010M Na₂H₂Y. Assume that the concentration of NH₃ remains 0.100M throughout. Calculate values for pNi after the addition of (a) 49.00mL, (b) 50.00mL,(c) 51.00mL. (20 points) K_f(NiY^2-) = 4.2 x 10¹⁸,   ¥á₄ at pH = 0.35,  For nickel ammine complexes, logK₁= 2.8, logK₂ = 2.2, logK₃ = 1.7, logK₄ = 1.2, logK₅ = 0.8, and logK₆ = 0.0

4. Calculate potentials after 50.0mL of 0.050M U⁴⁺ solution is added with (a)10.00mL, (b)49.90mL, (c) 50.00mL, (d) 50.10mL, and (e) 60.00mL of 0.100M V(OH)₄⁺. Assume that the pH is 1.00 throughout. (20 points)      UO₂²⁺ + 4H⁺ + 2e = U⁴⁺ + 2H₂O  E¢ª= 0.334V, V(OH)₄⁺ + 2H⁺ + e = VO²⁺ + 3H₂O   E¢ª= 1.000V

5. Air in the vicinity of a paper mill was analyzed for its sulfur dioxide content by drawing a quantity through 50.0mL of 0.01182M Ce(SO₄)₂ at the rate of 3.30L/min.
   Reaction: SO₂(g) + 2Ce⁴⁺ + 2H₂O ¡æ SO₄^2- + 2Ce³⁺ + 4H⁺

Upon completion of a 50.0min. sampling period, the excess Ce⁴⁺ was titrated with 11.32mL of 0.03264M Fe²⁺. Does the air meet the federal standard of 2ppm (or less) SO₂ ? (O = 16.000, S = 32.060)  Use 1.20 x 10^-3g/mL for the density of air. (15 points)

6. Survival of trout requires water in which the oxygen concentration is greater than 5ppm. A 100.0mL sample of lake water was analyzed according to the method using the following equation,    O₂(g) +4Mn(OH)₂(s) +2H₂O ¡æ 4Mn(OH)₃(s),   2Mn(OH)₃(s) +2I^- +6H⁺¡æ 2Mn²⁺+I₂ +6H₂O,    I₂ + 2S₂O₃^2- ¡æ 2I^- + S₄O₆^2-  with the liberated iodine requiring a 10.52mL titration with 0.01123M thiosulfate. Is the oxygen concentration sufficient to warrant the stocking of  trout in this lake. (10 points)

7. A 5.00mL portion of table wine was diluted to 100.0mL in volumetric flask. The ethanol in a 25.0mL aliquot was distilled into 100.0mL of 0.05232M K₂Cr₂O₇. Heating completed oxidation of the alcohol to acetic acid:   CH₃CH₂OH + 2Cr₂O₇^2- +16H⁺ ¡æ 3CH₃COOH + 4Cr³⁺ + 11H₂O following which the excess dichromate was titraed with 14.42mL of 0.03215M Fe²⁺. Calculate the weight-volume percentage of ethanol in the wine. (10 points) ( H = 1.008, C = 12.011, O = 16.000 )

8. Why is zinc amalgam rather than pure used in a Jones reductor ? (5 points)

9. What are your plans for this summer vacation ? (10 points)

Analytical Chemistry ¥° ('93 1st term, 3rd Exam. 93.5.24)

(H = 1.008, C = 12.011, N = 14.000, O = 16.000, S = 32.060, Cl = 35.453 )

1. Explain how the buffer solution of any desired pH is prepared. (10 points)

2. Why does an indicator have a transition range which extends over approximately 2 pH units ? (10 points)

3. The action of an alkaline I2 solution upon the rodencide warfarin, C₁₉H₁₆O₄, results in the formation of 1 mole of iodoform, CHI₃, for each mole of the parent compound reacted. Analysis for warfarin can then be based upon the following reaction between CHI₃ and Ag⁺
            CHI₃ + 3Ag⁺ + H₂O ¡æ 3AgI(s) + 3H⁺ + CO(g)
A 15.24g sample was treated with 20.00mL of 0.03010M AgNO₃, and the excess Ag⁺ was then titrated with 2.75mL of 0.05325M KSCN. Calculate the percentage of warfarin in the sample. (10 points)

4. A 5.272g sample containing NH₄Cl, (NH₄)₂SO₄, and inert materials was diluted to exactly 500.00mL. The Cl^- in a 50.00mL aliquot of this solution required 24.04mL of 0.0682M AgNO₃. The NH₄⁺ in a 25.00mL aliquot was converted to NH₃ and collected in 100.00mL of a 0.03120M sodium  tetraphenylborate solution.
         NH₃(g) + NaB(C₆H₅)₄ + H⁺ ¡æ NH₄B(C₆H₅)₄(s) + Na⁺
After the solid had been removed by filtration, titration of the filtrate and washings was required 8.30mL of 0.0682M AgNO₃ solution.
                 Ag⁺ + NaB(C₆H₅)₄ ¡æ AgB(C₆H₅)₄ + Na⁺
Calculate the percentage of NH₄Cl and (NH₄)₂SO₄ in the sample. (12 points)

5. (a) Calculate the pH of the solution that is 0.600M in glycolic acid and 0.500M in sodium glycolate. (10 points)
         HOCH₂COOH + H₂O  =  HOCH₂COO^-  + H₃O⁺         K_a = 1.48 x 10^-4

(b) Calculate the change in pH that occurs when 0.500mmole of HCl is added to 100.0mL of buffer solution (a). (10 points)

6. Calculate  ¥á-value for species derived from phosphoric acid in a solution buffered to a pH of 3.00. (10 points)
   ( H₃PO₄ : K_a1 = 7.11 x 10^-3,  K_a2 = 6.34 x 10^-8,  K_a3 = 4.20 x 10^-13 )

7. 50.00mL of a 0.100M ethylenediamine solution is titrated with 0.200M HCl solution. Calculate the pH after the addition of (a) 0.00, (b) 10.00, (c) 25.00, (d) 40.00, (e) 50.00, (f) 52.00mL of the acid. (18 points)    ethylenediamine( NH₂C₂H₄NH₂ )   K_b1 = 8.5 x 10^-5,    K_b2 = 7.1 x 10^-8

8. A Kjeldahl analysis was performed upon a 0.0600g sample of impure biguanide, C₂H₇N₅. The liberated ammonia, collected in 40.00mL of 4 % boric acid, was titrated with 20.23mL of 0.1124M HCl.  Calculate the percentage of biguanide in the sample. (10 points)

Analytical Chemistry ¥° ('93 1st term, 2nd Exam. 93.4.20)

1. Briefly describe the following terms.(20 points)
 (a) back-titration
 (b) primary standard
 (c) weight or gravimetric titrimetry
 (d) mass-action effect

2. Assume that an sample of sea water has a density of 1.030 and contains 1100 ppm Na⁺ and 300 ppm SO₄^2-.  Calculate the molar concentration of the two ions.  (atomic weight  Na: 22.9898   S: 32.0600   O: 15.9994) (10 points)

3. When a solution of HClO₄ was standardized against the solution of KBr dissolved 0.4385g of primary-standard grade HgO, the liberated OH- was neutralized with 42.49mL of this acid.           HgO(s) + 4Br^- + H₂O  ¡æ HgBr₄^2-  +  2OH^-
Calculate the molarity of the HClO₄ solution. (Hg: 200.5900,  O: 15.9994)  (10 points)

4. Write the following four compounds in order of decreasing molar solubility in water. (15 points)
   AgIO₃ (K_sp = 3.0 x 10^-8),             Sr(IO₃)₂ (K_sp = 3.3 x 10^-7),
   La(IO₃)₃ (K_sp = 6.2 x 10^-12),         Ce(IO₃)₄ (K_sp = 4.7 x 10^-17)

5. Use activities to calculate the molar solubility of La(OH)₃ in the solution that results when 50.0mL of 0.0300M LaCl₃ is mixed with 50.0mL of 0.100M KOH.
   (K_sp for La(OH)₃ = 2.0 x 10^-21,   ¥á(La³⁺) = 9.0¡Ê,   ¥á (OH^-) = 3.5¡Ê) (15 points)

6. Calculate the molar solubility of CdCO₃ in water by the systematic method.
   ( K_sp(CdCO₃) = 2.5 x 10^-14,   K_a1 for H₂CO₃ = 4.45 x 10^-7,
     K_a2 for H₂CO₃ = 4.70 x 10^-11)  (20 points)

7. A solution is 0.050M in Na₂SO₄ and 0.050M in NaIO₃. To this is added a solution containing Ba²⁺.  What is the concentration of anion that forms the less soluble barium salt when the more soluble precipitate begins to form ? (10 points)
   (K_sp for BaSO₄ = 1.30 x 10^-10,   K_sp for Ba(IO₃)₂ = 1.57 x 10^-9)

 Analytical Chemistry ¥° ('93 1st term, 1st Exam. 93.3.23)

1. What are the considerations in selecting a method of analysis ?(5 points)

2. Explain how the determinate method errors can be detected.(12 points)

3. Consider the following sets of replicate measurements.
   A : 0.842, 0.829, 0.834, 0.826, 0.830, 0.832
   B : 0.853, 0.849, 0.852, 0.856
  (1) For A set, calculate the (a) mean(2 points) (b) median(2 points) (c) spread (2 points) (d) standard deviation(3 points), and (e) coefficient of variation (3 points).
  (2) Calculate the 95% confidence limit for A set, and explain the meaning of this confidence limit.(5 points)
  (3) The first value in A set is anomalous. Determine whether that value is rejected or retained by     (a) Q test (96% confidence level)(4 points)  and (b) T_n test (95% confidence level)(4 points).
  (4) Calculate the pooled standard deviation, based upon data in A set and B set. (5 points)
  (5) Is there significant difference between the means of A set and B set at the 95% confidence level ?   Why ? (5 points)

4. Estimate the absolute standard deviation and the coefficient of variation for the result of the following calculation and round result.(5 points)
            y = 18.32( ¡¾ 0.03) - 1.37( ¡¾ 0.03) + 0.0052( ¡¾ 0.0002)

5. The following data were obtained in calibrating a cadmium ion electrode for the determination of pCd. There is a linear relationship between potential E and pCd.

       pCd              5.00     4.00       3.00      2.00       1.00
       E(mV)       - 53.8   - 27.7     + 2.7      31.9       65.1

   (1) Derive a least-squares expression for the best straight line through the points.(8 points)
   (2) Calculate the standard deviation for the slope and the intercept of the least-squares line.(6 points)
   (3) Calculate the pCd of a waste water in which the electrode potential was 18.7mV.(5 points)
   (4) Calculate the absolute standard deviation for the result if electrode potential were the mean of six measurements.(5 points)

 ¨ç S_xx = ¥Ò(x_i - x)² =  ¥Òx_i² - (¥Òx_i)²/N            ¨è   S_yy = ¥Ò (y_i - y)² = ¥Ò y_i² - ( ¥Òy_i)²/N
 ¨é S_xy =  ¥Ò(x_i - x)(y_i - y) = ¥Ò x_iy_i - ¥Ò x_iy_i/N       ¨ê   x  =  ¥Ò x_i/N      ¨ë   y  =  ¥Ò y_i/N
 ¨ì slope    m = S_xy/S_xx      ¨í  intercept   b = y - mx     ¨î s_y = [(S_yy - m²S_xx)/(N-2)]^1/2       ¨ï s_m  = s_y/(S_xx)^1/2      ¨ð  s_b = s_y[1/{N-{(¥Òx_i)²/¥Òx_i²}}]^1/2
 ¨ñ s_c = (s_y/m)[(1/M) + (1/N) + {(y_c - y)²/m²S_xx}]^1/2

6. Explain the conditions that larger particles can be obtained.(10 points)

7. A 0.6280g sample that contains NaCl, NaBr, plus impurities gives a precipitate of AgCl and AgBr that weighs 0.5064g. Another 0.6280g sample is titrated with 0.1050M AgNO₃ requiring 28.34mL. Calculate the percentage of NaCl and NaBr in the sample. (9 points)
 [formula weight(g) : AgCl = 143.32, AgBr = 187.77, NaCl = 58.44, NaBr = 102.90]

           Table.1 Value of t for various levels of probability

            Table.2 Critical Values for Rejection Quotient Q

            Table.3 Critical Values for Rejection Quotient T_n

ºÐ¼®È­ÇÐ ¥° ( 92Çг⵵ 1Çбâ 4Â÷½ÃÇè, 92.6.12 )

1. Nonaqueous neutralization titrationÀ» Çϱâ À§ÇØ amphiprotic solvent¸¦ ¼±ÅÃÇÒ ¶§ °í·ÁÇØ¾ß ÇÒ Á¡À» ethanolÀ» ÀÌ¿ëÇÏ¿© ¼³¸íÇ϶ó. (10Á¡)

2. ¿¡Åº¿ÃÀÇ autoprotolysis constant´Â 8.0 x 10^-20 ÀÌ°í, ¿¡Åº¿Ã¿¡¼­ aniline(C₆H₅NH₂)ÀÇ ¿°±âÇظ®»ó¼ö´Â 4.0 x 10^-14 ÀÌ´Ù. (a) anhydrous ethanol (pH = -log[C₂H₅OH₂⁺])°ú (b) ¹° ¿¡¼­ 0.01M C₆H₅NH₃⁺ ¿Í 0.02M C6H5NH2¸¦ Æ÷ÇÔÇÏ°í ÀÖ´Â ¿ë¾×ÀÇ pH¸¦ °¢°¢ ±¸Ç϶ó. ¹°¿¡¼­ÀÇ K_b = 3.94 x 10^-10 ÀÌ´Ù. (10Á¡)

3. Masking agent¶õ ¾î¶² ¿ªÇÒÀ» ÇÏ´Â °ÍÀΰ¡ ? ¿¹¸¦ µé¾î ¼³¸íÇ϶ó. (7Á¡)

4. Ȳ»êÀÌ¿Â(SO₄^2-)ÀÌ µé¾îÀÖ´Â 1.515g ½Ã·á ¼Ó¿¡ ÀÏÁ¤ °ú·®ÀÇ BaY^2-¿ë¾×À» ÷°¡ÇÏ°í »êÀÇ ³óµµ¸¦ Áõ°¡½ÃÅ°¹Ç·Î¼­ Ba²⁺°¡ Çظ®µÇ°Ô ÇÏ¿© ¸ðµç Ȳ»êÀÌ¿ÂÀ» BaSO₄ ħÀüÀ¸·Î ¸¸µé¾ú´Ù. ħÀüÀ» ¿©°úÇÏ°í ¾ÄÀº ÈÄ ¿©°ú¾×°ú ¾ÄÀº ¾×À» 250mL volumetric flask¿¡ ¸ð¾Æ pH10.0 buffer solutionÀ» ÀÌ¿ëÇÏ¿© ¿ë¾×À» Ç¥¼±±îÁö ¹±Çû´Ù. ÀÌ ¿ë¾× 25.0mL¸¦ ÃëÇÏ¿© 0.01545M Mg²⁺¿ë¾×À¸·Î ÀûÁ¤ÇÏ¿´´õ´Ï 28.73mL°¡ ¼ÒºñµÇ¾ú´Ù. ½Ã·áÁßÀÇ Na₂SO₄.10H₂OÀÇ ¹«°Ô ÇÔ·® %´Â ¾ó¸¶Àΰ¡ ? (10Á¡)

5. Pb, Mg, ZnÀ» Æ÷ÇÔÇÏ°í ÀÖ´Â ½Ã·á 0.4085gÀ» ³ì¿©¼­ CN^-À¸·Î ó¸®ÇÏ¿© Âø¹°À» ¸¸µé¾î ZnÀ» °¡¸®¿ö ÁÖ¾ú´Ù: Zn²⁺ + 4CN^- ¡æ Zn(CN)₄^2-. Pb¿Í Mg¸¦ ÀûÁ¤Çϴµ¥ 0.02064M EDTA°¡ 42.22mL°¡ ¼ÒºñµÇ¾ú´Ù. ±× ´ÙÀ½ Pb¸¦ BAL(2,3-dimercaptopropanol)·Î °¡¸®¿ö ÁÖ¾î EDTA°¡ ºÐ¸®µÇ¾î ³ª¿À°Ô ÇÏ¿© 0.007657M Mg¿ë¾×À¸·Î ÀûÁ¤ÇÏ¿´´õ´Ï 19.35mL°¡ Àû°¡µÇ¾ú´Ù. ¸¶Áö¸·À¸·Î formaldehyde¸¦ ³Ö¾îÁÖ¾î ZnÀ» demaskÇÏ¿´´Ù: Zn(CN)₄^2- + 4HCHO +4H2O ¡æ Zn²⁺ + 4HOCH₂CN + 4OH^-. ÀÌ°ÍÀ» 0.02064M EDTA·Î ÀûÁ¤ÇÏ¿´´õ´Ï 28.63mL°¡ Àû°¡µÇ¾ú´Ù. ½Ã·á ¼Ó¿¡ µé¾î ÀÖ´Â ¼¼ ±Ý¼ÓÀÇ ÇÔ·® ¹«°Ô %¸¦ ±¸Ç϶ó. (12Á¡)

6. 0.0500M U⁴⁺¿ë¾× 50.00mL¿¡ 0.0200M MnO₄^-¿ë¾×À» (a) 25.00mL  (b) 49.00mL (c)49.90mL
(d) 50.00mL (e) 50.10mL (f) 51.00mL¸¦ Àû°¡ÇÏ¿´À» ¶§ÀÇ Àü±ØÀüÀ§¸¦ ±¸ÇÏ°í, ÀûÁ¤°î¼±À» ±×·Á¶ó. (3Á¡ x 7)   ÀûÁ¤ÇÏ´Â µ¿¾È Ç×»ó [H⁺] = 1.00 À̶ó°í ÇÏÀÚ.
   ( E¢ª(UO₂²⁺/U⁴⁺) = + 0.334V,   E¢ª(MnO₄^-/Mn²⁺) = + 1.51V )

7. ´ÙÀ½ ¹ÝÀÀÀÇ Á¾¸»Á¡¿¡¼­ÀÇ Àü±ØÀüÀ§¸¦ ±¸Ç϶ó. (10Á¡)
   2Ce⁴⁺ + H₂SeO₃ + H₂O = SeO₄^2- + 2Ce³⁺ + 4H⁺   (in 1M H₂SO₄)
   ( E¢ª(SeO₄^2-/H₂SeO₃) = + 1.15V,  E^f(Ce⁴⁺/Ce³⁺) = 1.44V )

8. Ant-control¿¡ ÀÇÇØ ¸¸µé¾îÁø ½Ã·á 8.13gÀ» H2SO4¿Í HNO3¸¦ ÀÌ¿ëÇÏ¿© ºÐÇظ¦ ÇÏ°í ÀÌ ºÐÇع°¿¡ Á¸ÀçÇÏ´Â As¸¦ hydrazineÀ» ÀÌ¿ëÇÏ¿© ¸ðµÎ +3°¡·Î ¸¸µé¾ú´Ù. °ú·®ÀÇ È¯¿øÁ¦¸¦ Á¦°ÅÇÑ µÚ As³⁺ÀÇ ¾çÀ» ¾Ë±â À§ÇÏ¿© ¾àÇÑ ¿°±â¼º ¿ë¾×¿¡¼­ 0.02425M I₂·Î ÀûÁ¤ÇÏ¿´´õ´Ï 23.77mL°¡ µé¾î°¬´Ù. ¿ø·¡ ½Ã·á ¼Ó¿¡ µé¾îÀÖ´Â As₂O₃ÀÇ ÇÔ·® %´Â ¾ó¸¶Àΰ¡ ? (10Á¡)
   ( E¢ª(H₃AsO₄/H₃AsO₃) = + 0.559V,  E¢ª(I₂/I^-) = + 0.615V )

9. ´Ü´ÜÈ÷ ´ÝÇôÁ® ÀÖ´Â flask¿¡¼­ 0.01194M I2 50.00mL¿Í 1.657gÀÇ ½Ã·á¸¦ Àß Èçµé¾î ¹ÝÀÀ½ÃÄÑ È¥ÇÕ¹°¿¡ µé¾î ÀÖ´Â ethyl mercaptanÀÇ ¾çÀ» ±¸ÇÏ¿´´Ù.: 2C₂H₅SH + I₂ ¡æ 2I^- + C₂H₅SSC₂H₅ + 2H⁺,  °ú·®ÀÇ I₂ ¸¦ 0.01325M Na₂S₂O₃·Î ¿ªÀûÁ¤ÇÏ´Ï 16.77mL °¡ Àû°¡µÇ¾ú´Ù.  C₂H₅SH ÀÇ ÇÔ·® % ¸¦ ±¸Ç϶ó. (10Á¡)

   ¿øÀÚ·® : H: 1.008, C:12.011, O:16.000, Na: 22.990, Mg: 24.305, S:32.060, Zn: 65.380, As: 74.922, Pb: 207.2,

ºÐ¼®È­ÇÐ ¥° ( 92Çг⵵ 1Çбâ 3Â÷½ÃÇè, 92.5.27 )

1. Buffer solutionÀÇ definition, preparation ±×¸®°í buffer capacity¿¡ ´ëÇؼ­µµ ¼³¸íÇ϶ó. (12Á¡)

2. 39.9%(w/w) ÀÌ°í, ¹Ðµµ°¡ 1.200g/mL ÀÎ HCl ¼ö¿ë¾×ÀÇ pH´Â ¾ó¸¶Àΰ¡ ? (5Á¡)

3. 0.300M sodium mandelate(C₆H₅CH(OH)COONa) 250mL¿¡ 0.200M HClÀ» ÷°¡ÇÏ¿© pH 3.37ÀÎ ¿ÏÃæ¿ë¾×À» ¸¸µé·Á°í ÇÑ´Ù. 0.200M HClÀÇ ÇÊ¿äÇÑ ºÎÇÇ´Â ¾ó¸¶Àΰ¡? (7Á¡)
   K_a(C₆H₅CH(OH)COOH) = 3.88 x 10^-4

4. 0.05M H₂SO₄ ¿ë¾×ÀÇ ¼ö¼ÒÀÌ¿Â ³óµµ¸¦ °è»êÇ϶ó. ( K_a2 = 1.20 x 10^-2 ) (5Á¡)

5. Amphiprotic substance ÀÎ NaHA ÀÇ ¼ö¼ÒÀ̿³󵵰¡ [H3O⁺] ¡Ö  ( K_a1 x K_a2 )^1/2 ÀÓÀ» Áõ¸íÇ϶ó. (15Á¡)

6. 0.300M Na₂CO₃¿Í 0.200M HCl·ÎºÎÅÍ pH 9.60ÀÎ ¿ÏÃæ¿ë¾× 1.00L¸¦ ¸¸µå´Â ¹ý¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( H₂CO₃ ÀÇ K_a2 = 4.7 x 10^-11 ) (10Á¡)

7. 0.100M NaOH¿Í 0.080M hydrazine(H₂NNH₂)À¸·Î µÈ ¿ë¾× 50.0mL¸¦ 0.200M HClO₄·Î (a) 0.00mL  (b) 10.00mL  (c) 25.00mL  (d) 35.00mL  (e) 45.00mL  (f) 46.00mL ¸¸Å­ ÀûÁ¤ÇÏ¿´À» ¶§ °¢°¢ÀÇ pH¸¦ °è»êÇÏ°í ÀûÁ¤°î¼±À» ±×·Á¶ó. (3Á¡ x 7)
          H₂NNH₂ + H₂O = H₂NNH₃⁺ + OH^-      K_b = 1.3 x 10^-6

8. Kjeldahl method¿¡ ´ëÇØ ¼³¸íÇ϶ó. (12Á¡)

9. ¹éÆ÷µµÁÖ ½Ã·á 50.00mL¸¦ 0.03776M NaOH·Î ÀûÁ¤ÇÏ¿© Æä³îÇÁÅ»·» Á¾¸»Á¡¿¡ µµ´ÞÇÏ¿´À» ¶§ 21.48mL°¡ ¼ÒºñµÇ¾ú´Ù. Æ÷µµÁÖ 100mL´ç µé¾îÀÖ´Â tartaric acid ( H₂C₄H₄O₆, 2´ç·® )´Â ¸î gÀΰ¡ ?  (6Á¡) ( Æ÷µµÁÖ¿¡¼­ »êÀº tartaric acid ¹Û¿¡ ¾ø´Ù°í °¡Á¤ÇÑ´Ù. )

10. µµ½Ã°ø±â½Ã·á 3.00L¸¦ 0.0116M Ba(OH)₂ 50.0mL ¿ë¾×¿¡ Åë°ú½ÃÄÑ BaCO₃ ÇüÅ·ΠħÀü½ÃÄ×´Ù. °ú·®ÀÇ ¿°±â¸¦ 0.0108M HCl·Î ¿ªÀûÁ¤ÇÏ¿´´õ´Ï 23.6mL °¡ Àû°¡µÇ¾ú´Ù. ¸¸¾à CO2ÀÇ ¹Ðµµ°¡ 1.98g/L ¿´´Ù°í ÇÑ´Ù¸é °ø±âÁßÀÇ CO₂ ÇÔ·®Àº ppm( Áï mL CO₂/10⁶mL °ø±â)À¸·Î ¾ó¸¶À̰ڴ°¡ ? (7Á¡)

* ¿øÀÚ·® H:1.008, C:12.011, O:16.000, Cl:35.453

ºÐ¼®È­ÇÐ ¥° ( 92Çг⵵ 1Çбâ 2Â÷½ÃÇè, 92.5.6 )

1. concentrated H₂SO₄( d = 1.84g/mL, 95% )¸¦ ÀÌ¿ëÇÏ¿© 0.120M H₂SO₄ 500mL¸¦ ¸¸µé·Á°í ÇÑ´Ù. ¾î¶»°Ô ÇÏ¸é µÉ±î? ( H:1.01, S:32.06, O:16.00 ) (7Á¡)

2. »ìÃæÁ¦ ½Ã·á 1.534g¼Ó¿¡ µé¾î ÀÖ´Â arsenicÀ» Àû´çÇÑ ¹æ¹ýÀ» ÀÌ¿ëÇÏ¿© ¸ðµÎ H₃AsO₄·Î º¯È­½ÃÄ×´Ù. ±× ÈÄ ÀÌ »êÀ» ÁßÈ­½ÃŲ ´ÙÀ½ 0.08578M AgNO₃¸¦ Á¤È®È÷ 42.25mL ÷°¡ÇÏ´Ï ¸ðµç arsenicÀÌ Ag₃AsO₄ ħÀüÀÌ µÇ¾ú´Ù. ¿©°ú¾×°ú ħÀüÀ» ¾ÄÀº ¾×¿¡ µé¾î ÀÖ´Â °ú·®ÀÇ Ag⁺ ¾çÀ» ¾Ë¾Æº¸±â À§ÇÏ¿© 0.1000M KSCN À¸·Î ÀûÁ¤ÇÏ¿´´õ´Ï 10.35mL°¡ Àû°¡µÇ¾ú´Ù. ÀûÁ¤¹ÝÀÀÀº ´ÙÀ½°ú °°´Ù.    Ag⁺ + SCN^- = AgSCN   ½Ã·á¼Ó¿¡ µé¾îÀÖ´Â arsenicÀÌ ¸ðµÎ As₂O₃ ÇüÅ·ΠÁ¸ÀçÇÑ´Ù°í ÇÏ°í As₂O₃ÀÇ ÇÔÀ¯·® %¸¦ °è»êÇ϶ó.(As:74.92, O:16.00) (7Á¡)

3. ´ÙÀ½°ú °°Àº ¿ë¾× 100mL¿¡¼­ ³ìÀ» ¼ö ÀÖ´Â PbI₂ÀÇ ¾ç(g)À» °è»êÇ϶ó.
(a) H₂O (4Á¡)    (b) 4.00 x 10^-2M NaI (4Á¡)     (c) 4.00 x 10^-2M Pb(NO₃)₂ (4Á¡)    (d) 0.100M KClO₄ (¿©±â¼­´Â È°µ¿µµ¸¦ ÀÌ¿ëÇ϶ó. ¥á (Pb²⁺) = 4.5¡Ê,  ¥á(I^-) = 3.0¡Ê) (7Á¡)         (Pb:207.20, I:126.90,  K_sp(PbI₂) = 7.1 x 10^-9 ) Debye-Hückel½Ä   - log f_A = { 0.51 Z_A²( ¥ì)^1/2 } / { 1 + 0.33 ¥á_A(¥ì )^1/2 }

4. ¹°¿¡¼­ÀÇ MnSÀÇ ¿ëÇصµ¸¦ ±¸Ç϶ó. (15Á¡)
   ( K_sp = 3 x 10^-13, H₂SÀÇ K₁ = 5.7 x 10^-8, K₂ = 1.2 x 10^-15 )

5. 0.10M In³⁺ ¿Í 0.10M Tl⁺·Î ÀÌ·ç¾îÁø ¿ë¾×¿¡ ħÀüÁ¦·Î¼­ IO₃^- ¸¦ ÀÌ¿ëÇÏ¸é µÎ ÀÌ¿ÂÀ» Á¤·®ÀûÀ¸·Î ºÐ¸®ÇÒ ¼ö ÀÖÀ»±î? ¾øÀ»±î? ±× ÀÌÀ¯´Â?   K_sp( In(IO₃)₃ ) = 3.3 x 10^-11, K_sp( TlIO₃ ) = 3.1 x 10^-6 (7Á¡)

6. Cl^- ¿Í ClO₄^- ¸¦ Æ÷ÇÔÇÏ°í ÀÖ´Â ½Ã·á 2.215gÀÇ ½Ã·á¸¦ ¹°¿¡ ³ì¿© 250.0mL ¿ë¾×À» ¸¸µé¾ú´Ù. ÀÌ ¿ë¾× 50.0mL¸¦ ÃëÇÏ¿© 0.08625M AgNO₃ ¿ë¾×À¸·Î ÀûÁ¤ÇÏ¿´´õ´Ï 12.43mL°¡ Àû°¡µÇ¾ú´Ù. ¶Ç µû·Î ÀÌ ½Ã·á¿ë¾× 50.00mLÀ» ´Ù½Ã ÃëÇÏ¿© »ê¼º¿ë¾×ÇÏ¿¡¼­ V₂(SO₄)₃ ¸¦ ÀÌ¿ëÇÏ¿© ClO₄^-¸¦ ¸ðµÎ Cl^- ·Î ȯ¿ø½ÃÄ×´Ù.(ÀÌ ¶§ ¹Ù³ªµãÀº VO²⁺·Î »êÈ­µÇ°í SO₄^2-´Â º¯ÇÔÀÌ ¾ø´Ù.) ÀÌ È¯¿ø½ÃŲ ½Ã·á¸¦ 0.08625M AgNO₃ ¿ë¾×À¸·Î ÀûÁ¤ÇÏ¿´´õ´Ï 42.34mL °¡ Àû°¡µÇ¾ú´Ù. ½Ã·á¼ÓÀÇ Cl^- ¿Í ClO₄^-ÀÇ ¹«°Ô % ¸¦ °¢°¢ °è»êÇ϶ó. ( Cl:35.453, O:16.000 )(10Á¡)

7. 0.0400M KBr ¿ë¾× 50.0mL ¿¡ 0.0500M AgNO₃ ¸¦ (a) 5.00mL  (b) 35.00mL  (c) 39.00mL (d) 40.00mL  (e) 41.00mL  (f) 45.00mL¸¦ ÷°¡ÇÑ ÈÄÀÇ Ag⁺ ³óµµ¸¦ ±¸Ç϶ó. ±×¸®°í ÀÌ data¸¦ ÀÌ¿ëÇÏ¿© ÀûÁ¤°î¼±À» ±×·Á¶ó.  K_sp(AgBr) = 5.2 x 10^-13  (3Á¡ x 7)

8. ÀϹÝÀûÀ¸·Î ¿ë¾×¿¡¼­ ÀüÇØÁúÀÇ ³óµµ°¡ Áõ°¡ÇÔ¿¡ µû¶ó ħÀü¹ÝÀÀÀ̳ª Âø¹°¹ÝÀÀ µîÀÇ Çظ® ÆòÇüÀÌ Áõ°¡ÇÏ°Ô µÈ´Ù´Â ÀÌ·ÐÀû ±Ù°Å¸¦ ¹àÇô¶ó. (7Á¡)

9. Àº¹ý ÀûÁ¤ÁßÀÇ ÇϳªÀÎ Fajans¹ý¿¡ ´ëÇØ ¼³¸íÇ϶ó. (7Á¡)

ºÐ¼®È­ÇÐ ¥° ( 92Çг⵵ 1Çбâ 1Â÷½ÃÇè, 92.3.25 )
1. ´ÙÀ½ ¹Ýº¹ ÃøÁ¤µÈ °¢ sets¿¡ ´ëÇØ ¾Ë¾Æ º¸±â·Î ÇÏÀÚ.
       A: 0.325,  0.333,  0.328,  0.315,  0.335
       B: 0.341,  0.338,  0.336,  0.340,  0.335
a) ¿©±â¼­ A set¿¡ ´ëÇØ ©± mean (2Á¡) ©² median (2Á¡) ©³ spread or range (2Á¡) ©´ standard deviation    (3Á¡) ©µ coefficient of variation (3Á¡)À» ±¸Ç϶ó.
b) A set¿¡¼­ÀÇ 95% confidence intervalÀ» ±¸ÇÏ°í, ±× Àǹ̸¦ ¼³¸íÇ϶ó.(4Á¡)
c) A set¿¡ ÀÖ´Â 0.315¸¦ ¹ö¸± °ÍÀΰ¡ ¶Ç´Â ³õ¾ÆµÑ °ÍÀΰ¡¸¦ ©± Q test (96% confidence level) (3Á¡) ©²     T_n test(95% confidence level) (3Á¡)¸¦ ÀÌ¿ëÇÏ¿© °áÁ¤Ç϶ó.
d) set A¿Í set BÀÇ data¿¡ ´ëÇÑ pooled standard deviationÀ» ±¸Ç϶ó. (4Á¡)
e) set A¿Í set BÀÇ dataÀÇ Æò±Õ°ªÀº 95% confidence level¿¡¼­ ´Ù¸¥°¡? ÀÌÀ¯´Â? (4Á¡)
f) B setÀÇ data°¡ A setÀÇ dataº¸´Ù 95% confidence level¿¡¼­ precisionÀÌ ´õ ÁÁÀº°¡? ÀÌÀ¯´Â? (4Á¡)

           Table.1 Value of t for various levels of probability

            Table.2 Critical Values for Rejection Quotient Q

            Table.3 Critical Values for Rejection Quotient T_n

Table.4  Critical Values for F at the 5% level

2. ´Ü¹éÁúÀÇ ÇÔ·®À» ±¸Çϱâ À§ÇÏ¿© Èí±¤µµ¸¦ ÃøÁ¤ÇÏ¿© ¾òÀº calibration curve¸¦ ÀÌ¿ëÇÏ·Á°í ÇÑ´Ù.
   ----------------------------------------------------------------------
       ´Ü¹éÁú ( g)      0.00      9.36      18.72     28.08     37.44
         Èí±¤µµ         0.466     0.676     0.883     1.086     1.280
   ----------------------------------------------------------------------
 a) method of least square¸¦ ÀÌ¿ëÇÏ¿© °¡Àå ÁÁÀº Á÷¼±½ÄÀ» ±¸Ç϶ó. ÀÌ ¶§ slopeÀÇ standard deviationµµ ±¸ÇÏ°í À¯È¿¼ýÀÚ °³³ä¿¡µµ À¯ÀÇÇ϶ó. (15Á¡)
 b) À§ÀÇ ½ÇÇè data¿Í a)¿¡¼­ ±¸ÇÑ Á÷¼±½ÄÀÌ ³ªÅ¸³»¾îÁø graph¸¦ ±×·Á¶ó. (3Á¡)
 c) 3¹ø ÃøÁ¤ÇÏ¿© Æò±ÕÇÏ¿© ¾òÀº unknown protein sampleÀÇ Èí±¤µµ°¡ 0.973 À̾ú´Ù°í ÇÏ°í unknown¼Ó¿¡ µé¾îÀÖ´Â proteinÀÇ ÇÔ·®°ú uncertainty¸¦ ³ªÅ¸³»¾î¶ó. (7Á¡)

 ¨ç S_xx = ¥Ò(x_i - x)² =  ¥Òx_i² - (¥Òx_i)²/N            ¨è   S_yy = ¥Ò (y_i - y)² = ¥Ò y_i² - ( ¥Òy_i)²/N
 ¨é S_xy =  ¥Ò(x_i - x)(y_i - y) = ¥Ò x_iy_i - ¥Ò x_iy_i/N       ¨ê   x  =  ¥Ò x_i/N      ¨ë   y  =  ¥Ò y_i/N
 ¨ì slope    m = S_xy/S_xx      ¨í  intercept   b = y - mx     ¨î s_y = [(S_yy - m²S_xx)/(N-2)]^1/2       ¨ï s_m  = s_y/(S_xx)^1/2      ¨ð  s_b = s_y[1/{N-{(¥Òx_i)²/¥Òx_i²}}]^1/2
 ¨ñ s_c = (s_y/m)[(1/M) + (1/N) + {(y_c - y)²/m²S_xx}]^1/2

3. ´ÙÀ½ °è»ê °á°ú¿¡ ´ëÇØ absolute standard deviation°ú coefficient of variationÀ» °è»êÇ϶ó.
   a) y = 243( ¡¾1) x 760( ¡¾2) ¡À 1.006( ¡¾0.006) =              (5Á¡)
   b) y = [ 1.30(¡¾ 0.04) x 10^-16 ]^1/3  =                    (5Á¡)

4. ºÒ¼ø¹°À» Æ÷ÇÔÇÑ magnesite, MgCO₃ ½Ã·á 0.8507gÀ» HCl·Î ºÐÇؽÃŲ ÈÄ CO₂¸¦ ³¯·Áº¸³»¾î calcium oxide¿¡ ¸ð¾Æ Áú·®À» ´Þ¾Æº¸´Ï 0.1934g À̾ú´Ù. ½Ã·á¼Ó¿¡ µé¾îÀÖ´Â MgÀÇ %¸¦ °è»êÇ϶ó. ( Mg: 24.312, C: 12.011, O: 16.000 )  (10Á¡)

5. 0.300gÀÇ BaCl₂.2H₂O¸¦ Æ÷ÇÔÇÏ°í ÀÖ´Â ¿ë¾× 50.0mL¿Í 0.200gÀÇ NaIO₃¸¦ Æ÷ÇÔÇÏ°í ÀÖ´Â ¿ë¾× 50.0mL¸¦ È¥ÇÕÇÏ¿´´Ù. ¹°¿¡¼­ Ba(IO₃)₂°¡ ¿ëÇصÇÁö ¾Ê´Â´Ù°í ÇÑ´Ù¸é (a) ħÀüµÈ Ba(IO₃)₂ ÀÇ ¹«°Ô´Â ¾ó¸¶Àΰ¡ ? (6Á¡) (b) ¿ë¾×¿¡ ¹ÝÀÀÇÏÁö ¾Ê°í ³²¾Æ ÀÖ´Â È­ÇÕ¹°ÀÇ  Áú·®Àº ¾ó¸¶Àΰ¡ ? (6Á¡)
   ( Ba: 137.34, Cl: 35.453, H: 1.008, Na: 23.000, I: 126.904, O: 16.000 )

6. a) Zeta potentialÀ̶õ ¹«¾ùÀ̸ç(5Á¡) b) ÀÌ°ÍÀ» ÀÌ¿ëÇÏ¿© colloid°¡ ¾ÈÁ¤ÇÑ ÀÌÀ¯(2Á¡) ¿Í  c) coagulationÀÌ ÀϾ´Â ÀÌÀ¯(2Á¡)¿¡ ´ëÇØ ¼³¸íÇ϶ó.

ºÐ¼®È­ÇÐ ¥° ( 91Çг⵵ 1Çбâ 4Â÷½ÃÇè, 91.6.17 )

1. ÀϹÝÀûÀÎ redox indicatorÀÇ half-reactionÀº ´ÙÀ½°ú °°ÀÌ ¾µ ¼ö ÀÖ´Ù;
             In + 2H⁺ + 2e = InH₂     E¢ª= 0.600V.
 pH°¡  a) 6.00   ±×¸®°í   b) 8.00 ÀÎ ¿ë¾×¿¡¼­ÀÇ  ÀÌ Áö½Ã¾àÀÇ transition potential range¸¦ °¢°¢ °è»êÇ϶ó. ( 10Á¡ )

2. ´ÙÀ½ ¹ÝÀÀ¿¡ ´ëÇÑ ´ç·®Á¡¿¡¼­ÀÇ Àü±ØÀüÀ§¸¦ °è»êÇ϶ó. ´ç·®Á¡¿¡¼­ÀÇ [H⁺]= 0.100M ÀÌ´Ù.  ( 15Á¡ )
    MnO₄^- + HNO₂ = Mn²⁺ + NO₃^-   [ E¢ª(MnO₄^-/Mn²⁺)= 1.51V,  E¢ª(NO₃^-/HNO₂)= 0.94V ]

3. 0.100M V²⁺ 50mL¿ë¾×¿¡ 0.05M Sn⁴⁺·Î ÀûÁ¤ÇÒ ¶§ 0.05M Sn⁴⁺°¡  a) 10mL  b) 49mL  c) 50mL    d) 51mL   µé¾î °¬À» ¶§ÀÇ electrode potentialÀ» °è»êÇ϶ó. ( 15Á¡ )
   [ E¢ª(V³⁺/V²⁺)= -0.256V,  E¢ª(Sn⁴⁺/Sn²⁺)= 0.94V ]

4. Karl Fisher titration¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( ½Ã¾à, ¼ºÁú, ´ç·®Á¡ °üÂû, ÀÀ¿ëµî ) ( 15Á¡ )

5. Ce⁴⁺´Â ¿°±â¼º¿ë¾×ÇÏ¿¡¼­ ȯ¿øÁ¦¸¦ Àý´ë·Î ÀûÁ¤ÇÏÁö ¾Ê´Â´Ù. ±× ÀÌÀ¯´Â ? ( 5Á¡ )

6. Na₂S₂O₃¿ë¾×À» Ç¥Á¤Çϱâ À§ÇÏ¿© K₂Cr₂O₇ 0.1518gÀ» ¹±Àº HCl¿ë¾× ¼Ó¿¡¼­ ³ìÀÎ ÈÄ °ú·®ÀÇ I^-¸¦ °¡ÇÏ°í À̶§ »ý¼ºµÈ I₂¸¦ ÀÌ Na₂S₂O₃·Î ÀûÁ¤ÇÏ¿´´õ´Ï 39.75mL°¡ µé¾î°¬´Ù. Na₂S₂O₃ÀÇ M³óµµ´Â ¾ó¸¶Àΰ¡ ? ( 10Á¡ )   ( K: 39.098   Cr: 51.996  O: 15.999 )
 [ E¢ª(Cr₂O₇^2-/Cr³⁺)= 1.33V,   E¢ª(I₂/I^-)= 0.536V,   E¢ª(S₄O₆^2-/S₂O₃^2-)= 0.08V ]

7. Alkali metal sulfateÀÇ ½Ã·á 0.6490gÀ» ³ì¿© 100mL°¡ µÇ°Ô ÇÏ¿´´Ù. ÀÌ Áß 25.00mL¸¦ ÃëÇÏ¿© ¿©±â¿¡ µé¾îÀÖ´ø Na⁺¸¦ NaZn(UO₂)₃(OAc)₉.6H₂O ·Î ħÀü½ÃÅ°°í °É·¯¼­ ¾ÄÀº ÈÄ »êÀ¸·Î ³ì¿© Jones reductor¿¡¼­ uraniumÀ» U³⁺·Î ȯ¿ø½ÃÄ×´Ù. ´Ù½Ã °ø±â¸¦ ÅëÇÏ¿© U³⁺¸¦ U⁴⁺·Î ¸¸µç ÈÄ 0.1086M Ce⁴⁺·Î ÀûÁ¤ÇÏ¿´´õ´Ï 33.33mL°¡ Àû°¡µÇ¾ú´Ù. ½Ã·á ¼Ó¿¡ µé¾î Àִ Na₂SO₄ÀÇ ¹«°Ô %¸¦ °è»êÇ϶ó. ( 10Á¡ )  ( Na: 22.990,  O: 15.999,  S: 32.06 )     [ E¢ª(Ce⁴⁺/Ce³⁺)= 1.44V,   E¢ª(UO₂²⁺/U⁴⁺)= 0.334V ]

8. Cu²⁺´Â chelating agent H₂L°ú ¹ÝÀÀÇÏ¿© Âø¹° CuL₂^2-¸¦ ¸¸µé°í CHCl₃¿¡ ½±°Ô ³ì°Ô µÈ´Ù. Cu²⁺ÀÇ 1.00 x 10^-4M ¼ö¿ë¾×À» 0.01M H₂L ÀÌ ³ì¾ÆÀÖ´Â CHCl₃·Î extractÇÏ¿´À» ¶§ µÎ »ó »çÀÌ¿¡¼­ÀÇ Cu³óµµ´Â pH 5.65ÀÏ ¶§ ¶È°°¾Ò´Ù. (*** »ç½ÇÀº ÃßÃâµÇÁö ¾ÊÀ½. ¹®Á¦°¡ À߸øµÇ¾ú´Ù. )
a) CuL₂^2-°¡ À¯±â»ó¿¡¼­ Çظ®µÇÁö ¾Ê´Â´Ù°í Çϸé ÀÌ system¿¡¼­ ÆòÇüÀ» ³ªÅ¸³»´Â ½ÄÀ» ¸ðµÎ ½á¶ó. ( 5Á¡ )
b) K_ex¸¦ °è»êÇ϶ó. ( 5Á¡ )
c) pH 6.00 ÀÏ ¶§ ÀÌ systemÀÇ distribution ratio¸¦ °è»êÇ϶ó. ( 5Á¡ )
d) ¸¸¾à pH 6.00À¸·Î ¿ÏÃæµÈ 5.00 x 10^-5M Cu²⁺ 50mL¸¦ 0.0100M H2LÀÌ Æ÷ÇÔµÈ CHCl₃ 25.00mL·Î extractÇÏ·Á¸é ¼ö¿ë¾×»ó¿¡ ÀÖ´Â Cu¸¦ 99.9 % Á¦°ÅÇϱâ À§ÇØ ÇÊ¿äÇÑ extractionÀÇ ¼ö´Â ¾ó¸¶Àΰ¡ ? ( 5Á¡ )

ºÐ¼®È­ÇÐ ¥° ( 91Çг⵵ 1Çбâ 3Â÷½ÃÇè, 91.6.3 )

1. Buffer solutionÀ» °¡Àå Àß ¸¸µé ¼ö ÀÖ´Â ¹æ¹ýÀ» buffer capacity¿Í ionic strengthÀÇ ÃøÁ¤ °ï¶õÀ» °í·ÁÇÏ¿© ¼³¸íÇ϶ó. ( 10Á¡ )

2. 2.0 x 10^-7M HCl ¼ö¿ë¾×ÀÇ pH´Â ? ( 5Á¡ )

3. pH°¡ 3.50ÀÎ ¿ÏÃæ¿ë¾×À» ¸¸µé±â À§ÇØ 1.00M formic acid( HCOOH ) 400mL¿¡ ÷°¡Çؾ߸¸ ÇÏ´Â NaOH´Â ¸î g Àΰ¡ ? ( HCOOHÀÇ K_a= 1.77 x 10^-4, Na: 23.00, O: 16.00, H: 1.00, NaOH¸¦ ÷°¡Çصµ ºÎÇÇÀÇ º¯È­°¡ ¾ø´Ù°í °¡Á¤ÇÑ´Ù. ) ( 10Á¡ )

4. H₂L⁺´Â À̾缺ÀÚ»êÀ¸·Î¼­ÀÇ ¿ªÇÒÀ» ÇÏ¿© HL, L^-µîÀ¸·Î µÇ±âµµ Çϴµ¥ À̶§ HL¸¸ÀÌ Á¸ÀçÇÒ ¶§ÀÇ [H⁺]´Â ´ÙÀ½°ú °°À½À» Áõ¸íÇ϶ó. ( 15Á¡ )
          [H⁺] ¡Ö  {(K₁K₂C_HL + K₁K_w)/(K₁ + C_HL)}^1/2 ¡Ö  (K₁K₂)^1/2
 ¿©±â¼­ K₁,K₂´Â H₂L⁺ÀÇ »ê Çظ®»ó¼öÀÌ°í C_HLÀº HL ÀÇ Ã³À½³óµµ(Áï formal ³óµµ)ÀÌ´Ù.

5. 0.100M Na₂CO₃¿ë¾× 50mL¸¦ 0.200M HCl¿ë¾×À¸·Î ÀûÁ¤ÇÏ·Á°í ÇÑ´Ù. ÀÌ HClÀÇ Àû°¡ºÎÇÇ°¡  a) 0.00mL   b) 12.50mL   c) 24.00mL   d) 25.00mL   e) 26.00mL   f) 37.50mL g) 49.00mL  h) 50.00mL i) 51.00mL  ÀÏ ¶§ÀÇ °¢°¢ pH¸¦ ±¸ÇÏ°í, ÀûÁ¤°î¼±À» ±×·Á¶ó.
   H₂CO₃ÀÇ K_a1= 4.45 x 10^-7, K_a2 = 4.7 x 10^-11 ÀÌ´Ù ( 15Á¡ )

6. Carbonate error¿¡ ´ëÇØ ¼³¸íÇÏ°í ÀÌ°ÍÀ» ÇÇÇÒ ¼ö ÀÖ´Â ¹æ¹ýÀ» ½á¶ó. ( 5Á¡ )

7. Kjeldahl method¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 10Á¡ )

8. Amphiprotic solvent»ó¿¡¼­ neutralizationÇÒ·Á°í ÇÒ ¶§ °í·ÁÇØ¾ß ÇÒ Á¡¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 15Á¡ )

9. BrO₃^-¿Í Br^-¸¦ Æ÷ÇÔÇÏ°í ÀÖ´Â ½Ã·á 3.650g À» ¹°¿¡ ³ì¿© 250mL°¡ µÇ°Ô ÇÏ¿´´Ù. »ê¼ºÈ­½ÃŲ ÈÄ µû·Î ÃëÇÑ ½Ã·á 25mL¿¡ AgNO₃¸¦ ³Ö¾î AgBrħÀüÀ» ½ÃÄÑ ¿©°úÇÏ°í ¾ÄÀº ÈÄ Ni(CN)₄^2- ¿ë¾×¿¡¼­ ³ì¿´´Ù; Ni(CN)₄^2- + 2AgBr(S) ¡æ 2Ag(CN)₂^- + Ni²⁺ + 2Br^- . À̶§ ºÐ¸®µÇ¾î  ³ª¿Â Ni²⁺À» 0.02089M EDTA·Î ÀûÁ¤ÇÏ¿´´õ´Ï 26.73mL°¡ ¼Ò¸ðµÇ¾ú´Ù. ´Ù½Ã ½Ã·á 10mL¸¦ ÃëÇÏ¿© As³⁺¸¦ ³Ö¾î ¸ðµç BrO₃^-¸¦ ȯ¿ø½ÃÄÑ Br^-¸¦ ¸¸µç ÈÄ AgNO₃¸¦ ÷°¡ÇÏ¿© AgBrħÀüÀ» ½ÃŲ ÈÄ À§¿Í °°Àº ¹æ¹ýÀ¸·Î ³ìÀÌ°í ³ª¿Â Ni²⁺¸¦ °°Àº ³óµµÀÇ EDTA·Î ÀûÁ¤ÇÏ¿´´õ´Ï 21.94mL°¡ ÷°¡ µÇ¾ú´Ù. ½Ã·áÁßÀÇ NaBr°ú NaBrO₃ÀÇ % ¸¦ ±¸Ç϶ó. ( 15Á¡ )
 ( Na: 23.00,  Br: 79.90,  O: 16.00 )

ºÐ¼®È­ÇÐ ¥° ( 91Çг⵵ 1Çбâ 2Â÷½ÃÇè, 91.5.14 )

1. Concentrated H₂SO₄ ( 95% w/w, ¹Ðµµ 1.84g/mL )¸¦ ÀÌ¿ëÇÏ¿© 0.08M H₂SO₄ 2.0L¸¦ ¸¸µé·Á°í ÇÑ´Ù. ¾î¶»°Ô ÇÏ¸é µÉ±î ? ( H: 1.0079, S: 32.06, O: 15.9994 )  ( 10Á¡ )

2. 1.223gÀÇ »ìÃæÁ¦½Ã·á¿¡ µé¾î ÀÖ´Â ¸ðµç As¸¦ Àû´çÇÑ Ã³¸®¹æ¹ýÀ¸·Î H3AsO4·Î ¸¸µç ÈÄ ÀÌ »êÀ» ÁßÈ­½ÃÅ°°í 0.07891MÀÇ AgNO₃¸¦ Á¤È®È÷ 40.00mL¸¦ ÷°¡ÇÏ¿© ¸ðµÎ Ag₃AsO₄·Î ħÀü½ÃÄ×´Ù.  Ä§ÀüÀ» ºÐ¸®Çس½ ÈÄ ¿©ºÐÀ¸·Î ³²¾ÆÀÖ´ø Ag⁺¸¦ 0.100M KSCNÀ¸·Î ÀûÁ¤ÇÏ¿´´õ´Ï 11.27mL°¡ ÷°¡µÇ¾ú´Ù. ÀÌ ¶§ »ý±ä ħÀü»ý¼º¹°Àº AgSCNÀÌ´Ù. ½Ã·áÁßÀÇ As₂O₃ÀÇ ÇÔ·® % ¸¦ ±¸Ç϶ó. ( As: 74.9216 )  ( 10Á¡ )

3. 1 titer°¡ 11.0mg Fe₂O₃/mLÀÎ KMnO₄ ¿ë¾×ÀÇ molarity´Â ¾ó¸¶Àΰ¡ ?  ¹ÝÀÀÀº ´ÙÀ½°ú °°´Ù. ( Fe: 55.847 ) ( 10Á¡ )
      MnO₄^- +  5Fe²⁺  +  8H⁺  ¡æ  Mn²⁺  +  5Fe³⁺  +  4H₂O

4. ¹°Àº ¼¾ ¹«±â»ê¿¡ ´ëÇØ leveling solvent¶ó°í ÇÏ°í ¹Ý¸é¿¡ ethanolÀº differentiating solvent¶ó°í ÇÏ´Â ÀÌÀ¯´Â ¹«¾ùÀΰ¡ ? ( 5Á¡ )

5. 0.0333MÀÇ Mg(ClO₄)₂¿ë¾×¿¡ µé¾îÀÖ´Â Cd₂Fe(CN)₆ÀÇ ¿ëÇصµ¸¦  a) activity  b) molar concentrationÀ» ÀÌ¿ëÇÏ¿© °è»êÇ϶ó. ( À̶§ Cd²⁺¿Í Fe(CN)₆^4-ÀÇ ¼öÈ­¹Ý°æÀº 5¡ÊÀÌ´Ù.) ( 15Á¡ )                Cd₂Fe(CN)₆(S)   =   2Cd²⁺   +   Fe(CN)₆^4-       K^sp = 3.2 x 10^-17

6. H₃O⁺ÀÇ ³óµµ°¡ 1.0 x 10^-4M ÀÏ ¶§ÀÇ CuSÀÇ molar solubility¸¦ °è»êÇ϶ó. ( 15Á¡ )
   (  k_sp(CuS)= 6.0 x 10^-36, H₂S ÀÇ K_a1= 5.7 x 10^-8,  K^a2= 1.2 x 10^-15 )

7. CuCl₂^-ÀÇ formation constant´Â ´ÙÀ½°ú °°ÀÌ ³ªÅ¸³¾ ¼ö ÀÖ´Ù;   Cu⁺+ 2Cl^- = CuCl₂^-, K_f = 7.9 x 10⁴. NaClÀÇ ³óµµ°¡ 1.0 x 10^-3MÀÎ ¿ë¾×¿¡¼­ÀÇ CuCl( K_sp = 1.2 x 10^-6 )ÀÇ ¿ëÇصµ¸¦ °è»êÇ϶ó. ( 15Á¡ )

8. ¿ë¾× 200mL¿¡ Pb(OH)₂ 0.200gÀ» ³ìÀ̱â À§ÇÏ¿© À¯ÁöµÇ¾î¾ß ÇÏ´Â [OH^-]ÀÇ ³óµµ´Â ¾ó¸¶ÀÎ
   °¡ ? ( Pb: 207.2 )   Pb(OH)₂(S) + OH^-  =  Pb(OH)₃^-    K_f = 5.0 x 10^-2    ( 5Á¡ )

9. Volhard method·Î Cl^-¸¦ Á¤·®ÇÒ ¶§´Â Br^-¸¦ Á¤·®ÇÒ ¶§º¸´Ù ´õ ¸¹Àº ´Ü°è°¡ ÇÊ¿äÇÑ°¡ ? ¾Æ´Ñ°¡ ?  ±× ÀÌÀ¯´Â ?  ( 5Á¡ )

10. Cl^- ¿Í ClO₄^-¸¦ Æ÷ÇÔÇÑ ½Ã·á 1.998gÀ» ¹°¿¡ ³ì¿© 250mL ¿ë¾×À¸·Î ¸¸µé¾ú´Ù. ÀÌ ¿ë¾× 50mL¸¦ ÃëÇÏ¿© Cl^-ÀÇ ¾çÀ» ¾Ë¾Æº¸±â À§ÇÏ¿© 0.08551M AgNO₃·Î ÀûÁ¤ÇÏ¿´À»¶§ 13.97mL°¡ ÇÊ¿äÇÏ¿´´Ù. ¶Ç´Ù½Ã ¿ë¾× 50mL¸¦ ÃëÇÏ¿© V₂(SO₄)₃¸¦ ÀÌ¿ëÇÏ¿© ClO₄^- ¸¦ Cl^-·Î ȯ¿ø½ÃŲ ÈÄ À§ÀÇ AgNO₃·Î ÀûÁ¤ÇÏ¿´À» ¶§ 40.12mL°¡ µé¾î°¬´Ù. ȯ¿ø¹ÝÀÀ½ÄÀº ´ÙÀ½°ú °°´Ù.
          ClO₄^-  +  4V₂(SO₄)₃  +  4H₂O  ¡æ  Cl^-  +  12SO₄^2-  +  8VO²⁺  +  8H⁺
½Ã·á ¼ÓÀÇ Cl^- ¿Í ClO₄^-ÀÇ Áú·® % ¸¦ °¢°¢ ±¸Ç϶ó.  ( Cl: 35.453 )  ( 10Á¡ )

ºÐ¼®È­ÇÐ ¥°( 91Çг⵵ 1Çбâ 1Â÷½ÃÇè, 91.4.10 )

1. ´ÙÀ½ ¹Ýº¹ ÃøÁ¤µÈ °¢ sets¿¡ ´ëÇØ ¾Ë¾Æº¸±â·Î ÇÏÀÚ.
       A:  0.624,  0.613,   0.596,   0.607,   0.582
       B:  0.628,  0.622,   0.627,   0.635
a) ¿©±â¼­ A set¿¡ ´ëÇØ ©± mean ( 3Á¡ ) ©² median ( 3Á¡ ) ©³ spread or range ( 3Á¡ ) ©´ standard    deviation ( 6Á¡ ) ©µ coefficient of variation ( 5Á¡ ) À» ±¸Ç϶ó.
b) A set¿¡¼­ÀÇ 95% confidence intervalÀ» ±¸ÇÏ°í, ±× Àǹ̸¦ ¼³¸íÇ϶ó. ( 7Á¡ )
c) A set¿¡¼­ ¸¶Áö¸·¿¡ ÀÖ´Â 0.582´Â ¹ö¸± °ÍÀΰ¡ ¶Ç´Â ¾Æ´Ñ°¡¸¦ ©± Q test ( 96% confidence level) ( 5Á¡      ) ©² T_n test ( 95% confidence level) ( 5Á¡ )À» ÅëÇØ °áÁ¤Ç϶ó.
d) set A¿Í set B¸¦ ÀÌ¿ëÇÏ¿© pooled standard deviationÀ» ±¸Ç϶ó. ( 7Á¡ )
e) set AÀÇ data¿Í set BÀÇ dataÀÇ Æò±Õ°ªÀº 95% confidence level¿¡¼­ ´Ù¸¥ °ÍÀΰ¡ ? ( 7Á¡ )
f) B setÀÇ data°¡ A setÀÇ dataº¸´Ù 95% confidence level¿¡¼­ precisionÀÌ ´õ ÁÁÀº°¡? ( 7Á¡ )

           Table.1 Value of t for various levels of probability

            Table.2 Critical Values for Rejection Quotient Q

            Table.3 Critical Values for Rejection Quotient T_n

Table.4  Critical Values for F at the 5% level

2. Á¶Á¦ ¾àÇ°¼Ó¿¡ µé¾î ÀÖ´Â codeineÀ» ºÐ¼®ÇÏ´Â ¹æ¹ýÀ» ÀÌ¿ëÇÏ¿© codeineÀÌ µé¾î ÀÖÁö ¾ÊÀº ½Ã·á¿¡ ´ëÇØ ¾òÀº °á°ú´Â ´ÙÀ½°ú °°´Ù; 0.2, -0.1, 0.0, 0.3, 0.1, 0.3mg.  4¹ø ºÐ¼®ÇÑ °á°úÀÇ Æò±Õ¿¡ ±Ù°ÅÇÏ°í 99% confidence level¿¡¼­ÀÇ detection limit( codeineÀÇ mgÀ¸·Î )¸¦ °è»êÇ϶ó. ( 10Á¡ )

3. peptizationÀ̶õ ¹«¾ùÀÌ¸ç ¾î¶»°Ô ÀÌ°ÍÀ» ÇÇÇÒ ¼ö ÀÖÀ»±î ? ( 5Á¡ )

4. 0.6447gÀÇ ÀÌ»êÈ­¸Á°£(manganese dioxide)À» ¿°È­ÀÌ¿ÂÀ» Æ÷ÇÔÇÏ°í ÀÖ´Â ½Ã·á 1.1402gÀÌ ³ì¾ÆÀÖ´Â »ê¼º¿ë¾×¿¡ ÷°¡ÇÏ¿´´Ù. ¿°¼Ò±âüÀÇ ¹ß»ýÀÌ ¿Ï·áµÈ ÈÄ ³²¾Æ ÀÖ´Â ÀÌ»êÈ­¸Á°£À» ¿©°úÇÏ¿© ¸ðÀ¸°í ¾Ä°í ´Þ¾Ò´õ´Ï 0.3521gÀ̾ú´Ù. ÀÌ ºÐ¼®°á°ú¸¦ ¿°È­ÀÌ¿ÂÀ» Æ÷ÇÔÇÏ°í ÀÖ´Â ½Ã·á ÁßÀÇ ¿°È­ ¾Ë·ç¹Ì´½ÀÇ %·Î ³ªÅ¸³»¾î¶ó. ( Mn: 54.938,  O: 15.999,  Cl: 53.453, Al: 26.982 ) ( 20Á¡ )

5. ¿°È­À̿°ú ¿ä¿Àµå ÀÌ¿ÂÀ» Æ÷ÇÔÇÏ°í ÀÖ´Â ½Ã·á 0.6407gÀ» ÇҷΰÕÈ­Àº ħÀüÀ¸·Î ¸¸µé¾î Áú·®À» ´Þ¾Ò´õ´Ï 0.4430g ÀÌ µÇ¾ú´Ù. ±× ´ÙÀ½ ÀÌ Ä§ÀüÀ» Cl₂±âü¼Ó¿¡¼­ ¼¼°Ô °¡¿­ÇÏ¿© ¸ðµç AgI¸¦ AgCl·Î º¯È­½ÃÄÑ Ä§ÀüÀÇ Áú·®À» ´Þ¾Æº» °á°ú 0.3181gÀÌ µÇ¾ú´Ù. ½Ã·á ¼Ó¿¡ µé¾îÀÖ´Â ¿°È­À̿°ú ¿ä¿Àµå ÀÌ¿ÂÀÇ %¸¦ °è»êÇ϶ó. (  Ag: 107.868,   I: 126.904, Cl: 35.453 ) ( 10Á¡ )

6. ħÀüÀÔÀÚ¸¦ Å©°Ô ÇÏ´Â ¹æ¹ýÀ» »ó´ë °úÆ÷È­µµ¿Í ¿¬°üÁö¾î ¼³¸íÇ϶ó. ( 7Á¡ )

ºÐ¼®È­ÇÐ ¥°( 90Çг⵵ 1Çбâ 4Â÷½ÃÇè, 90.6.14 )

1. 0.0200M NH₄Cl 10mL¿Í 0.0320M (CH₃)₃N 10mL¸¦ ÇÔ²² È¥ÇÕÇÏ¿´À» ¶§ÀÇ pH´Â ¾ó¸¶Àΰ¡ ?
   ( 10Á¡ ) ( NH₄⁺ ÀÇ K_a= 5.70 x 10^-10,  (CH₃)₃N ÀÇ K_b= 6.31 x 10^-5 )

2. Chelate effect¿¡ ´ëÇØ ¼³¸íÇ϶ó.( 10Á¡ )

3. 0.0500M Mg²⁺ 50mL¸¦ 0.0500M EDTA·Î ÀûÁ¤ÇÑ´Ù°í ÇÏÀÚ.  Mg²⁺¿ë¾×Àº pH 10À¸·Î ¿ÏÃæµÇ¾î Àִµ¥ ÀÌ ¶§  ¥á(Y^4-)´Â 0.36ÀÌ´Ù.  EDTA°¡ ´ÙÀ½ °¢ ºÎÇǸ¸Å­ Àû°¡µÇ¾úÀ» ¶§ ¿ë¾×¿¡ ÀÖ´Â [Mg²⁺]ÀÇ ³óµµ´Â ¾ó¸¶Àΰ¡ ?   ÀÌ ¶§ K_f(MgY^2-)= 6.2 x 10⁸ ÀÌ´Ù. ( 15Á¡ )
   a) 5.0mL     b) 50mL      c) 51.0mL

4. ´ÙÀ½ ¿ë¾îµéÀ» °£´ÜÈ÷ ¼³¸íÇ϶ó. ( 15Á¡ )
   a) chelating ligand             b) conditional formation constant
   c) auxiliary complexing agent   d) blocking ion     e) masking agent

5. Back titration°ú  displacement titration¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 10Á¡ )

6. 1.00M HCl¿ë¾×¿¡¼­ 0.0100M Tl⁺ 100mL¸¦ 0.0100M IO₃^-·Î ÀûÁ¤ÇÏ¿´À» ¶§ ´ç·®Á¡¿¡¼­ÀÇ potentialÀ» °è»êÇ϶ó. ÀÌ ¶§ potential ÃøÁ¤Àº PtÀü±Ø°ú SCEÀü±Ø(E = 0.241V)»çÀÌ¿¡¼­ ÃøÁ¤µÈ´Ù. ( 10Á¡ )
         IO₃^- +  2Cl^- + 6H⁺  +  4e  ¡æ  ICl₂^-  +  3H₂O      E¢ª= 1.24V
                    Tl³⁺  +  2e     ¡æ    Tl⁺               E¢ª= 0.77V

7. Jones reductor¿Í Walden reductor¸¦ ºñ±³ ¼³¸íÇ϶ó. ( 10Á¡ )

8. Redox indicatorÀÇ color°¡ º¯ÇÏ´Â potential range¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 10Á¡ )

9. ±Ý¼ÓÀÌ¿ÂÀ» extractionÇϴµ¥ ´ëÇÑ distribution coefficient°¡ ¼ö¿ë¾×ÀÇ pH¿Í À¯±â»ó¿¡¼­ÀÇ ligandÀÇ ³óµµ¿¡ µû¶ó ´Þ¶óÁüÀ» º¸¿©¶ó. ( 10Á¡ )

ºÐ¼®È­ÇÐ ¥°( 90Çг⵵ 1Çбâ 3Â÷½ÃÇè, 90.5.24 )

1. Phosphoric acid (H₃PO₄)ÀÇ K_a2¹ÝÀÀ ¹× disodium oxalate(Na₂C₂O₄)ÀÇ K_b2¹ÝÀÀÀ» ½á¶ó. ( 10Á¡ )

2. ½ÇÁ¦·Î pH 9.5ÀÎ ¿ÏÃæ¿ë¾×À» 1 L ¸¸µé·Á°í ÇÑ´Ù. ¾î¶»°Ô ÇÏ¸é °¡Àå ÁÁÀ»±î ? ( 10Á¡ )

3. 0.10M ethylamine(CH₃CH₂NH₂)ÀÇ pH´Â 11.80 ÀÌ´Ù.  ethylamineÀÇ K_b´Â ¾ó¸¶Àΰ¡ ? ( 5Á¡ ) ÀÌ K_b°ªÀ» ÀÌ¿ëÇÏ¿© 0.10M ethylammonium chloride(CH₃CH₂NH₃⁺Cl^-)ÀÇ pH¸¦ °è»êÇ϶ó. ( 5Á¡ )

4. a) ¹° 500mL¿¡ K₂CO₃(F.W= 138.206) 4.00g ÀÌ ³ì¾Æ Àִµ¥ ÀÌ ¿ë¾×À» pH 10.80 ÀÌ µÇ°Ô Çϱâ À§Çؼ­´Â NaHCO₃(F.W= 84.007)À» ¸î g ÷°¡ÇØ¾ß Çϴ°¡ ? ( 5Á¡ ) ¿©±â¼­ H₂CO₃ÀÇ  K_a1= 4.45 x 10^-7, K_a2= 4.69 x 10^-11  ÀÌ´Ù.
b) ¸¸¾à a)¿ë¾×¿¡ 0.100M HCl 10mL¸¦ ÷°¡Çϸé pH´Â ¾ó¸¶³ª µÉ±î ? ( 5Á¡ )
c) pH 10ÀÎ ¿ë¾×À» ¸¸µé±â À§Çؼ­´Â 4.00gÀÇ K₂CO₃¸¦ ³ìÀÎ ¿ë¾× 250mL¿¡ 0.320M HNO₃ ¸î mL ¸¦ ÷°¡Çؾ߸¸ Çϴ°¡ ? ( 5Á¡ )

5. K_a1= 1.00 x 10^-4 °ú K_a2= 1.00 x 10^-8 ÀÎ diprotic acid, H₂A ¿¡ ´ëÇØ »ý°¢ÇØ º¸ÀÚ.
   ´ÙÀ½ °¢ ¿ë¾×¿¡¼­ÀÇ pH, [H₂A], [HA^-] ¿Í [A^2-]ÀÇ ³óµµ¸¦ ±¸Ç϶ó. ( 15Á¡ )
   a) 0.100M H₂A      b) 0.100M NaHA       c) 0.100M Na₂A

6. 3.42gÀÇ oxalic acid(HOOC-COOH, F.W= 90.036, pK_a1= 1.252, pK_a2= 4.266)¸¦ ³ì¿© ¿ë¾×À» 500mL°¡ µÇ°Ô ÇÑ ÈÄ pH°¡ 4.50ÀÌ µÇµµ·Ï Çϱâ À§Çؼ­´Â 0.900M KOH´Â ¸î mL¸¦ °¡Çؾ߸¸ Çϴ°¡ ? ( 10Á¡ )

7. Gran's plot¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 10Á¡ )

8. NaOH¿ë¾×À» standardizationÇÏ´Â ¹æ¹ý¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 10Á¡ )

9. ´ÙÀ½À» °£´ÜÈ÷ ¼³¸íÇ϶ó. ( 10Á¡ )
   a) leveling effect
   b) buffer capacity ( or buffer intensity )

ºÐ¼®È­ÇÐ ¥°( 90Çг⵵ 1Çбâ 2Â÷½ÃÇè, 90.5.3 )

1. ´ÙÀ½À» °£´ÜÈ÷ ¼³¸íÇ϶ó. ( 10Á¡ )
   1) disproportionation       2) common ion effect     3) digestion
   4) coprecipitation          5) blank titration

2. ¸¸¾à 0.1M Cl^-, Br^-, CO₃^2-, CrO₄^2-¸¦ Æ÷ÇÔÇÏ°í ÀÖ´Â ¿ë¾×¿¡ Ag⁺¸¦ ³ÖÀ¸¸é ¾î´À ¼ø¼­·Î ħÀüÀÌ »ý±â°Ú´Â°¡ ?    [ ÀÌ ¶§  K_sp(AgCl)= 1.8 x 10^-10,    K_sp(AgBr)= 5.0 x 10^-13,
   K_sp(Ag₂CrO₄) = 1.2 x 10^-12,    K_sp(Ag₂CO₃) = 8.1 x 10^-12 ]        ( 10Á¡ )

3. ÀüÇØÁúÀÇ ³óµµ°¡ Áõ°¡ÇÒ¼ö·Ï ħÀü¹°ÀÇ ¿ëÇصµ°¡ Áõ°¡ÇÏ´Â ÀÌÀ¯´Â ¹«¾ùÀΰ¡ ? ( 5Á¡ )

4. Ba(IO₃)₂·Î Æ÷È­µÈ 0.05M Mg(NO₃)₂ ¿ë¾×¿¡¼­ÀÇ Ba²⁺ ³óµµ¸¦ °è»êÇ϶ó. ÀÌ ¶§ È°µ¿µµ °è¼ö¸¦ °í·ÁÇ϶ó. K_sp[Ba(IO₃)₂] = 1.5 x 10^-9,   ¼öÈ­¹Ý°æ   Ba²⁺: 500pm, IO₃^-: 450pm   ( 15Á¡ )

5. pH= 4.0À¸·Î À¯ÁöµÈ ¿ë¾×¿¡¼­ÀÇ CaC₂O₄( F.W = 128.098 )ÀÇ ¿ëÇصµ(g/L)¸¦ °è»êÇ϶ó.( 15Á¡ )  ´ÙÀ½°ú °°Àº ÆòÇüÀÌ »ý±ä´Ù.
    CaC₂O₄(S)       =  Ca²⁺ +  C₂O₄^2-        K_sp= 1.3 x 10^-8
    C₂O₄^2-  +  H₂O  =  HC₂O₄^-  + OH^-         K_b1= 1.8 x 10^-10
    HC₂O₄^-  +  H₂O  =  H₂C₂O₄  + OH^-         K_b2= 1.8 x 10^-13

6. ħÀüÀÔÀÚ¸¦ Å©°Ô ¸¸µå´Â ¹ýÀ» ¸ðµÎ ¿­°ÅÇÏ°í ÀÌÀ¯¸¦ ¹àÇô¶ó. ( 10Á¡ )

7. ´ÜÁö ferrous ammonium sulfate, FeSO₄.(NH₄)₂SO₄.6H₂O ( F.W = 392.13 )¿Í ferrous chloride, FeCl₂.6H₂O ( F.W = 234.84 )¸¸À» Æ÷ÇÔÇÏ°í ÀÖ´Â °íüȥÇÕ¹°À» 0.6734g ÃëÇÏ¿´´Ù. ÀÌ ½Ã·á¸¦ 1M H₂SO₄¿¡ ³ìÀÌ°í H₂O₂¸¦ ÀÌ¿ëÇÏ¿© ¸ðµÎ Fe(¥±)¸¦  Fe(¥²)À¸·Î »êÈ­½ÃÅ°°í CupferronÀ¸·Î ħÀü½ÃŲ ÈÄ ferric cupferron complex¸¦ Å¿ö¼­ ferric oxide, Fe₂O₃ (F.W = 159.69)·Î¼­ 0.1864gÀ» ¾ò¾ú´Ù. ¿ø·¡ ½Ã·á¿¡¼­ÀÇ  Cl(A.W = 35.45)ÀÇ weight %¸¦ °è»êÇ϶ó. ( 15Á¡ )

8. Volhard titrationÀÇ ¿ø¸®¿¡ ´ëÇØ ½á¶ó. ( 5Á¡ )

9. 0.100M Ag⁺¿Í 0.100M Hg²⁺¸¦ Æ÷ÇÔÇÏ°í ÀÖ´Â 10mLÀÇ È¥ÇÕ¹°À» 0.100M KCNÀ¸·Î ÀûÁ¤ÇÏ¿© Hg₂(CN)₂¿Í  AgCNÀ¸·Î ħÀü½ÃÄ×´Ù. K_sp(AgCN)= 2.2 x 10^-16, K_sp[Hg₂(CN)₂]= 5 x 10^-40
   1) °¢°¢ ´ÙÀ½ ºÎÇÇÀÇ KCNÀ» ÷°¡ÇßÀ» ¶§ÀÇ [CN^-] ¸¦ °è»êÇ϶ó. ( 10Á¡ )
      15.00,  19.90,  20.00,  20.10,  25.00mL
   2) À§ÀÇ 19.90mL¿Í ´ç·®Á¡¿¡¼­ °¢°¢ AgCN ħÀüÀÌ »ý±â°Ú´Â°¡ ?  ÀÌÀ¯´Â ? ( 5Á¡ )

ºÐ¼®È­ÇÐ ¥°( 90Çг⵵ 1Çбâ 1Â÷½ÃÇè, 90.3.26 )

1. ´ÙÀ½ quantityÀÇ SI unit¸¦ ½á¶ó. ( 14Á¡ )
   1) mass   2) time   3) temperature   4) amount of substance   5) frequency
   6) pressure   7) energy

2. 53.4% NaOH ¼ö¿ë¾×À» 16.7mL ÃëÇÏ¿© 2.00L·Î ¹±ÇûÀ» ¶§ 0.169M NaOH°¡ ¸¸µé¾îÁ³´Ù¸é óÀ½ ¿ë¾×ÀÇ ¹Ðµµ´Â ¾ó¸¶À̰ڴ°¡ ? ( 11Á¡ )

3. ´ÙÀ½ °¢°¢À» °è»êÇÏ°í absolute uncertainty¿Í percent relative uncertaintyµµ ³ªÅ¸³»¾î¶ó. ( 10Á¡ )
   1) 83.2(¡¾1.0) x 32.3(¡¾0.2)/ 20.3(¡¾0.2) =
   2) [ 5.84(¡¾0.05) - 1.86(¡¾0.01) ]/ 22.3(¡¾0.2) =

4. Powdered mineral sampleÀÇ Ca contentÀ» 2°¡Áö ¹æ¹ýÀ¸·Î °¢°¢  5¹ø¾¿ ÃøÁ¤ÇÏ¿´´Ù. Æò±Õ°ªÀº 90% confidence level¿¡¼­ »ó´çÈ÷ ´Ù¸¦±î ?   ( 15Á¡ )
        -----------------------------------------------------------------
                                     Ca ( %Á¶¼º )
  -----------------------------------------------------------------
       ¹æ¹ý 1       0.0271    0.0282    0.0279    0.0271    0.0275
       ¹æ¹ý 2       0.0271    0.0268    0.0263    0.0274    0.0269
 -----------------------------------------------------------------

           Table.1 Value of t for various levels of probability

5. ´Ü¹éÁúÀÇ ÇÔ·®À» ±¸Çϱâ À§ÇÏ¿© Èí±¤µµ¸¦ ÃøÁ¤ÇÏ¿© ¾òÀº calibration curve¸¦ ÀÌ¿ëÇÏ·Á°í ÇÑ´Ù.
   ----------------------------------------------------------------------
       ´Ü¹éÁú ( g)      0.00      9.36      18.72     28.08     37.44
         Èí±¤µµ         0.466     0.676     0.883     1.086     1.280
   ----------------------------------------------------------------------
1) method of least square¸¦ ÀÌ¿ëÇÏ¿© °¡Àå ÁÁÀº Á÷¼±½ÄÀ» ±¸Ç϶ó. ÀÌ ¶§ slope¿Í interceptÀÇ standard deviationµµ ±¸ÇÏ°í À¯È¿¼ýÀÚ °³³ä¿¡µµ À¯ÀÇÇ϶ó. ( 20Á¡ )
2) À§ÀÇ ½ÇÇè data¿Í 1)¿¡¼­ ±¸ÇÑ Á÷¼±½ÄÀÌ ³ªÅ¸³»¾îÁø graph¸¦ ±×·Á¶ó. ( 5Á¡ )
3) unknown protein sampleÀÇ Èí±¤µµ°¡ 0.973 À̾ú´Ù°í ÇÏ°í unknown¼Ó¿¡ µé¾îÀÖ´Â proteinÀÇ ÇÔ·®°ú uncertainty¸¦ ³ªÅ¸³»¾î¶ó. ( 10Á¡ )

6. Random compositionÀ» °®´Â bulk materialÀ» ºÐ¼®Çϴµ¥ À־ sampling standard deviationÀº 5% À̾ú´Ù.  mean ¿¡¼­ÀÇ error°¡ 4% À̳»°¡ µÇ´Âµ¥ 95% confidence¸¦ °®µµ·Ï ÇÏ·Á¸é ºÐ¼®ÇؾßÇÒ ½Ã·á´Â ¸î °³À̾î¾ß Çϴ°¡ ? ( 15Á¡ )

ºÐ¼®È­ÇÐ ¥° ( 89Çг⵵ 1Çбâ 4Â÷½ÃÇè, 89.?.? )

1. Gran's plot¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 15Á¡ )

2. Isoelectric focusing(µîÀü ÁýÁßÁ¡)¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 8Á¡ )

3. Chelate effect¿¡ ´ëÇØ ¿¹¸¦ µé¾î ¼³¸íÇ϶ó. ( 10Á¡ )

4. Iodometry(°£Á¢ ¿ä¿Àµå¹ýÀûÁ¤, ¿ä¿ÀµåÈ­¹ý)¿Í iodimetry(Á÷Á¢ ¿ä¿Àµå¹ýÀûÁ¤, ¿ä¿Àµå¹ý)¸¦ ºñ±³ ¼³¸íÇ϶ó. ( 10Á¡ )

5. 0.074M pyridine( C₅H₅N:, K_b= 1.69 x 10^-9 ) 25.00mL¸¦ 0.096M HCl·Î ÀûÁ¤ÇÑ´Ù°í ÇÏÀÚ. HClÀ» ´ÙÀ½ ºÎÇǸ¸Å­ ÀûÁ¤ÇÏ¿´À» ¶§ÀÇ pH´Â ¾ó¸¶Àΰ¡ ?  ( 12Á¡ )
   ÀûÁ¤¹ÝÀÀÀº ´ÙÀ½°ú °°´Ù;      C₅H₅N: +  H⁺  ¡æ   C₅H₅NH⁺
   a) 16.40mL      b) Ve         c) 20.24mL

6. 0.100M ÀÌ¿°±âÈ­ÇÕ¹°, B ( pK_b1= 5.00, pK_b2= 10.00 ) 100mL¸¦ 1.00M HCl·Î ÀûÁ¤ÇÑ´Ù°í ÇÏÀÚ.  Ã·°¡µÈ HClÀÇ ºÎÇÇ°¡ ´ÙÀ½°ú °°À» ¶§ pH¸¦ ±¸Ç϶ó. ( 15Á¡ )
   a) 4.0mL    b) 10.0mL    c) 16.0mL    d) 20.0mL    e) 23.0mL

7. 0.050M chloroacetic acid( ClCH₂COOH, K_a= 1.36 x 10^-3 ) 100.0mL¿¡ 0.060M methylamine      ( CH₃NH₂, K_b= 4.4 x 10^-4 ) 60.0mL¸¦ ÷°¡ÇÑ´Ù¸é pH´Â ¾ó¸¶°¡ µÇ°Ú´Â°¡? ( 10Á¡ )

8. pH 5.5 ¿¡¼­ xylenol orange Áö½Ã¾àÀ» ÀÌ¿ëÇÏ°í Zn²⁺ Ç¥ÁØ¿ë¾×À» ÀÌ¿ëÇÏ¿© ¿ªÀûÁ¤À¸·Î 25.00mLÀÇ Ni²⁺¿ë¾×À» ºÐ¼®ÇÏ·Á°í ÇÑ´Ù. ÀÌ ¿ë¾×¿¡ 0.053M Na₂EDTA 25.00mL¸¦ °¡ÇÑ ÈÄ 0.023M Zn²⁺Ç¥ÁØ¿ë¾×À¸·Î ¿ªÀûÁ¤ÇÏ¿´À» ¶§ 17.50mL°¡ µé¾î°£ ÈÄ Á¾¸»Á¡ÀÌ ³ªÅ¸³µ´Ù. Ni²⁺¿ë¾×ÀÇ ³óµµ´Â ¾ó¸¶Àΰ¡ ? ( 10Á¡ )

9. 1M HCl¿ë¾×¿¡¼­ ÀϾ´Â ´ÙÀ½ »êÈ­-ȯ¿ø¹ÝÀÀ¿¡ ´ëÇÑ ´ç·®Á¡¿¡¼­ÀÇ ÀüÀ§¸¦ °è»êÇ϶ó.( 15Á¡ )
   a) Ce⁴⁺ + Fe²⁺ = Fe³⁺ + Ce³⁺   [ E¢ª(Fe³⁺/Fe²⁺) = 0.767V, E¢ª(Ce⁴⁺/Ce³⁺) = 1.70V ]
   b) IO₃^- + 2Tl⁺ + 2Cl^- + 6H⁺   =   ICl2- + 2Tl3+ + 3H2O
         [ E¢ª(IO3-/ICl2-) = 1.24V,   E¢ª(Tl³⁺/Tl⁺) = 0.77V ]
   c) UO₂²⁺ + Sn²⁺ + 4H⁺  =  U⁴⁺ +  Sn⁴⁺  + 2H₂O
         [ E¢ª(UO₂²⁺/U⁴⁺) = 0.334V,   E¢ª(Sn⁴⁺/Sn²⁺) = 0.139V ]

10. Jones reductor¿Í Walden reductor ¿¡ ´ëÇØ ºñ±³ ¼³¸íÇ϶ó. ( 8Á¡ )

11. ÂøÈ­¹ý ÀûÁ¤¿¡¼­ displacement titration°ú Indirect titration ¿¡ ´ëÇØ ¿¹¸¦ µé¾î ¼³¸íÇ϶ó. ( 10Á¡ )

12. Nonaqueous ¿ë¾×¿¡¼­ ÀûÁ¤Çؾ߸¸ ÇÏ´Â °æ¿ì¿¡´Â ¾î¶² °ÍÀÌ Àִ°¡ ? ( 7Á¡ )

ºÐ¼®È­ÇÐ ¥° ( 89Çг⵵ 1Çбâ 3Â÷½ÃÇè, 89.5.15 )

1. ´ÙÀ½ ¿ë¾îµé¿¡ ´ëÇØ °£´ÜÈ÷ ¼³¸íÇ϶ó.
   a) Electrical double layer ( 3Á¡ )    b) Coprecipitation ( 3Á¡ )
   c) Back titration ( 3Á¡ )             d) Buffer capacity ( 3Á¡ )
   e) Turbidimetry ( 4Á¡ )               f) Nephelometry ( 3Á¡ )

2. Gravimetric Analysis¸¦ ÇÒ·Á¸é ÀÔÀÚ°¡ Å« ħÀüÀ» ¾ò¾î¾ß ÇÑ´Ù. ÀÔÀÚ°¡ Å« ħÀüÀ» ¾òÀ» ¼ö ÀÖ´Â À̷аú ÀÌ¿Í °ü·ÃµÈ ¹æ¹ý¿¡ ´ëÇØ ½á¶ó. ( 20Á¡ )

3. Aluminum tetrafluoroborate, Al(BF₄)₃ (F.W = 287.39)¿Í Magnesium nitrate,Mg(NO₃)₂ (F.W = 148.31)¸¸À» Æ÷ÇÔÇÏ°í Àִ ȥÇÕ¹°ÀÇ ¹«°Ô°¡ 0.5656g À̾ú´Ù. ÀÌ°ÍÀ» 1% HF ¼ö¿ë¾×À¸·Î ³ìÀÎ ÈÄ °ú·®ÀÇ nitron (F.W = 312.37) ¿ë¾×À¸·Î ó¸®ÇÏ¿© nitron tetrafluoroborate (F.W = 400.18)°ú nitron nitrate (F.W = 375.39)ÀÇ È¥ÇÕ¹° ÇüÅÂÀÇ Ä§ÀüÀ¸·Î ¾òÀº ÈÄ ¹«°Ô¸¦ ´Þ¾Æº¸´Ï 2.644g À̾ú´Ù. ( nitron Àº tetrafluoroborate ¶Ç´Â nitrate ¿Í 1:1 ·Î ¹ÝÀÀÇÑ´Ù. )  Ã³À½ °íü È¥ÇÕ¹°¿¡ µé¾î ÀÖ´ø Mg (A.W = 24.312)ÀÇ ¹«°Ô%´Â ¾ó¸¶Àΰ¡ ? ( 15Á¡ )

4. 0.072M KI 25.00mL¸¦ 0.048M AgNO₃·Î ÀûÁ¤ÇÑ °Í¿¡ ´ëÇØ ¾Ë¾Æº¸ÀÚ. ÷°¡µÈ AgNO₃ÀÇ ºÎÇÇ°¡ ´ÙÀ½°ú °°À» ¶§ °¢°¢¿¡¼­ÀÇ [Ag⁺]¸¦ °è»êÇ϶ó. ( 10Á¡ )
   a) 36.8mL      b)  V_e      c) 38.1mL

5. Argentometric titration¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 20Á¡ )

6. pH°¡ 8.0ÀÎ buffer solutionÀ» ¸¸µé·Á°í ÇÑ´Ù. ¾î¶»°Ô Çϸé Àß ¸¸µé ¼ö ÀÖÀ»±î ? ( 15Á¡ )

7. HL₂⁺ÀÎ À̾缺ÀÚ»êÀÇ Áß°£ÇüÅÂÀÎ HLÀÇ  pH°¡  (pK₁ + pK₂)/2 °¡ µÊÀ» º¸¿©¶ó. ( 15Á¡ )

8. Malonic acid, CH₂(COOH)₂ ¸¦ °£·«ÇÏ°Ô H₂M À̶ó°í ¾²±â·Î ÇÏÀÚ. ´ÙÀ½ °¢ ¿ë¾×¿¡ µé¾î ÀÖ´Â [H₂M], [HM^-], [M^2-] ÀÇ ³óµµ¿Í pH¸¦ ±¸Ç϶ó. ( 15Á¡ )
   a) 0.100M H₂M     b) 0.100M NaHM     c) 0.100M Na₂M
   H₂M  =  H⁺ +  HM^-        K_a1 = 1.42 x 10^-3
   HM^-  =  H⁺ +  M^2-        K_a2 = 2.01 x 10^-10

ºÐ¼®È­ÇÐ ¥° ( 89Çг⵵ 1Çбâ 2Â÷½ÃÇè, 89.4.4 )

1. Explain the following terms. ( 20Á¡ )
   a) Gibbs free energy       b) Le Chatelier's principle
   c) Common ion effect       d) Formation constant

2. Which will be more soluble in water ?  Suggest the reason. ( 8Á¡ )
    TlIO₃ ( K_sp = 3.1 x 10^-6 ),   Sr(IO₃)₂ ( K_sp = 3.3 x 10^-7 )

3. As [Cl^-] is increased in Ag⁺ aqueous solution, the amounts of AgCl is increased. However at high value of [Cl^-], the amounts of AgCl is decreased. Why so ? ( 8Á¡ )

4. Explain why the solubility of salts is increased by the addition of ions to the solution. ( 8Á¡ )

5. How does the pH affect the solubility of CaF₂ ?  Explain the reason. ( 8Á¡)

6. It is desired to perform complete separation of 0.02M Ca²⁺ and 0.02M Ce³⁺ by precipitation with oxalate ion ( C₂O₄^2-). Decide whether this is feasible ( 15Á¡ )
     CaC₂O₄ ( K_sp= 1.3 x 10^-8 ),  Ce₂(C₂O₄)₃ ( K_sp= 3.0 x 10^-29 )

7. For a solution of Ni²⁺ and ethylenediamine(en), the following equilibrium constants apply at 20¡É.
               Ni²⁺    +  en   =  Ni(en)²⁺        log K₁ = 7.52
             Ni(en)²⁺  +  en   =  Ni(en)₂²⁺       log K₂ = 6.32
             Ni(en)₂²⁺ +  en   =  Ni(en)₃²⁺       log K₃ = 4.49
Calculate the concentration of free Ni²⁺ in a solution prepared by mixing 0.200mole of en plus 1.0 mL of 0.0200M Ni²⁺ and diluting to 1.00 L.  ( 15Á¡ )

8. Suggest a reason why the hydrate radii decrease in the order;
   Sn⁴⁺¡µ In³⁺¡µ Cd²⁺¡µ Rb⁺     ( 8Á¡ )

9. Consider the following simultaneous equilibria.
        Ag₃PO₄(S)       =   3Ag⁺    +  PO₄^3-    K_sp = 2.8 x 10^-18
        PO₄^3-   +  H₂O  =   HPO₄^2-  +  OH^-      K_b1 = 2.3 x 10^-2
        HPO₄^2-  +  H₂O  =   H₂PO₄^-  +  OH^-      K_b2 = 1.6 x 10^-7
        H₂PO₄^-  +  H₂O  =   H₃PO₄   +  OH^-      K_b3 = 1.4 x 10^-12
Calculate the molarity of Ag⁺ in a saturated aqueous solution of Ag₃PO₄ at pH 5.00 ( 20Á¡ )

10. When ammonium sulfate dissolves, both the anion and cation have acid-base reactions in water.
         (NH₄)₂SO₄(S)  =   2NH₄⁺    +  SO₄^2-       K_sp = 276
         NH₄⁺          =   NH₃(aq)  +  H⁺          K_a = 5.70 x 10^-10
         SO₄^2- +  H₂O  =   HSO₄^-    +  OH^-         K_b = 9.80 x 10^-13
    a) Write a charge balance for this system. ( 5Á¡ )
    b) Write a mass balance for this system. ( 5Á¡ )
    c) Find the concentration of NH₃(aq) if the pH fixed at 9.00  ( 10Á¡ )

ºÐ¼®È­ÇÐ ¥° ( 89Çг⵵ 1Çбâ 1Â÷½ÃÇè, 89.?.? )

1. ´ÙÀ½ ¹°¸®Àû ¾çÀÇ SI unit¸¦ ½á¶ó. ( 6Á¡ )
   ±æÀÌ,  Áú·®,  ½Ã°£,  ¿Âµµ,  ¾Ð·Â,  ¹°ÁúÀÇ ¾ç

2. ´ÙÀ½À» Á¤ÀÇÇ϶ó;  1 mole,  1 M,  1 F   ( 6Á¡ )

3. 48.0% HBr ( MW= 80.9g )¼ö¿ë¾×ÀÇ ¹Ðµµ´Â 1.50g/mL ÀÌ´Ù. ( 20Á¡ )
  a) ÀÌ ¿ë¾×ÀÇ formality´Â ¾ó¸¶Àΰ¡ ?
  b) 40.0gÀÇ HBrÀ» Æ÷ÇÔÇÏ°í ÀÖ´Â ¿ë¾×ÀÇ Áú·®Àº ¾ó¸¶Àΰ¡ ?
  c) 450 mmol ÀÇ HBrÀ» Æ÷ÇÔÇÏ°í ÀÖ´Â ¿ë¾×ÀÇ ºÎÇÇ´Â ¾ó¸¶Àΰ¡ ?
  d) 0.250M HBrÀ» 0.300L ¸¸µå´Âµ¥ ÀÌ ¿ë¾×Àº ¾ó¸¶³ª ÇÊ¿äÇÑ°¡ ?

4. ´ÙÀ½ ±â±¸µéÀº ¹«¾ùÀ» ÇÒ ¶§ »ç¿ëÇÏ´Â °ÍÀÎÁö °£´ÜÈ÷ ±â¼úÇ϶ó. ( 12Á¡ )
  a) buret   b) pipet   c) volumetric flask  d) aspirator  e) rubber policeman  f) desiccator

5. Àú¿ï·Î ½Ã·áÀÇ Áú·®À» ´Ù´Âµ¥ À־ ºÎ·Â¿¡ ÀÇÇÑ ¿ÀÂ÷¸¦ º¸Á¤ÇÏ´Â ÀÌÀ¯¿¡ ´ëÇØ ¼³¸íÇ϶ó. ( 5Á¡ )

6. ´ÙÀ½ ½ÄÀÇ °á°ú¿¡ ´ëÇØ  absolute uncertainty¿Í percent relative uncertainty±îÁö °è»êÇÏ¿© ¹àÇô¶ó. ƯÈ÷ À¯È¿¼ýÀÚ °³³ä¿¡ À¯ÀÇÇ϶ó. ( 10Á¡ )
              [ 4.97(¡¾0.05) - 1.86(¡¾0.01) ]/ 21.1(¡¾0.2)

7. 135.2, 142.3, 137.5, 139.3, 141.7¿¡ ´ëÇÑ Æò±Õ°ª, Ç¥ÁØÆíÂ÷, medianÀ» °è»êÇ϶ó.(6Á¡)

8. °¡¼Ö¸°¿¡ µé¾îÀִ ÷°¡Á¦ÀÇ % ¸¦ ¿©¼¸ ¹ø ÃøÁ¤ÇÑ °á°ú ´ÙÀ½ °ªÀ» ¾ò¾ú´Ù. ; 0.12,  0.17,0.14,  0.16,  0.20,  0.11% .   Ã·°¡Á¦ÀÇ %¿¡ ´ëÇÑ 90% ¿Í 99% confidence intervalÀ» ±¸Ç϶ó.  ( 10Á¡ )

           Table.1 Value of t for various levels of probability

9. ´Ù¼¸°¡ÁöÀÇ ´Ù¸¥ ±¤¼®½Ã·á(°¢°¢ ´Ù¸¥ Ti ÇÔ·®À» °¡Áö°í ÀÖ´Ù.)ÀÇ Ti ÇÔ·®À» µÎ °¡Áö ¹æ¹ý¿¡ ÀÇÇØ ÃøÁ¤ÇÑ °á°ú´Â ´ÙÀ½°ú °°´Ù.
    ---------------------------------------------------
         ½Ã·á         A         B         C         D         E
    ---------------------------------------------------
        ¹æ¹ý 1      0.0134    0.0144    0.0126    0.0125    0.0137
        ¹æ¹ý 2      0.0135    0.0156    0.0137    0.0137    0.0136
   ----------------------------------------------------
   ÀÌ µÎ ºÐ¼®¹æ¹ýÀÌ 90% confidence level¿¡¼­ »ó´çÈ÷ ´Ù¸¥Áö¸¦ ¹àÇô¶ó. ( 15Á¡ )

10. ¾î¶² ½Ã·á ¼Ó¿¡ µé¾îÀÖ´Â proteinÀÇ ¾çÀ» Á¶»çÇÏ¿´´Âµ¥ ÀÌ proteinÀÇ ¾ç°ú absorbance(Èí±¤µµ)°¡ ºñ·Ê°ü°è¿¡ ÀÖÀ½À» ¾Ë¾Ò´Ù. µ¥ÀÌŸ´Â ´ÙÀ½°ú °°´Ù.
                  protein(¥ìg)             0.00      9.50      19.10     29.30     39.54
       absorbance(at 595nm)      0.468     0.678     0.887     1.110     1.333
    a) method of least square¸¦ ÀÌ¿ëÇÏ¿© °¡Àå ÁÁÀº Á÷¼±½ÄÀ» ±¸Ç϶ó. À¯È¿¼ýÀÚ¿¡ À¯ÀÇÇÏ°í                 y = [ m(¡¾¥ò _x) ] x + [ b(¡¾¥ò_b) ] ÇüÅ·Π³ªÅ¸³»¾î¶ó. ( 25Á¡ )
    b) ½ÇÇè data¸¦ ÀÌ¿ëÇÏ¿© ±¸ÇÑ Á÷¼±½ÄÀ» ±×·¡ÇÁ·Î ±×·Á º¸¾Æ¶ó. ( 5Á¡ )
    c) unknown protein ½Ã·áÀÇ Èí±¤µµ°¡ 0.973 À̶ó¸é ÀÌ ½Ã·á¿¡ Æ÷ÇÔµÈ proteinÀÇ ¾ç°ú Ç¥ÁØÆíÂ÷·Î ³ªÅ¸³½ ºÒÈ®Á¤µµ´Â ¾ó¸¶Àΰ¡ ? ( 10Á¡ )
                                                                 
    m = |  ¥Òx_iy_i  ¥Òx_i |¡À   D           D = | ¥Ò(x_i²)  ¥Òx_i  |
           | ¥Òy_i      n   |                         | ¥Òx_i        n   |
                             
    b = | ¥Ò(x_i²)  ¥Òx_iy_i  | ¡À  D         ¥ò_y ¡Ö  s_y = [¥Ò(d_i-d)2/(degree of freedom)]^1/2
          | ¥Ò x_i      ¥Òy_i    |

     ¥ò_m² = ¥ò_y²n/D         ¥ò_b² = ¥ò_y²¥Ò(x_i²)/D