Problem Set 1. 2] What is the molar concentration of 0.78 % (w/v) NaCl(aq) (MW = 58.4 g/mol). (5 points)

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1 Problem Set 1 1] Express the answer for the following operation the proper number of significant digits and absolute uncertainty: (5 points) (36.2 ± 0.4)/(27.1 ± 0.6) = 2] What is the molar concentration of 0.78 % (w/v) NaCl(aq) (MW = 58.4 g/mol). (5 points) 3] How many milliliters of M HCl are required to make ml of mm HCl? (5 points) 4] You have obtained the following values for the analysis of Cu in an ore sample. (10 points) 2.53% 2.47% 2.51% 2.99% 2.49% 2.54% Using valid statistical methods show how one of the values can be rejected. 5] The analysis of phosphate in fertilizer was made using a reliable method. Seven measurements were conducted. The mean value of phosphate in the sample is 1.72 mg/g with a standard deviation, s of Express the sample concentration (with uncertainty) assuming a 95% confidence level. (10 points) 6] Using the information from problem 5 estimate the chance that the true mean will be 2.20 mg/g or greater. Hint you will need to calculate z for this one, and think about the light bulb example from lecture. (10 points) 7] A blood sample was sent to two different labs for cholesterol analysis. The results are: Lab 1 x = 221 mg/dl s = 11 n = 10 Lab 2 x = 233 mg/dl s = 14 n = 10 Are the two standard deviations different significantly different at the 95% confidence limit? (10 points) 8] You have carefully followed an analytical procedure with n = 6 and found a mean of 6.37 mm with s = Meanwhile, Joe Cutcorners used a modified procedure with n = 4, x = 6.87 mm with s = Assuming that the standard deviations are not statistically different from each other, does Joe s method have a systematic error, i.e. statistically different at the 95% confidence limit? (10 points) 9] What is the ph of a solution of M HCl? (5 points) 10) Calculate the ph of a solution of M acetic acid, a = (5 points) 1

2 11) Calculate the ph of a solution of M acetic acid and M sodium acetate. (5 points) 12) Calculate the sp of barium sulfate (MW 233) if its solubility is measured as mg/ml. (5 points) 13) Part of the labeling of a class A pipet is the letters TD. What does this mean? (5 points) a) The correct liquid delivery process should have entire contents of the pipet should be blown out with the pipet blub. b) The pipet should be acid washed between usages. c) The pipet is defective and only semi- quantitative d) The pipet is coated with an inert agent. e) The solution delivery process will leave behind a small amount of liquid in the tip. 14) Calculate the solubility of PbCl 2 ( sp = 1.7 x 10-5 ) in the presence of M NaCl. (10 points) Answers to Problem Set 1 1] 1.34 ± ] ] ] Q = / = Q table = 0.56 Q > Q table so it can be rejected 5] 1.72±(2.447*0.17/7 1/2 )=1.72±0.16mg/g 6] z=( /0.17)=2.8 Table 4-1 z=2.8;area=.4974 chance = = , 0.26% chance 7] F=14 2 /11 2 =1.62 F- Table = 3.18 so they are not different from each other 8] S pooled =( * *3/64-2) 1/2 =0.332 t=( /0.332)(6*4/64) 1/2 =2.41 t so they are different from each other. 9] ) ) ) 9.7 x

3 13) The solution delivery process will leave behind a small amount of liquid in the tip. 14) x(0.1222x) 2 =1.7e- 5; x=1.7e- 5/(0.1222x) 2 ; let 2x=0; x 1 =1.14e- 3 (10pts) x 2 =1.10e- 3; x 3 =1.10e- 3 3

4 Problem Set 2 1) Detection limit of any instrumental method is defined as (5 points) a) signal/background = 4/1 b) background/signal = 2/1 c) signal/background = 3/2 d) signal/background = 3/1 2) Express the answer for the following calculation the proper number of significant figures: (5 points) (2.772± ±0.05) = Answer 3) The concentration of H in a ph solution is (5 points) a) b) c) d) e) ) Standard deviation is expression of (5 points) a) precision b) background c) sensitivity d) accuracy e) dynamic range 5) The analysis of Mn (m/m) was conducted on a Martian rock sample. The following values were obtained: 4.77% 4.82% 5.22% 4.92% 5.82% 4.99% Using valid statistics, which if any of the values can be rejected? Show your work for credit (5 points) 6) Sketch a plot of a calibration curve. Label the axes and the following: (10 points) a) background b) dynamic range c) sensitivity d) limit of detection 7) A sample solution with an unknown concentration of herbicide (λ max = 636 nm) was analyzed by absorption spectroscopy. A 10.0 ml sample was diluted to ml and the absorbance was measured as Another 10.0 ml sample was mixed with 10.0 ml of M then diluted to ml. The absorbance of this solution was measured as The absorbance of the blank was zero. What is the concentration of this herbicide? 4

5 8) The analysis of Pb in a drinking water was repeated 6 times and yielded a mean of ppm with a standard deviation of ppm. What are the limits for the concentration assuming a 95% confidence level? (5 points) a) 0.245±0.065 ppm b) 0.245±0.011 ppm c) 0.245±0.013 ppm d) 0.245±0.007 ppm 9) The ph of a solution of a M weak acid (HA) a = is (5 points) a) b) c) d) ) The solubility of the salt MA 2 ( sp = ) is (5 points) MA(s) = M 2 (aq) 2A - (aq) a) b) c) d) ) What is the solubility of a salt, AB ( sp = ) in the presence of 0.10 M B -? (5 points) a) b) c) d) ) The mass of 37.1% (m/m) HCl(aq) (d = 1.19 g/ml, MW HCl = 36.46) required to make 2.00 L of 1.00 M HCl is (5 points) a) 83.9 g b) 128 g c) 72.9 g d) 197 g 13) A ml sample was diluted to 2.00 L. A subsequent analysis revealed that the concentration of analyte revealed in the diluted sample was M. What is the concentration of this analyte in the original undiluted sample? (5 points) a) M b) M c) M d) M 14) The a of a weak acid (HA) is What is b for the following reaction? (5 points) A - H 2 O = HA OH - a) b) M c) M d) M 15) The Rope- A- Dope fishing line company guarantees that their Jaws- Max nylon line will haul in at least an 80 lbs gilled monster. Their chief statistician, Myron numbers has 200 samples of the Jaws- Max line tested and finds that the mean weight for line 5

6 breakage is 120 lbs with a standard deviation of 60 lbs. What are the chances that the hooked 80 pounder will get away if you were using Jaws- Max and end up being another fishing story? (10 points) Problem Set 2 Answers 1] d 2] 11.04/ ] b 4] a 5] Q=( )/( ) = 0.57 df =5 Q table = 0.56 < 0.57 the number can be rejected. 6] see book and lecture notes 7] = kc x (10.0/500.0) k 5.00e- 3*(10.00/500.0) = - kc x (10.0/500.0) = k 5.00e- 3*(10.00/500.0) k = 1930 use = kc x (10.0/500.0) = 1930C x (10.0/500.0) C x = 9.48e- 3 M 8] /- (2.571*0.011/6 1/2 ) = / ppm* *b was the best answer. 9] HA = H A x x x x 2 / x = 2.7e- 6 x = 5.2e- 4 M ph = ] MA 2 = M 2 2A x 2x (2x) 2 x = 8.9e- 17 x = 2.8e- 6 M 11] AB = A B x 0.10x x(0.10x) 0.10x = 7.2e- 12 x = 7.2e- 11 M 12] 2.00L*1.00 mol/l*36.46g/mol*1/0.371 = 197 g 13] 1.00e- 3 M * 2.00/ = 2.00e- 2 M 14] a b = w b = 1.00e- 14/7.2e- 6 = 1.4e- 9 15] z = [x x- bar]/s = [80 120]/60 = z 0.7 area = Area from 0 to 80 is = % 6

7 Problem Set 3 1) The background of a method based on Beer s law is best described as a) concentration b) absorbance c) molar absorptivity d) noise 2) Express the answer for the following calculation the proper number of significant figures: (5 points) (2.75cm ±0.03 cm 4.28cm ±0.05 cm) = All answers are in units of cm 2 a) ± 1.6 b) ± 1.60 c) ± 0.20 d) 11.8 ± 0.2 e) 12 ± 0.2 3) The concentration of H in a ph 8.55 solution is (5 points) a) 2.82e- 9 M b) 2.818e- 9 M c) 3.5e8 M d) 3.55e8 M e) 2.8e- 9 M 4) Linear Range is an expression of (5 points) a) signal to noise ratio b) analyte concentration range over which the c signal c) analyte detection limits of method d) accuracy and precision of method or technique e) precision of repeated experiment results 5) What is the ph of a 2.11 M solution of HNO 3 (aq)? a) b) c) d) 7.76e- 4 e) ) Which of the following best describes the reason for using the method standard addition over the calibration curve? a) Limited dynamic range of technique b) To compensate for nonlinear effects c) To accommodate the effects of a complex matrix d) To increase the detection limit of the method e) To decrease the detection limit of the method 7

8 7) What is the 95% confidence interval for 5 measurements whose average is 3.44 and with a standard deviation of 0.04? a) ± 0.04 b) ± 0.4 c) ± 0.05 d) ± 0.1 e) ± ) What is the relative population that lies above the value of 55.1 for a Gaussian distribution whose mean is 33.8 and with a standard deviation of 11.8? a) 0.50% b) 3.6% c) 1.8% d) 46% e) 0. 18% 9) Which of the following values may be discarded with 90% confidence? a) 9.77 b) 9.11 c) 8.89 d) none of the above 10) What is the volume of M HCl(aq) required to make a solution of ml of M HCl(aq)? a) ml b) 622- ml c) ml d) 180- ml e) 233- ml 11) The concentrated HCl(aq) is 37.1% (m/m) HCl(aq) (d = 1.19 g/ml, MW HCl = 36.46). What is the molarity of this solution? a) 6.22 M b) 8.44 M c) 12.1 M d) 3.77 M e) 6.71 M 12) The molality of a solution of HX is What is the molarity of that solution of the density is measured as 1.33 g/ml and the MW of the solute is 88.2 g/mol? a) M b) 1.24 M c) 2.81 M d) 1.82 M e) M 13) A sample solution of an analyte has an absorbance of A solution of standard has an absorbance of when that analyte has a concentration of 3.44e- 3M. Assuming that Beer s law applies to both solutions what is the concentration of analyte in the sample? Also assume that the absorbance of the blank solution is zero. a) 3.77e- 3 M b) 1.81e- 3 M c) 8.19e- 4 M d) 4.66e- 3M e) 2.41e- 3 M 8

9 14) Two methods of analyses were compared. Method A had a mean of 23.2 with a standard deviation of 4.4. Method B had a mean of 24.1 with a standard deviation of 4.8. Both sets of measurements were done 6 times. What is the F ratio and are the standard deviations significantly different from each other at the 95% confidence level? a) 1.19, no b) 1.19, yes c) 0.840, no d) 0.840, yes e) 1.09, no 15) Standard deviation can be best described as a measure of a) detection limit b) accuracy c) sensitivity d) linearity e) precision 16) The method of least squares fits a line (L) to a set of x,y data by a) maximizing Σ (x i - x L ) b) minimizing Σ (x i - x L ) 2 c) minimizing Σ (y i - y L ) 2 d) maximizing Σ (y i - y L ) 2 e) minimizing Σ (y i - y L ) 17) When does Beer s law typically fail? a) when c < 1 M b) when A > 1 c) when e > 1e4 d) when b = 1 cm e) when eb < 0 18) Sensitivity in a Beer s law analysis can be best described as a) The value of A when c = 0 b) The product of e b c) The minimum c detectable by the method d) The concentration range in which A c e) The precision achieved when the method is repeated several times 19) (10 points) A sample solution was analyzed by the standard addition method using its absorbance characteristic at 455 nm. a) In the first experiment a ml of aqueous sample was diluted to ml with water. Its measured absorbance is b)in the second experiment ml was mixed with µl of 3.22e- 5M and diluted to ml with water. The measured absorbance of this solution is What is the concentration of the analyte in the sample? 9

10 Answers to Problem Set 3 Answers 1) d 2) d (2.75 cm ±0.03 cm 4.28cm ±0.05 cm) = (2.75 cm ±1.1% 4.28 cm ± 1.2%) = ± (1.1% 2 1,2% 2 ) 1/2 = 11.8 cm 2 ± 1.6% = 11.8 cm 2 ± 0.2 cm 2 3) e 4) b 5) a 6) c 7) c ± (0.04)/(5) 1/2 8) b z = ( )/ use table 4-1 area = above = % 9) a Q = / = 0.75 Q table = 0.64 for n = 5 so 9.77 can be discarded 10) d 11) c 12) d 1.56 mol * 88.2 g HX = g 1.56 mol HX/1137.5g soln * 1.33 g/ml * 1000 ml/l = mol/l 13) e 0.229/0.327 = c/3.44e- 3 14) a 15) e 16) c 17) b 18) b 19) Part a = k (10.00/500.0) c Part b = k (10.00/500.0) c k (1.00e- 6/0.5000) 3.22e = k (1.00e- 6/0.5000) 3.22e- 5 k = 3.73e = 3.73e8 (10.00/500.0) c c = 5.1e- 8 M 10

11 Problem Set 4 1] What is the ph of a solution of M Na 2 HA solution given the following: H 3 A = H 2 A - H a = 2.8e- 2 H 2 A - = HA 2- H a = 7.7e- 5 HA 2- = A 3- H a = 9.3e- 11 2] What is the MBE for 1.00e- 3 M [Ag(NH 3 ) 2 ]Cl for the following reaction sequence? [Ag(NH 3 ) 2 ]Cl à Ag(NH 3 ) 2 Cl - Ag(NH 3 ) 2 = Ag(NH 3 ) NH 3 Ag(NH 3 ) = Ag NH 3 3] What is the CBE for the follow reaction sequence? H 2 S = H HS - HS - = H S 2- H 2 O = H OH - 4] What is the ph of a solution containing 0.25 M sodium acetate, and 0.25 M CH 3 COOH? a = 1.75e- 5 5] Which of the following monoprotic acids would be best for creating a buffer system at ph 7.00? acid A a = 5.6e- 4 acid B a = 7.7e- 6 acid C a = 1.9e- 8 acid D a = 7.3e- 11 6] What is pag when of ml of M AgNO 3 is mixed with ml of M NaCl? AgCl sp = 1.8e- 10 7] The weak acid, HA has a = 1.0e- 5. What is the fraction, α A- at ph 7.00? 8] What is the MBE for the following sequence of reactions? MgF 2 = Mg 2 2F - sp F - H 2 O = HF OH - b Mg 2 H 2 O = Mg(OH) H β 9] What is for this reaction? 2- H 2 SO 3 = SO 3 2H =? H 2 SO 3 = HSO - 3 H = 1.23e- 2 HSO = SO 3 H = 6.6e- 8 11

12 10] What is the difference between the end point and the equivalence point? 11] What is the solubility of SrF 2 ( sp = 2.8e- 9) at ph 4.00? HF a = 6.76e- 4. For partial credit clearly express what you are doing, e.g. substitutions label equations as #1, #2. (20 points) 12] A mixture of AgCl (MW , sp =1.8e- 10) and AgBr (MW 187.9, sp =5.0e- 13) weighs g. This mixture is reduced to silver metal (AW 107.9), which weighs g. Calculate the mass of AgCl in the original sample. (20 points) 13] What is the concentration Cl - required to remove 99% of Ag in a solution of F AgNO 3? (10 points) Answers to Problem Set 4 1] What is the ph of a solution of M Na 2 HA solution given the following: H 3 A = H 2 A - H a = 2.8e- 2 H 2 A - = HA 2- H a = 7.7e- 5 HA 2- = A 3- H a = 9.3e- 11 ph = ½(- log7.7e- 5 - log9.3e- 11) = ] What is the MBE for 1.00e- 3 M [Ag(NH 3 ) 2 ]Cl for the following reaction sequence? [Ag(NH 3 ) 2 ]Cl à Ag(NH 3 ) 2 Cl - Ag(NH 3 ) 2 = Ag(NH 3 ) NH 3 Ag(NH 3 ) = Ag NH e- 3 M = [Ag(NH 3 ) 2 ] [Ag(NH 3 ) ] [Ag ] 3] What is the CBE for the follow reaction sequence? H 2 S = H HS - HS - = H S 2- H 2 O = H OH - [H ] = [OH - ] [HS - ] 2[S 2- ] 4] What is the ph of a solution containing 0.25 M sodium acetate, and 0.25 M CH 3 COOH? a = 1.75e- 5 ph = pa log [base]/[acid] =

13 5] Which of the following monoprotic acids would be best for creating a buffer system at ph 7.00? acid A a = 5.6e- 4 acid B a = 7.7e- 6 acid C a = 1.9e- 8 acid D a = 7.3e- 11 acid C 6] What is pag when of ml of M AgNO 3 is mixed with ml of M NaCl? AgCl sp = 1.8e ] The weak acid, HA has a = 1.0e- 5. What is the fraction, α A- at ph 7.00? ] What is the MBE for the following sequence of reactions? MgF 2 = Mg 2 2F - sp F - H 2 O = HF OH - b Mg 2 H 2 O = Mg(OH) H β 2[Mg 2 ] 2[Mg(OH) ] = [F - ] [HF] 9] What is for this reaction? 2- H 2 SO 3 = SO 3 2H =? H 2 SO 3 = HSO - 3 H = 1.23e- 2 HSO = SO 3 H = 6.6e e ] What is the difference between the end point and the equivalence point? Equivalence point is where the titrant added stoichiometrically consumes all of the analyte. The end point is where some physical property indicates that the equivalence point is reached. The two volumes are quite often different from each other. 11] What is the solubility of SrF 2 ( sp = 2.8e- 9) at ph 4.00? HF a = 6.76e- 4. For partial credit clearly express what you are doing, e.g. substitutions label equations as #1, #2. (20 points) 13

14 SrF 2 (s) = Sr 2 2F - sp 2 pts F - H 2 O = HF OH - b = w / a 2 pts MBE: 2[Sr 2 ] = [F - ] [HF] 4 pts ph = 4.00 è [OH- ] = 1.0e- 10 M b = [HF][OH - ]/[F - ] è [HF] = [F - ] 2 pts next few steps 8 pts 2[Sr 2 ] = [F - ] [HF] = [F - ] #1 sp = [Sr 2 ][F - ] 2 è [F - ] = (sp/[sr 2 ]) 1/2 #2 Sub 2 into 1 2[Sr 2 ] = (sp/[Sr 2 ]) 1/2 [Sr 2 ] 3 = 9.52e- 10 [Sr 2 ] = s = 9.8e- 4 M 2 pts 12] A mixture of AgCl (MW , sp =1.8e- 10) and AgBr (MW 187.9, sp =5.0e- 13) weighs g. This mixture is reduced to silver metal (AW 107.9), which weighs g. Calculate the mass of AgCl in the original sample. (20 points) x g AgCl y g AgBr = g 4 pts x g AgCl*(mol AgCl/ g)*(mol Ag/mol AgCl)*(107.9 g/mol) = x g Ag 4 pts y g AgCl*(mol AgBr/187.9 g)*(mol Ag/mol AgBr)*(107.9 g/mol) = y g Ag 4 pts x g Ag y g Ag = g 4 pts y = x sub into above x (2.000 x) = g mass Ag = g 4pts 13] What is the concentration Cl - required to remove 99% of Ag in a solution of F AgNO 3? (10 points) [Ag ] = (1-0.99) F = 1.00e- 3 M sp = [Ag ][Cl - ] = 1.8e- 10 = 1.00e- 3 M*[Cl - ] è [Cl - ] = 1.80e- 7 M 14

15 Problem Set 5 1] What is the ph of a solution of a 1.0 M phthalic acid solution? (5 points) COOH COOH a1 = 1.12e- 3 a2 = 3.90e- 6 a) 5.92e- 4 b) 8.36 c) 7.22 d) 4.18 e) ] A g sample contained only NaBr (MW ) and NaCl (MW 58.44). It was dissolved into water and precipitated with excess AgNO 3. The precipitate (AgBr(s) (MW ) & AgCl(s) (MW )) was dried. The mass of this precipitate weighed g. What are the two equations necessary to solve for the masses of each equation? Let x = grams of NaBr and y = grams of NaCl a) x y = & (x/102.89) (y/58.44) = b) x y = & {x(187.80/102.89)} {y(143.35/58.44} = c) x y = & {x(187.80/102.89)} {y(143.35/58.44} = d) x y = & {x(187.80/102.89)} {y(143.35/58.44} = e) x y = & {x(102.89/187.80)} {y(58.44/143.35} = ] What is the mole fraction of HA - at ph 3.00 given H 2 A = H HA - a1 = 1.0e- 3 HA - = H A 2- a2 = 1.0e- 9 a) 0.25 b) 0.50 c) 0.88 d) 0.15 e) ] What is the ph of a solution of 0.10 M NaHCO 3? H 2 CO 3 = HCO - 3 H HCO - 3 = CO 2-3 H a1 = 4.45e- 7 a2 = 4.69e- 11 a) 8.34 b) 6.22 c) 7.84 d) 9.71 e) ] What is or are the simplifying assumption(s) that allow for the use of the Henderson- Hasselbalch equation? ph = p a log([a - ]/[HA]) 15

16 a) p a > ph b) p a = ± 1 ph & [A - ] = [HA] c) formal concentrations of A - & HA are the c) formal concentrations of A - & HA are not same as equilibrium concentrations the same as equilibrium concentrations 6] Write a charge balance equation of for a solution of 0.10 phthalic acid. Neglect the ionization of water. See problem 1 for structure. CBE: 7] Write a mass balance equation for the M acetic acid. CH 3 COOH = H CH 3 COO - a = 1.75e- 5 8] What is the ph of a solution consisting of M CH 3 COONa and M CH 3 COOH? See also problem 7. a) 1.75 b) 4.76 c) d) e) ] A common primary standard for the standardization of bases is a) HCl(aq) b) Potassium Hydrogen Phthalate c) Bromine d) NaOH(aq) e) DNA 10] Solutions of NaOH(aq) titrant must be restandardized frequently because of a) solvation of NO 2 from the atmosphere producing HNO 3 b) solvation of Cl 2 from the atmosphere producing HCl c) solvation of SO 2 from the atmosphere producing H 2 SO 3 d) solvation of N 2 from the atmosphere producing HNO 3 e) solvation of CO 2 from the atmosphere producing H 2 CO 3 11] What is the definition of titrant a) it is the reagent solution whose concentration is known and is added b) it is the analyte whose concentration in sample is unknown c) it is the concentration of sample at the end point d) it is the volume difference between the end and equivalence points 16

17 to the sample in small increments 12] Write down the hydrolysis reaction for HCO 3 - demonstrating that it is a weak base. 13] The b for dichloroacetate, Cl 2 CHCOO - is Cl 2 CHCOOH a = 5.0e- 2 a) 1.0e- 14 b) 2.0e- 12 c) 4.0e- 8 d) 2.0e- 13 e) 5.0e- 9 14] A ml solution of M NaI(aq) is titrated with M AgNO 3 (aq). If ml of the AgNO 3 solution is added, which of the statements below is true? a) This is the equivalence point. b) The solution is pink in color. c) There is a solid precipitate and an excess of Ag. d) There is no solid precipitate but there is excess of Ag relative to I -. e) There is a solid precipitate and an excess of I -. Problems are based on the following: A ml sample of M SCN is titrated with M CuNO 3. The sp of CuSCN is 4.8e ] Write down the reaction that takes place during this titration. 16] What is pcu when ml of the M CuNO 3 is added to the ml sample of M SCN solution? The sp of CuSCN is 4.8e- 15. a) 1.33 b) c) d) 5.87 e) ] What is pcu when ml of the M CuNO 3 is added to the ml sample of M SCN solution? 17

18 a) 7.93 b) 6.44 c) 9.13 d) 7.16 e) ] What is pcu when ml of the M CuNO 3 is added to the ml sample of M SCN solution? a) 3.22 b) 4.44 c) 7.18 d) e) ] Which diagram best describes the curve for the titration of ml of M SCN with M CuNO 3? Vol SCN Vol SCN [Cu ] pcu Vol CuNO 3 [Cu ] Vol SCN Vol CuNO 3 Answers to Problem Set 5 1] e) 1.48 only a1 is important. x 2 /(1.0- x) = 1.12e- 3; x = ] c) x y = & {x(187.80/102.89)} {y(143.35/58.44} = ] b) 0.50, D = [1.0e- 3] 2 [1.0e- 3] 2 [1.0e- 3*1.0e- 9] = 2.0e- 6, N = [1.0e- 3] 2 = 1.0e- 6, α = ] a) 8.34 = ½ (p a1 p a2 ) 5] c) formal concentrations of A - & HA are the same as equilibrium concentrations 6] [H ] = 2[A 2- ] [HA - ] 7] 0.10 M = [CH 3 COOH] [CH 3 COO - ], also [H ] = [CH 3 COO - ] 8] c) ph = p a watch S.F. 9] b) Potassium Hydrogen Phthalate 10] solvation of CO 2 from the atmosphere producing H 2 CO 3 11] a) it is the reagent solution whose concentration is known and is added to the sample in small increments 12] HCO 3 - H 2 O à H 2 CO 3 OH - 13] b = w / a = 2.0e ] There is a solid precipitate and an excess of Ag. Pink is O (Fajan s method) 15] Cu SCN - = CuSCN(aq) 16] Mol SCN - = L*0.100 mol/l = 5.00e- 3 Mol Cu = L*0.100 mol/l = 2.50e- 3 18

19 Excess mol SCN - = 5.00e e- 3 = 2.50e- 3 [SCN - ] = 2.50e- 3 mol/ L = 3.33e- 2 M CuSCN(s) = Cu SCN x 3.33e- 2 x (3.33e- 2 x)x (3.33e- 2 )x = 4.8e- 15 x = 1.44e- 13 M pcu = ] Mol SCN - = L*0.100 mol/l = 5.00e- 3 Mol Cu = L*0.100 mol/l = 5.00e- 3 CuSCN(s) = Cu SCN x x x 2 = 4.8e- 15 x = 6.9e- 8 pcu = ] Mol SCN - = L*0.100 mol/l = 5.00e- 3 Mol Cu = L*0.100 mol/l = 7.50e- 3 Excess mol Cu = 7.50e e- 3 = 2.5e- 3 [Cu ] = 2.50e- 3 mol/0.125 L = M pcu = ] Which diagram best describes the curve for the titration of ml of M SCN with M CuNO 3? pcu Vol CuNO 3 19

20 Problem Set 6 1] The ph of solution of M of a weak acid, HA is What is a for this acid? 2] What is the aqueous solubility of AgCl at ph 4.00 ( sp = 1.8e- 11)? 3] The two a s for salicylic acid (H 2 A) are 1.07e- 3 and 1.82e- 14. What is b for sodium salicylate (NaHA)? 4] What is the charge balance for a solution of 0.10 M NaHCO 3? a1 H 2 CO 3 = a2 HCO 3 - = HCO 3 - H CO 3 2- HCO 3 - H 2 O H 2 CO 3 OH - 5] Which of the following is a valid mass balance for a solution for 0.10 M NaHCO 3? 6] Which of the following would best explain the solubility of Ag 2 SO 4? 7] A g silver ore sample was treated with HNO 3 and then with excess NaCl(aq). A precipitate was dried and weighed g. What is percent silver in the ore? AW: Ag 107.9, H 1.008, O 16.00, N 14.01, Cl ] Which of the following is not a primary standard? a) Potassium Hydrogen Phthalate (HP) b) Benzoic Acid c) Potassium Hydrogen Iodate d) NaOH e) NaHCO 3 Problems 9-11: A solution of M AgNO 3 is used to titrate a ml solution of M Cl. The sp of AgCl is 1.8e- 11 9] What is pag if ml of the titrant is added to the Cl solution? 10] What is pag if ml of the titrant is added? 20

21 21 12] What is ph of solution containing M HOCl ( a = 3.0e- 8) and M NaOCl. 13] The useful ph range of most buffering systems is a) within 1 ph unit of p a b) within 0.5 ph unit of p a c) within 10 ph units of p a d) within 5 ph units of p a e) within 0.1 ph unit of p a 14] What is the ph of solution that is 0.10 M NaH 2 PO 3? H 3 PO 3 H 2 PO 3 - H a = 3e- 2 H 2 PO 3 - HPO 3 2- H a = 1.62e- 7 15] The fraction (relative concentration) of H 2 PO 3 - from a 0.10 F H 3 PO 3 at ph 5.00 can be calculated from which formula? a) ] [ ] [ ] [ a a a H H H b) ] [ ] [ ] [ a a a a H H H c) ] [ ] [ a a a a a H H d) ] [ ] [ ] [ a a a H H H e) ] [ ] [ ] [ a a a a H H H 16] Which of the follow is an amphoteric species? a) H 2 CO 3 b) HF

22 c) F - d) HCO 3 - e) CO ] What is the mass balance equation for the following sequence of reactions? CaC 2 O 4 (s) Ca 2 2- C 2 O 4 C 2 O 2-4 H 2 O HC 2 O - 4 OH - b1 HC 2 O - 4 H 2 O H 2 C 2 O 4 OH - b2 sp = 1.3e- 8 18] What is the charge balance equation for the reaction sequence of problem 17? 19] What is the molar solubility of BaF 2 ( sp = 1.7e- 6) at ph 7.20? a (HF) = 6.8e- 4 Answers to Problem Set 6 1] [H ] = 2.0e- 6 M HA = H A x x x a = x 2 / ( x) 2.0e- 6 2 / = 8.3e- 11 Answer d) 2] AgCl Ag Cl x x sp = x 2 = 1.8e- 11 x = 4.2e- 6 Answer a) 3] HA - H 2 O H 2 A OH - a b = w b = w / a = 1.00e- 14 / 1.07e- 3 = 9.35e- 12 Answer c) 4] Answer b: [Na ] [H ] = [HCO 3 - ] 2[CO 3 2- ] [OH - ] 5] Answer e: 0.10 M = [H 2 CO 3 ] [HCO 3 - ] [CO 3 2- ] 6] Answer d: [Ag ]/2 7] g AgCl (mol AgCl / g) (mol Ag / mol AgCl) ( g / mol Ag) (100 / g) = 5.37 % 8] Answer d): NaOH 9] Titration Rxn: Ag Cl - AgCl(s) 22

23 Moles of excess Cl - = L * M Cl L * M AgNO 3 = 5.00e- 3 mol Cl - [Cl - ] = 5.00e- 3 mol Cl - / L = 3.33e- 2 AgCl(s) Ag Cl x x3.33e- 2 x(3.33e- 2) 1.8e- 11 x = 5.4e- 10 M pag = 9.27 answer a) 10] This is the equivalence point. AgCl(s) Ag Cl x x x 2 = 1.8e- 11 x = 4.24e- 6 pag = 5.37 answer b) 11] We are past the equivalence point. ( ) ml * M Ag ( ml) = M Ag or pag = answer d) 12] a = [H ][OCl - ] / [HOCl] [H ] = a [HOCl] / [OCl - ] [H ] = 3.0e- 8 ph = 7.52 Answer a) 13] Answer a) 14] ph = ½ (p a1 p a2 ) = ½ ( ) = 4.1 answer e) 15] answer: b b) [ H ] 2 a1 a1 [ H [ H ] ] a1 a2 16] HCO 3 - Answer d 17] Answer b 18] 2[Ca 2 ] = 2[C 2 O 4 2- ] [HC 2 O 4 - ] [OH - ] Answer e 23

24 19] MBE: 2[Ba 2 ] = [F - ] [HF] & [H ] = 6.31e- 8 M; [OH - ] = 1.58e- 7 sp = [Ba 2 ][F - ] 2 = 1.7e- 6 a (HF) = [H ][F - ]/[HF] = 6.8e- 4 3 variables: [Ba 2 ], [F - ], [HF] F - H 2 O HF OH - b = w / a = 1.00e- 14/6.8e- 4 = 1.47e- 11 Using b solve for [HF] [HF] = b [F - ]/[OH - ] will sub into MBE 2[Ba 2 ] = [F - ] [HF] 2[Ba 2 ] = [F - ] b [F - ]/[OH - ] Sub all knowns into above 2 [Ba 2 ] = [F - ] 1.47e- 11*[F - ]/1.58e- 7 2 [Ba 2 ] [F - ] sub into sp sp = [Ba 2 ][F - ] 2 = [Ba 2 ](2[Ba 2 ]) 2 [Ba 2 ] = (sp/4) 1/3 = 7.5e- 3M 24

25 Problem Set 7 (starts with question 3) 2] What is the ph after ml of M HClO 4 is added to 50.0 ml of M NaCN? 3] The selection of a visual acid- base indicator should be based on. a) its transition ph which should not overlap the steepest portion of the titration curve. b) its transition ph which should overlap the excess H portion of the titration curve. c) its transition ph which should overlap the buffer portion of the titration curve. d) its transition ph which should overlap the flattest portion of the titration curve. e) its transition ph which should overlap the steepest portion of the titration curve. 4] The conditional formation constant f for CaY 2- is related to f through which of the relationships? a) f = f at ph =0 b) f = α y4- f c) f = α y4- f d) f = 1 / f e) f = f 2 5] In reference to EDTA titrations the symbol, α y4-, indicates which of the following? a) The fraction of metal chelated by EDTA b) The concentration of EDTA in the Y 4- form. c) The fraction of EDTA in the Y 4- form. d) The analytical concentration of metal. e) The fraction of EDTA not in the Y 4- form. 6] A spontaneous electrochemical cell would have which of the following? a) E cell = 0 b) E cell > 0 c) E cell < 0 d) E cell 0 e) E cell 0 7] If A e - = B has E 0 = V then the E 0 for 2A 2e - = 2B is. 25

26 8] The standard cell potential for the following is Fe(s)/Fe 2 (aq)//sn 2 (aq)/sn(s) Fe 2 2e - = Fe(s) E 0 = V Sn 2 2e - = Sn(s) E 0 = V 9] The E 0 for the following is FeCO 3 (s) 2e - 2- = Fe(s) CO 3 Fe 2 2e - = Fe(s) sp {FeCO 3 (s)} = 2.1e- 11 E 0 =? E 0 = V 10] It is advantageous to conduct EDTA titrations of metal ions in a) acidic ph s to assist metal ion hydrolysis b) basic ph s to prevent metal ion hydrolysis c) basic ph s to maximize Y 4- fraction d) basic ph s to minimize Y 4- fraction e) acidic ph s to maximize Y 4- fraction 11] Which is true of the equivalence point for the redox titration of Fe 2 with Ce 4? a) only [Fe 2 ] = [Fe 3 ] b) [Fe 2 ] = [Ce 3 ] and [Ce 4 ] = [Fe 3 ] c) [Fe 2 ] = [Ce 4 ] and [Fe 3 ] = [Ce 3 ] d) [Fe 2 ] = [Fe 3 ] and [Ce 4 ] = [Ce 3 ] e) [Fe 2 ] = 0 12] A ph electrode was found to have a potential of V in a ph 4.01 buffer solution. A sample solution was found to have a potential of V. What is the ph of that sample? E = const ph 13] MnO 4 can be standardized with which of the following? a) H 2 O 2 b) CH 4 c) H 2 O d) NaC 2 O 4 e) C 2 H 4 14] Calculate pca if 20.0 ml of M of EDTA is added to 15.0 ml of M Ca 2 at ph

27 15] Calculate the cell potential when 25.0 ml of M Ce 4 is added to 15.0 ml of M Fe 2. Ce 4 e - = Ce 3 Fe 3 e - = Fe 2 E 0 = 1.70 V E 0 = V 16] How many grams of 2 C 2 O 4 (MW ) must be added 25.0 ml of M HCl to give a ph of when this solution is diluted to ml. a1 = 5.60e- 2 a2 = 5.42e- 5 Answers to Problem Set 7 2] [H ] = 5.00 mmol / ml = 3.33e- 2 ph = ] its transition ph which should overlap the steepest portion of the titration curve. 4] f = α y4- f 5] The fraction of EDTA in the Y 4- form. 6] E cell > 0 7] V 8] E = (- 0.44) = 0.30 V 9] E = (0.0592/2) log 1/ sp = V 10] basic ph s to maximize Y 4- fraction 11] [Fe 2 ] = [Ce 4 ] and [Fe 3 ] = [Ce 3 ] 12] ] NaC 2 O 4 27

28 14] mol EDTA = 20.0 ml * M = 1.0 mmol mol Ca 2 = 15.0 ml * M = 0.75 mmol excess EDTA region where, [CaY 2- ] = 0.75 mmol / 35.0 ml = 2.1e- 2 M [EDTA] = 0.25 mmol / 35.0 ml = 7.1e- 3 M f = [CaY 2- ] / [Ca 2 ]*[Y 4- ] [Y 4- ] = α Y4- [EDTA] f *α Y4- = f = [CaY 2- ] / [Ca 2 ]*[EDTA] f = 4.9e10 f = 5.4e- 2*4.9e10 = 2.6e9 2.6e9 = 2.1e- 2 M / [Ca 2 ]*7.1e- 3 M [Ca 2 ] = 1.1e- 9 M pca = ] Mol Ce 4 = 25.0 ml * M = 0.25 mmol Mol Fe 2 = 15.0 ml * M = 0.15 mmol Excess Ce 4 region Ce 4 Fe 2 = Fe 3 Ce E = 1.70 (0.0592) log (0.15/0.10) = 1.69 V 16] p a1 = 1.25 p a2 = 4.27 Therefore only a2 is important. HC 2 O 4 - = H C 2 O 4 2- [C 2 O 2-4 ] i = x / L [H ] i = 25.0*0.500 mmol / ml = M [H ] f = = 3.16e- 5 M 2- x = mol C 2 O 4 28

29 H 2- C 2 O 4 = - HC 2 O x / 0.500L 0 - y - y y 3.16e y = 3.16e- 5 y = M a = [H ][C 2 O 4 2- ] / [HC 2 O 4 - ] 5.42e- 5 = [3.16e- 5][C 2 O 4 2- ] / [0.025] [C 2 O 4 2- ] = 4.29e- 2 M 4.29e- 2 M = x / 0.500L 3.16e- 5 x = 2.15e- 2 mol g 2 C 2 O 4 = 2.15e- 2 mol * g/mol = 3.57 g 29

30 Problem Set 8 Problems 1-2 involve the following titration: 1] ml of M CH 3 COOH(aq) (pa = 4.757) is titrated with M NaOH(aq). What is the volume of NaOH solution required to reach the equivalence point? (5 points) a) 150- ml b) ml c) ml d) 100- ml e) 125- ml 2] What is the ph of the equivalence point in problem 1? (5 points) a) b) c) d) e) ] What is the final ph if solutions of ml of M NaOH(aq) and ml of M CH 3 COOH(aq) are added together? (5 points) a) b) c) d) e) ] What is the difference between the end point and equivalence point for a monobasic- monoacid titration? (5 points) a] The equivalence point is where mol acid = mol base and the end point is where indicator changes color. b] The end point is where mol acid = mol base and the equivalence point is where indicator changes color. c] There are no differences between the concepts of the end and equivalence points. d] The equivalence point is where mol acid = mol base and the end point is where the ph is 14. 5] ml of M oxalic acid (H 2 C 2 O 4, pk a1 = 1.252, p a2 = 4.266) is titrated with ml of M NaOH. What is the ph of that titrated solution? (5 points) a) 7.00 b) c) d) e) ] What is f for SrEDTA 2- at ph 11? a) 0.85 b) c) d) e) 1.55e

31 7] The formal concentration of EDTA is 1.00 mm. What is the concentration of the Y 4- form at ph 4? a) 1.00 mm b) mm c) mm d) 0.85 mm e) Problems 8-11 involve the following titration. 8] A solution of ml of M NiCl 2 (aq)is titrated with M EDTA in a solution of M NH 3 at ph What is pni if ml of the titrant solution is added? Note that α Ni2 = at M NH 3. (5 points) a) b) c) d) e) ] What is f for the NiEDTA 2- complex in NH 3 at ph 11? (5 points) a) (0.85)18.62 b) c) (0.85) d) (0.85) ] What is [NiEDTA 2- ] if ml of titrant is added to the NiCl 2 solution in problem 8? (5 points) a) 2.00e2 b) 5.23e- 6 c) 2.00e- 9 d) 7.99e- 4 e) 4.00e- 4 11] Which is true if ml of M EDTA titrant is added to the ml of M NiCl 2 solution in 0.1M NH 3? Assume equilibrium conditions. (5 points) a) [Ni 2 ] = [EDTA] b) [NiEDTA 2- ] > [EDTA] c) [NiEDTA 2- ] = [EDTA] d) [Ni 2 ] > [EDTA] 12] An electrochemical cell will discharge spontaneously if (5 points) a) E cell < 0 b) E cell > 0 c) E cell = 0 d) does not depend on E cell 13] The reductions take place at which electrode? (5 points) a) anode b) toade c) cathode d) alkaline e) amphiprotic 31

32 14] The purpose of a reference electrode is to provide (5 points) a) to prevent mixing of the electrolyte solution. b) to provide a means of ionic transport between the anode and cathode c) to enable the reductive process at the anode. d) to provide a stable potential in which an electrode reaction can be compared to. e) to provide comic relief. 15] What is E 0 cell for the reaction below? (5 points) F 2 2Fe 2 = 2F - 2Fe 3 F 2 2e - = 2F - E 0 red = V Fe 3 e - = Fe 2 E 0 red = V a) V b) V c) V d) V e) V 16] What is E 0 cell for the reaction below? (5 points) Hg 2 SO 4 (s) 2e- = 2Hg(l) SO 4 2- Hg 2 2 2e - = 2Hg(l) Hg 2 SO 4 (s) = Hg SO 4 E 0 red = V sp = 7.4e- 7 a) E 0 cell = log = log 2 1 = log sp 1 = log 0 b) Ecell sp c) d) E E 0 cell 0 cell sp sp 17] What are the final concentrations of each ion when ml of M Ce 4 is mixed with ml of M Cu? (5 points) Ce 4 e = Ce 3 E 0 red = 1.44 V Cu 2 e = Cu E 0 red = V 18] What is the potential of the final solution when ml of M Ce 4 is mixed with ml of M Cu? (5 points) 32

33 Answers to Problem Set 8 1 & 2] eq. pt. mol CH 3 COOH = mol OH - mol CH 3 COOH = ml*0.100 M = 7.50 mmol vol NaOH = 7.50 mmol/ M = 150- ml total volume = 150- ml75.0- ml = 225- ml [CH 3 COO - ] = 7.50 mmol/225- ml = 3.33e- 2 M CH 3 COO - H 2 O = CH 3 COOH OH e x - - x x b = w / a = 5.71e- 10 = x 2 /(3.33e- 2- x) x 2 /(3.33e- 2) x = 4.36e- 6 poh = ph = ] mol CH 3 COOH = ml*0.100 M = 7.50 mmol mol OH - = ml* M = 10.0 mmmol excess OH- = mmol = 2.50 mmol [OH - ] = 2.50 mmol/275- ml = 9.09e- 3 poh = ph = ] a] The equivalence point is where mol acid = mol base and the end point is where indicator changes color. 5] mol of OH - = ml* M = 2.50 mmol mol of H 2 C 2 O 4 = ml* M = 2.50 mmol Therefore we have only HC 2 O 4 - which is an amphiprotic species. ph is ph = ½(pk a1 p a2 ) = ½( ) = ] f = α y4- f = 0.85*5.4e8 = 4.6e8 7] [Y 4- ] = 3.8e- 9*1.00e- 3 M = 3.8e- 12 M b and c are correct. 33

34 8] Initial mol Ni 2 = ml*1.00e- 3 M = mmol Added mol EDTA = ml*1.00e- 3 M = mmol Excess Ni 2 = mmol = mmol C Ni2 = mmol / ml = 3.33e- 4 M Free [Ni 2 ] = α Ni2 C Ni2 = 1.34e- 4*3.33e- 4 = 4.47e- 8 M pni = ] f = α Ni2 α Y4- * f = 1.34e- 4*0.85* = 4.7e14 10&11]Initial mol Ni 2 = ml*1.00e- 3 M = mmol Added mol EDTA = ml*1.00e- 3 M = mmol 12] b) E cell > 0 13] cathode [NiEDTA] = mmol / ml = 4.00e- 4 M Excess EDTA = mmol / ml = 2.00e- 4 M f = [NiEDTA]/C Ni *[EDTA] = 4.00e- 4/C Ni *2.00e- 4 = 4.7e14 C Ni = 4.3e- 15 [Ni 2 ] = 1.34e- 4*4.3e- 14 = 5.8e- 18 M pni = Therefore [NiEDTA 2- ] > [EDTA] 14] d) to provide a stable potential in which the cathode reaction can be compared. 15] Ecell = Ecath Eanod = = V 16] E = /2 log 1/[Hg 2 2 ] sp = 7.4e- 7 = [Hg 2 2 ][SO 4 2- ] [Hg 2 2 ] = 7.4e- 7/[SO 4 2- ] E = /2 log [SO 4 2- ]/7.4e- 7 = V 34

35 17&18] Reaction: Ce 4 Cu = Ce 3 Cu 2 Initial mol Ce 4 = ml* M = 1.25 mmol Initial mol Cu = ml* M = 0.75 mmol More Ce 4 than Cu therefore Mol Ce 3 = 0.75 mmol [Ce 3 ] = 0.75 mmol / ml = 1.9e- 2 M Mol Ce 4 = 0.50 mmol [Ce 4 ] = 0.50 mmol / ml = 1.3e- 2 M Mol Cu = 0.00 mmol [Cu ] = 0.00 Mol Cu 2 = 0.75 mmol [Cu 2 ] = 0.75 mmol / ml = 1.9e- 2 M E = log 1.9e- 2 / 1.3e- 2 = 1.43 V 35

36 Problem Set 9 1] When titrating a weak acid with a strong base it is expected that the equivalence point will be a) Slightly acidic, since the equilibrium HA Ý H A - predominates. b) Neutral since [OH - ] = [H ] c) Slightly basic since the equilibrium A - H 2 O Ý HA OH - predominates. d) It is impossible to know since it could be acidic or basic depending on the a of the acid. e) Strongly basic since excess OH - is present. 2] A sample solution of ml M oxalic acid (H 2 C 2 O 4 ) is titrated with ml of M of NaOH. Which of the following is true after the two solutions are mixed? a) This is the first equivalence point. b) This a ph buffer region where [H 2 C 2 O 4 ] = [HC 2 O 4 - ] c) This is a metal buffer region d) This is the second equivalence point. e) This is the excess OH - region where the ph is strongly alkaline. 3] A sample solution of ml M oxalic acid (H 2 C 2 O 4 ) is titrated with ml of M of NaOH. Which of the following is true after the two solutions are mixed? a) There is 2.50 mmol of H 2 C 2 O 4 present. b) There is 1.50 mmol of OH - present. c) There is 2.50 mmol of HC 2 O 4 - present. d) There is 1.00 mmol of H 2 C 2 O 4 present. e) There is 1.00 mmol of C 2 O 4 2- present 4] What is the fraction of EDTA in the Y 4- form at ph 7.00? 5] What is the conditional formation constant of CaEDTA 2- at ph 10.00? 6] The fraction of free metal in the following equilibrium can be expressed as: M L Ý ML É = [ML] / [M][L] 36

37 [M ] a) α m = 1 β [ M ] b) α m = 1 β[ L] 1 c) α m = 1 β[ L] 1 d) α m = 2 1 β[ L] β[ L] 1 e) α m = 1 β 7] What is the ph of a titration solution that consists of M CH 3 COOH ( a = 1.75e- 5) and M NaOH? 8] What is E 0 for the following half reaction if E 0 for Zn 2 2e - Ý Zn(s) is V? ½ Zn 2 e - Ý ½ Zn(s) 9] What is E 0 cell for the following reaction? 2Na(s) 2H Ý 2Na H 2 (g) Na e - Ý Na(s) E 0 = V 2H 2e - Ý H 2 (g) E 0 = V 10] What is the half reaction potential for reduction of 1.00e- 5 M H? 11] The response of a F - selective electrode was found to be V in standardize 1.00e- 3 M solution. The response of this electrode in an unknown solution of F - is V. What is [F - ] for that unknown solution? 12] ml of M Sn 4 is titrated with M Ti 2 in the following reaction: Sn 4 2Ti 2 = Sn 2 2Ti 3 What is the added volume of titrant required to reach the equivalence point? 37

38 13] Which of the following is true at the equivalence point of problem 12? a) [Sn 4 ] = [Ti 2 ] & [Sn 2 ] = [Ti 3 ] b) [Sn 4 ] = [Ti 2 ] = [Sn 2 ] = [Ti 3 ] c) 4[Sn 4 ] = 2[Ti 2 ] & 2[Sn 2 ] = 3[Ti 3 ] d) 2[Sn 4 ] = [Ti 2 ] & 2[Sn 2 ] = [Ti 3 ] e) [Sn 4 ] = 2[Ti 2 ] & [Sn 2 ] = 2[Ti 3 ] 14] The linear ph range for the average ph electrode is about: a) 0 to 14 b) - 5 to 18 c) 2 to 10 d) 1 to 14 e) - 10 to 10 15] A very common interference for the glass ph electrode is a) F - b) Cl - c) Br - d) Na e) C 16] Calculate the ph of a mixture of ml of M NaOH and ml of M H 3 AsO 4 (arsenic acid). H 3 AsO 4 = H 2 AsO - 4 H a1 = 5.8e- 3 H 2 AsO - 4 = HAsO 2-4 H a2 = 1.10e- 7 HAsO 2-4 = AsO 3-4 H a3 = 3.2e ] Which of the following species is the strongest reducing agent? A e- = A B e- = B - D 2 e- = D E 0 = 0.75 V E 0 = 0.25 V E 0 = V a) A b) B - c) B d) D 2 e) D 38

39 18] Calculate the standard state cell potential for the following Cu(s)/Cu 2 (aq)// (aq)/(s) 19] What is the standard state reduction potential for the following reaction? AgBr(s) e - = Ag(s) Br - Ag e - = Ag(s) AgBr(s) = Ag Br - E 0 = V sp = 5.0e- 13 Answers To Problem Set 9 1] c 2] d 3] e: 1.00 mmol past 1st eq. pt. Initial HA- = 2.5 mmol HA - OH - = A 2- H mmol A 2-4] b 5] a: f = 0.36* = 1.8e- 10 6] c 7] e: ph = pa 8] e 9] d: E 0 cell = ( ) V 10] c: E = E log 1/[H ] = log 1/[1.00e- 5] = V 11] c: E = const log [F - ] = const log [1.00e- 3] const = = log [F - ] [F - ] = 1.30e- 4 M 12] a: ml*0.125 M Sn 4 *(2mol Ti 2 /mol Sn 4 )*1/0.100 M Ti 2 = 45.0 ml 13] d 14] c 15] d 16] b: 2 nd eq. pt where all H 3 AsO 4 is titrated to HAsO 4 2- which is intermediate of H 2 AsO 4 - and AsO These two equilibria become important: H 2 AsO - 4 = HAsO 2-4 H a2 = 1.10e- 7 HAsO 2-4 = AsO 3-4 H a3 = 3.2e- 12 Therefore ph = ½(p a2 p a3 ) = ½ ( ) =

40 17] e 18] a: E cell = E cath E anod = = V 19] Start with: E = E 0 (Ag /Ag) log 1/[Ag ] Realize that sp = [Ag ] [Br - ] [Ag ] = sp / [Br - ] sub into Nernst Eqn above E = E 0 (Ag /Ag) log [Br - ]/ sp let [Br- ] = 1 for standard state conditions E 0 = log 1/5.00e- 13 = V 40