problemset1( - university of idaho 253/homewor… · · 2013-05-29answers(to(problem(set(1(! 1]...

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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 0.1000 M HCl are required to make 100.0 mL of 25.00 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 0.17. 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 = 0.37. Meanwhile, Joe Cutcorners used a modified procedure with n = 4, x = 6.87 mM with s = 0.22. 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 0.100 M HCl? (5 points) 10) Calculate the pH of a solution of 0.025 M acetic acid, Ka = 1.8 × 10

-‐5. (5 points)

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11) Calculate the pH of a solution of 0.025 M acetic acid and 0.025 M sodium acetate. (5 points) 12) Calculate the Ksp of barium sulfate (MW 233) if its solubility is measured as 0.0023 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 PbCl2 (Ksp = 1.7 x 10

-‐5) in the presence of 0.122 M NaCl. (10 points) Answers to Problem Set 1 1] 1.34 ± 0.03 2] 0.13 3] 25.0 4] Q = 2.99-‐2.54/2.99-‐2.47 = 0.865 Qtable = 0.56 Q > Qtable so it can be rejected 5] 1.72±(2.447*0.17/71/2)=1.72±0.16mg/g 6] z=(2.20-‐1.72/0.17)=2.8 Table 4-‐1 z=2.8;area=.4974 chance = 0.5000-‐0.4974 = 0.0026, 0.26% chance 7] F=142/112=1.62 F-‐Table = 3.18 so they are not different from each other 8] Spooled=(0.37

2*5+0.222*3/6+4-‐2)1/2=0.332 t=(6.87-‐6.37/0.332)(6*4/6+4)1/2=2.41 ttable@95%=2.306 so they are different from each other. 9] 1.000 10) 3.17 11) 4.74 12) 9.7 x 10-‐11

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13) The solution delivery process will leave behind a small amount of liquid in the tip. 14) x(0.122+2x)2=1.7e-‐5; x=1.7e-‐5/(0.122+2x)2; let 2x=0; x1=1.14e-‐3 (10pts) x2=1.10e-‐3; x3=1.10e-‐3

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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±0.002 + 8.27±0.05) =

Answer__________ 3) The concentration of H+ in a pH 6.772 solution is (5 points) a) 1.6904 × 10-‐7 b) 1.69 × 10-‐7 c) 1.690 × 10-‐7 d) 1.7 × 10-‐7 e) 1.69044× 10-‐7 4) 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 (λmax = 636 nm) was analyzed by absorption spectroscopy. A 10.0 mL sample was diluted to 500.0 mL and the absorbance was measured as 0.366. Another 10.0 mL sample was mixed with 10.0 mL of 5.00×10-‐3 M then diluted to 500.0 mL. The absorbance of this solution was measured as 0.559. The absorbance of the blank was zero. What is the concentration of this herbicide?

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8) The analysis of Pb in a drinking water was repeated 6 times and yielded a mean of 0.245 ppm with a standard deviation of 0.011 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 0.100 M weak acid (HA) Ka = 2.7×10

-‐6 is (5 points) a) 4.7 × 10-‐5 b) 4.7 × 10-‐6 c) 2.7 × 10-‐5 d) 5.2 × 10-‐4 10) The solubility of the salt MA2 (Ksp = 8.9×10

-‐17) is (5 points) MA(s) = M2+(aq) + 2A-‐(aq) a) 3.9 × 10-‐5 b) 2.8 × 10-‐6 c) 4.5 × 10-‐6 d) 8.4 × 10-‐7 11) What is the solubility of a salt, AB (Ksp = 7.2×10

-‐12) in the presence of 0.10 M B-‐? (5 points) a) 7.2 × 10-‐12 b) 7.2 × 10-‐11 c) 5.7 × 10-‐13 d) 2.7 × 10-‐6 12) The mass of 37.1% (m/m) HCl(aq) (d = 1.19 g/mL, MWHCl = 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 100.0 mL sample was diluted to 2.00 L. A subsequent analysis revealed that the concentration of analyte revealed in the diluted sample was 1.00 × 10-‐3 M. What is the concentration of this analyte in the original undiluted sample? (5 points) a) 1.00 × 10-‐3 M b) 2.00 × 10-‐2 M c) 5.00 × 10-‐5 M d) 2.00 × 10-‐1 M 14) The Ka of a weak acid (HA) is 7.2×10

-‐6. What is Kb for the following reaction? (5 points)

A-‐ + H2O = HA + OH-‐

a) 1.4 × 10-‐9 b) 7.2 × 108 M c) 7.2 × 10-‐8 M d) 6.8 × 10-‐1 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 Knumbers has 200 samples of the Jaws-‐Max line tested and finds that the mean weight for line

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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+/-‐0.05 3] b 4] a 5] Q=(5.82-‐5.22)/(5.82-‐4.77) = 0.57 df =5 Qtable = 0.56 < 0.57 the number can be rejected. 6] see book and lecture notes 7] 0.559 = kCx(10.0/500.0) + k 5.00e-‐3*(10.00/500.0)

-‐0.366 = -‐ kCx(10.0/500.0) 0.193 = k 5.00e-‐3*(10.00/500.0) k = 1930 use 0.559 = kCx(10.0/500.0) 0.559 = 1930Cx(10.0/500.0) Cx = 9.48e-‐3 M

8] 0.245 +/-‐ (2.571*0.011/61/2) = 0.245 +/-‐ 0.0115 ppm* *b was the best answer. 9] HA = H+ + A-‐

0.100-‐x +x +x x2/0.100-‐x = 2.7e-‐6 x = 5.2e-‐4 M pH = 3.28

10] MA2 = M2+ + 2A-‐

-‐-‐ +x +2x (2x)2x = 8.9e-‐17 x = 2.8e-‐6 M 11] AB = A+ + B-‐ -‐-‐ x 0.10+x x(0.10+x) ≈ 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 * 2.00/0.1000 = 2.00e-‐2 M 14] KaKb = Kw Kb = 1.00e-‐14/7.2e-‐6 = 1.4e-‐9 15] z = [x – x-‐bar]/s = [80 – 120]/60 = 0.667 z ≈ 0.7 area = 0.258 Area from 0 to 80 is 0.500 – 0.258 = 0.242 ≈24%

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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 cm2 a) 11.77 ± 1.6 b) 11.770 ±

1.60 c) 11.77 ± 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 HNO3(aq)? a) -‐0.324 b) 0.324 c) -‐0.32 d) 7.76e-‐4 e) 0.32 6) 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

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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) ± 0.06 8) 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? 9.11 8.89 9.01 9.77 9.05 a) 9.77 b) 9.11 c) 8.89 d) none of the

above 10) What is the volume of 0.233 M HCl(aq) required to make a solution of 500.0-‐mL of 0.0840 M HCl(aq)? a) 90.0-‐mL b) 622-‐mL c) 33.4-‐mL d) 180-‐mL e) 233-‐mL 11) The concentrated HCl(aq) is 37.1% (m/m) HCl(aq) (d = 1.19 g/mL, MWHCl = 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 1.56. 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) 0.924 M b) 1.24 M c) 2.81 M d) 1.82 M e) 0.155 M 13) A sample solution of an analyte has an absorbance of 0.229. A solution of standard has an absorbance of 0.327 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

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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 (xi -‐ xL)

b) minimizing Σ (xi -‐ xL)

2 c) minimizing Σ (yi -‐ yL)

2 d) maximizing Σ (yi -‐ yL)

2 e) minimizing Σ (yi -‐ yL)

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 10.00-‐mL of aqueous sample was diluted to 500.0-‐mL with water. Its measured absorbance is 0.378. b)In the second experiment 10.00-‐mL was mixed with 1.00-‐µL of 3.22e-‐5M and diluted to 500.0-‐mL with water. The measured absorbance of this solution is 0.402. What is the concentration of the analyte in the sample?

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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%) = 11.77 ± (1.1%2 + 1,2%2)1/2 = 11.8 cm2 ± 1.6% = 11.8 cm2 ± 0.2 cm2

3) e 4) b 5) a 6) c 7) c ± 2.776 (0.04)/(5)1/2 8) b z = (33.8-‐55.1)/11.8 ≅ 1.80 use table 4-‐1 area = 0.4641

above = 0.5000 – 0.04641 ≅ 3.6% 9) a Q = 9.77-‐9.11/9.77-‐8.89 = 0.75 Qtable = 0.64 for n = 5 so 9.77 can be discarded 10) d 11) c 12) d 1.56 mol * 88.2 g HX = 137.5 g 1.56 mol HX/1137.5g soln * 1.33 g/mL * 1000 mL/L = 1.824 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 0.378 = k (10.00/500.0) c

Part b 0.402 = k (10.00/500.0) c + k (1.00e-‐6/0.5000) 3.22e-‐5 0.402 = 0.378 + k (1.00e-‐6/0.5000) 3.22e-‐5 k = 3.73e8 0.378 = 3.73e8 (10.00/500.0) c

c = 5.1e-‐8 M

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Problem Set 4 1] What is the pH of a solution of 0.100 M Na2HA solution given the following: H3A = H2A

-‐ + H+ Ka = 2.8e-‐2 H2A

-‐ = HA2-‐ + H+ Ka = 7.7e-‐5 HA2-‐ = A3-‐ + H+ Ka = 9.3e-‐11 2] What is the MBE for 1.00e-‐3 M [Ag(NH3)2]Cl for the following reaction sequence? [Ag(NH3)2]Cl à Ag(NH3)2

+ + Cl-‐ Ag(NH3)2

+ = Ag(NH3)+ + NH3

Ag(NH3)+ = Ag+ + NH3

3] What is the CBE for the follow reaction sequence? H2S = H

+ + HS-‐ HS-‐ = H+ + S2-‐ H2O = H

+ + OH-‐ 4] What is the pH of a solution containing 0.25 M sodium acetate, and 0.25 M CH3COOH? Ka = 1.75e-‐5 5] Which of the following monoprotic acids would be best for creating a buffer system at pH 7.00? acid A Ka = 5.6e-‐4 acid B Ka = 7.7e-‐6 acid C Ka = 1.9e-‐8 acid D Ka = 7.3e-‐11 6] What is pAg when of 50.00 mL of 0.100 M AgNO3 is mixed with 50.00 mL of 0.100 M NaCl? AgCl Ksp = 1.8e-‐10 7] The weak acid, HA has Ka = 1.0e-‐5. What is the fraction, αA-‐ at pH 7.00? 8] What is the MBE for the following sequence of reactions? MgF2 = Mg2+ + 2F-‐ Ksp F-‐ + H2O = HF + OH

-‐ Kb

Mg2+ + H2O = Mg(OH)+ + H+ β 9] What is K for this reaction? H2SO3 = SO3

2-‐ + 2H+ K = ? H2SO3 = HSO3

-‐ + H+ K = 1.23e-‐2 HSO3

-‐ = SO32-‐ + H+ K = 6.6e-‐8

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10] What is the difference between the end point and the equivalence point? 11] What is the solubility of SrF2 (Ksp = 2.8e-‐9) at pH 4.00? HF Ka = 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 143.35, Ksp=1.8e-‐10) and AgBr (MW 187.9, Ksp=5.0e-‐13) weighs 2.000 g. This mixture is reduced to silver metal (AW 107.9), which weighs 1.300 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 0.100 F AgNO3? (10 points) Answers to Problem Set 4 1] What is the pH of a solution of 0.100 M Na2HA solution given the following: H3A = H2A

-‐ + H+ Ka = 2.8e-‐2 H2A

-‐ = HA2-‐ + H+ Ka = 7.7e-‐5 HA2-‐ = A3-‐ + H+ Ka = 9.3e-‐11

pH = ½(-‐log7.7e-‐5 + -‐log9.3e-‐11) = 7.07 2] What is the MBE for 1.00e-‐3 M [Ag(NH3)2]Cl for the following reaction sequence? [Ag(NH3)2]Cl à Ag(NH3)2

+ + Cl-‐ Ag(NH3)2

+ = Ag(NH3)+ + NH3

Ag(NH3)+ = Ag+ + NH3

1.00e-‐3 M = [Ag(NH3)2

+] + [Ag(NH3)+] + [Ag+]

3] What is the CBE for the follow reaction sequence? H2S = H

+ + HS-‐ HS-‐ = H+ + S2-‐ H2O = H

+ + OH-‐ [H+] = [OH-‐] + [HS-‐] + 2[S2-‐] 4] What is the pH of a solution containing 0.25 M sodium acetate, and 0.25 M CH3COOH? Ka = 1.75e-‐5 pH = pKa + log [base]/[acid] = 4.757

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5] Which of the following monoprotic acids would be best for creating a buffer system at pH 7.00? acid A Ka = 5.6e-‐4 acid B Ka = 7.7e-‐6 acid C Ka = 1.9e-‐8 acid D Ka = 7.3e-‐11 acid C 6] What is pAg when of 50.00 mL of 0.100 M AgNO3 is mixed with 50.00 mL of 0.100 M NaCl? AgCl Ksp = 1.8e-‐10 4.87 7] The weak acid, HA has Ka = 1.0e-‐5. What is the fraction, αA-‐ at pH 7.00? 0.99 8] What is the MBE for the following sequence of reactions? MgF2 = Mg2+ + 2F-‐ Ksp F-‐ + H2O = HF + OH

-‐ Kb

Mg2+ + H2O = Mg(OH)+ + H+ β 2[Mg2+] + 2[Mg(OH)+] = [F-‐] + [HF] 9] What is K for this reaction? H2SO3 = SO3

2-‐ + 2H+ K = ? H2SO3 = HSO3

-‐ + H+ K = 1.23e-‐2 HSO3

-‐ = SO32-‐ + H+ K = 6.6e-‐8

8.1e-‐10

10] 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 SrF2 (Ksp = 2.8e-‐9) at pH 4.00? HF Ka = 6.76e-‐4. For partial credit clearly express what you are doing, e.g. substitutions label equations as #1, #2…. (20 points)

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SrF2(s) = Sr2+ + 2F-‐ Ksp 2 pts

F-‐ + H2O = HF + OH-‐ Kb = Kw/Ka 2 pts

MBE: 2[Sr2+] = [F-‐] + [HF] 4 pts pH = 4.00 è [OH-‐] = 1.0e-‐10 M 2 pts Kb = [HF][OH

-‐]/[F-‐] è [HF] = 0.1479 [F-‐] next few steps 8 pts 2[Sr2+] = [F-‐] + [HF] = 1.1479 [F-‐] #1 Ksp = [Sr

2+][F-‐]2 è [F-‐] = (Ksp/[Sr2+])1/2 #2 Sub 2 into 1 2[Sr2+] = 1.1479(Ksp/[Sr2+])1/2 [Sr2+]3 = 9.52e-‐10 [Sr2+] = s = 9.8e-‐4 M 2 pts 12] A mixture of AgCl (MW 143.35, Ksp=1.8e-‐10) and AgBr (MW 187.9, Ksp=5.0e-‐13) weighs 2.000 g. This mixture is reduced to silver metal (AW 107.9), which weighs 1.300 g. Calculate the mass of AgCl in the original sample. (20 points)

x g AgCl + y g AgBr = 2.000 g 4 pts x g AgCl*(mol AgCl/143.35 g)*(mol Ag/mol AgCl)*(107.9 g/mol) = 0.7527x g Ag 4 pts y g AgCl*(mol AgBr/187.9 g)*(mol Ag/mol AgBr)*(107.9 g/mol) = 0.5742y g Ag 4 pts 0.7527x g Ag + 0.5742y g Ag = 1.300 g 4 pts y = 2.000 – x sub into above 0.7527x + 0.5742(2.000 – x) = 1.300 g mass Ag = 0.849 g 4pts

13] What is the concentration Cl-‐ required to remove 99% of Ag+ in a solution of 0.100 F AgNO3? (10 points) [Ag+] = (1-‐0.99) 0.100 F = 1.00e-‐3 M Ksp = [Ag

+][Cl-‐] = 1.8e-‐10 = 1.00e-‐3 M*[Cl-‐] è [Cl-‐] = 1.80e-‐7 M

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Problem Set 5 1] What is the pH of a solution of a 1.0 M phthalic acid solution? (5 points)

Ka1 = 1.12e-‐3 Ka2 = 3.90e-‐6

a) 5.92e-‐4 b) 8.36 c) 7.22 d) 4.18 e) 1.48 2] A 0.5000g sample contained only NaBr (MW 102.89) and NaCl (MW 58.44). It was dissolved into water and precipitated with excess AgNO3. The precipitate (AgBr(s) (MW 187.80) & AgCl(s) (MW 143.35)) was dried. The mass of this precipitate weighed 1.500 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 = 1.500 & (x/102.89) + (y/58.44) = 0.5000 b) x + y = 1.500 & {x(187.80/102.89)} + {y(143.35/58.44} = 1.500 c) x + y = 0.500 & {x(187.80/102.89)} + {y(143.35/58.44} = 1.500 d) x + y = 1.500 & {x(187.80/102.89)} + {y(143.35/58.44} = 0.500 e) x + y = 0.500 & {x(102.89/187.80)} + {y(58.44/143.35} = 1.500 3] What is the mole fraction of HA-‐ at pH 3.00 given H2A = H

+ + HA-‐ Ka1 = 1.0e-‐3 HA-‐ = H+ + A2-‐ Ka2 = 1.0e-‐9 a) 0.25 b) 0.50 c) 0.88 d) 0.15 e) 0.010 4] What is the pH of a solution of 0.10 M NaHCO3? H2CO3 = HCO3

-‐ + H+ Ka1 = 4.45e-‐7 HCO3

-‐ = CO32-‐ + H+ Ka2 = 4.69e-‐11

a) 8.34 b) 6.22 c) 7.84 d) 9.71 e) 4.66 5] What is or are the simplifying assumption(s) that allow for the use of the Henderson-‐Hasselbalch equation? pH = pKa + log([A

-‐]/[HA])

COOH

COOH

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a) pKa > pH b) pKa = ± 1 pH & [A

-‐] = [HA] c) formal concentrations of A-‐ & HA are the same as equilibrium concentrations

c) formal concentrations of A-‐ & HA are not 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 0.100 M acetic acid. CH3COOH = H

+ + CH3COO-‐ Ka = 1.75e-‐5

8] What is the pH of a solution consisting of 0.100 M CH3COONa and 0.100 M CH3COOH? See also problem 7. a) 1.75 b) 4.76 c) 4.757 d) 5.717 e) 1.756 9] 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 NO2 from the atmosphere producing HNO3

b) solvation of Cl2 from the atmosphere producing HCl

c) solvation of SO2 from the atmosphere producing H2SO3

d) solvation of N2 from the atmosphere producing HNO3

e) solvation of CO2 from the atmosphere producing H2CO3

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

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to the sample in small increments 12] Write down the hydrolysis reaction for HCO3

-‐ demonstrating that it is a weak base. ___________________________________________________________ 13] The Kb for dichloroacetate, Cl2CHCOO

-‐ is

Cl2CHCOOH Ka = 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 50.00 mL solution of 0.100 M NaI(aq) is titrated with 0.100 M AgNO3(aq). If 75.00 mL of the AgNO3 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 15-‐18 are based on the following: A 50.0-‐mL sample of 0.100 M KSCN is titrated with 0.0500 M CuNO3. The Ksp of CuSCN is 4.8e-‐15. 15] Write down the reaction that takes place during this titration. _______________________________________________________________ 16] What is pCu when 25.0-‐mL of the 0.100 M CuNO3 is added to the 50.0-‐mL sample of 0.100 M KSCN solution? The Ksp of CuSCN is 4.8e-‐15. a) 1.33 b) 12.84 c) 10.68 d) 5.87 e) 7.00 17] What is pCu when 50.0-‐mL of the 0.100 M CuNO3 is added to the 50.0-‐mL sample of 0.100 M KSCN solution?

18

a) 7.93 b) 6.44 c) 9.13 d) 7.16 e) 8.52 18] What is pCu when 75.0-‐mL of the 0.100 M CuNO3 is added to the 50.0-‐mL sample of 0.100 M KSCN solution? a) 3.22 b) 4.44 c) 7.18 d) 12.70 e) 1.70 19] Which diagram best describes the curve for the titration of 50.0-‐mL of 0.100 M KSCN with 0.0500 M CuNO3? Answers to Problem Set 5 1] e) 1.48 only Ka1 is important. x2/(1.0-‐x) = 1.12e-‐3; x = 0.0335 2] c) x + y = 0.500 & {x(187.80/102.89)} + {y(143.35/58.44} = 1.500 3] 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 = 0.50 4] a) 8.34 = ½ (pKa1 + pKa2) 5] c) formal concentrations of A-‐ & HA are the same as equilibrium concentrations 6] [H+] = 2[A2-‐] + [HA-‐] 7] 0.10 M = [CH3COOH] + [CH3COO

-‐], also [H+] = [CH3COO-‐]

8] c) 4.757 pH = pKa watch S.F. 9] b) Potassium Hydrogen Phthalate 10] solvation of CO2 from the atmosphere producing H2CO3 11] a) it is the reagent solution whose concentration is known and is added to the sample in small increments 12] HCO3

-‐ + H2O à H2CO3 + OH-‐

13] Kb = Kw/Ka = 2.0e-‐13 14] There is a solid precipitate and an excess of Ag+. Pink is OK (Fajan’s method) 15] Cu+ + SCN-‐ = CuSCN(aq) 16] Mol SCN-‐ = 0.0500L*0.100 mol/L = 5.00e-‐3

Mol Cu+ = 0.02500-‐L*0.100 mol/L = 2.50e-‐3

Vol CuNO3

Vol KSCN

[Cu+]

Vol KSCN [Cu+]

Vol KSCN

pCu

Vol CuNO3

19

Excess mol SCN-‐ = 5.00e-‐3 -‐ 2.50e-‐3 = 2.50e-‐3 [SCN-‐] = 2.50e-‐3 mol/0.0750 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 = 12.84 17] Mol SCN-‐ = 0.0500L*0.100 mol/L = 5.00e-‐3 Mol Cu+ = 0.0500-‐L*0.100 mol/L = 5.00e-‐3 CuSCN(s) = Cu+ + SCN-‐ -‐-‐ x x x2 = 4.8e-‐15 x = 6.9e-‐8 pCu = 7.16 18] Mol SCN-‐ = 0.0500L*0.100 mol/L = 5.00e-‐3 Mol Cu+ = 0.0750-‐L*0.100 mol/L = 7.50e-‐3 Excess mol Cu+ = 7.50e-‐3 -‐ 5.00e-‐3 = 2.5e-‐3 [Cu+] = 2.50e-‐3 mol/0.125 L = 0.0200 M pCu = 1.70 19] Which diagram best describes the curve for the titration of 50.0-‐mL of 0.100 M KSCN with 0.0500 M CuNO3?

pCu

Vol CuNO3

20

Problem Set 6 1] The pH of solution of 0.050 M of a weak acid, HA is 5.69. What is Ka for this acid?

2] What is the aqueous solubility of AgCl at pH 4.00 (Ksp = 1.8e-‐11)? 3] The two Ka’s for salicylic acid (H2A) are 1.07e-‐3 and 1.82e-‐14. What is Kb for sodium salicylate (NaHA)? 4] What is the charge balance for a solution of 0.10 M NaHCO3? Ka1 H2CO3 = 6.352 Ka2 HCO3

-‐ = 10.329 HCO3

-‐ ⇄ H+ + CO32-‐

HCO3-‐ + H2O ⇄ H2CO3 + OH

-‐ 5] Which of the following is a valid mass balance for a solution for 0.10 M NaHCO3? 6] Which of the following would best explain the solubility of Ag2SO4? 7] A 0.9961 g silver ore sample was treated with HNO3 and then with excess NaCl(aq). A precipitate was dried and weighed 0.0711 g. What is percent silver in the ore? AW: Ag 107.9, H 1.008, O 16.00, N 14.01, Cl 35.45 8] Which of the following is not a primary standard?

a) Potassium Hydrogen Phthalate (KHP) b) Benzoic Acid c) Potassium Hydrogen Iodate d) NaOH e) NaHCO3

Problems 9-‐11: A solution of 0.100 M AgNO3 is used to titrate a 100.00 mL solution of 0.100 M KCl. The Ksp of AgCl is 1.8e-‐11 9] What is pAg if 50.00 mL of the titrant is added to the KCl solution? 10] What is pAg if 100.00 mL of the titrant is added?

21

12] What is pH of solution containing 0.100 M HOCl (Ka = 3.0e-‐8) and 0.100 M NaOCl. 13] The useful pH range of most buffering systems is

a) within 1 pH unit of pKa b) within 0.5 pH unit of pKa c) within 10 pH units of pKa d) within 5 pH units of pKa e) within 0.1 pH unit of pKa

14] What is the pH of solution that is 0.10 M NaH2PO3? H3PO3 ⇄ H2PO3

-‐ + H+ Ka = 3e-‐2 H2PO3

-‐ ⇄ HPO32-‐ + H+ Ka = 1.62e-‐7

15] The fraction (relative concentration) of H2PO3

-‐ from a 0.10 F H3PO3 at pH 5.00 can be calculated from which formula?

a) 211

2

2

][][][

aaa KKHKHH

++ ++

+

b) 211

21

][][][

aaa

a

KKHKHHK

++ ++

+

c) 211

221

][][ aaa

aa

KKHKHKK

++ ++

d) 21

21

2

2

][][][

aaa KKHKHH

++ ++

+

e) 211

2

21

][][][

aaa

a

KKHKHHK

++ ++

+

16] Which of the follow is an amphoteric species?

a) H2CO3 b) HF

22

c) F-‐ d) HCO3

-‐ e) CO3

2-‐ 17] What is the mass balance equation for the following sequence of reactions? CaC2O4(s) ⇄ Ca2+ + C2O4

2-‐ Ksp = 1.3e-‐8 C2O4

2-‐ + H2O ⇄ HC2O4-‐ + OH-‐ Kb1

HC2O4-‐ + H2O ⇄ H2C2O4 + OH

-‐ Kb2 18] What is the charge balance equation for the reaction sequence of problem 17? 19] What is the molar solubility of BaF2 (Ksp = 1.7e-‐6) at pH 7.20? Ka (HF) = 6.8e-‐4

Answers to Problem Set 6 1] [H+] = 2.0e-‐6 M HA = H+ + A-‐

0.050-‐x x x Ka = x

2 / (0.050-‐x) ≅ 2.0e-‐62 / 0.050 = 8.3e-‐11 Answer d) 2] AgCl ⇄ Ag+ + Cl-‐ -‐-‐ x x Ksp = x

2 = 1.8e-‐11 x = 4.2e-‐6 Answer a) 3] HA-‐ + H2O ⇄ H2A + OH

-‐ KaKb = Kw Kb = Kw/Ka = 1.00e-‐14 / 1.07e-‐3 = 9.35e-‐12

Answer c) 4] Answer b: [Na+] + [H+] = [HCO3

-‐] + 2[CO32-‐] + [OH-‐]

5] Answer e: 0.10 M = [H2CO3] + [HCO3

-‐] + [CO32-‐]

6] Answer d: [Ag+]/2 7] 0.0711 g AgCl (mol AgCl / 143.32 g) (mol Ag / mol AgCl) (107.868 g / mol Ag) (100 / 0.9961g) = 5.37 % 8] Answer d): NaOH 9] Titration Rxn: Ag+ + Cl-‐ ⇄ AgCl(s)

23

Moles of excess Cl-‐ = 0.10000 L * 0.100 M KCl – 0.05000 L * 0.100 M AgNO3 = 5.00e-‐3 mol Cl-‐ [Cl-‐] = 5.00e-‐3 mol Cl-‐ / 0.1500 L = 3.33e-‐2 AgCl(s) ⇄ Ag+ + Cl-‐ -‐-‐ x x+3.33e-‐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 x2 = 1.8e-‐11 x = 4.24e-‐6 pAg = 5.37 answer b)

11] We are past the equivalence point. (150.00 – 100.00) mL * 0.100 M Ag+ (250.00 mL) = 0.0200 M Ag+ or pAg = 1.699 answer d) 12] Ka = [H

+][OCl-‐] / [HOCl] [H+] = Ka [HOCl] / [OCl-‐] [H+] = 3.0e-‐8

pH = 7.52 Answer a) 13] Answer a) 14] pH = ½ (pKa1 + pKa2) = ½ (1.5 + 6.790) = 4.1 answer e)

15] answer: b b) 211

21

][][][

aaa

a

KKHKHHK

++ ++

+

16] HCO3

-‐ Answer d 17] Answer b

18] 2[Ca2+] = 2[C2O4

2-‐] + [HC2O4-‐] + [OH-‐] Answer e

24

19] MBE: 2[Ba2+] = [F-‐] + [HF] & [H+] = 6.31e-‐8 M; [OH-‐] = 1.58e-‐7 Ksp = [Ba

2+][F-‐]2 = 1.7e-‐6 Ka (HF) = [H+][F-‐]/[HF] = 6.8e-‐4

3 variables: [Ba2+], [F-‐], [HF] F-‐ + H2O ⇄ HF + OH-‐ Kb = Kw/Ka = 1.00e-‐14/6.8e-‐4 = 1.47e-‐11 Using Kb solve for [HF] [HF] = Kb[F

-‐]/[OH-‐] will sub into MBE

2[Ba2+] = [F-‐] + [HF]

2[Ba2+] = [F-‐] + Kb[F-‐]/[OH-‐]

Sub all knowns into above

2 [Ba2+] = [F-‐] + 1.47e-‐11*[F-‐]/1.58e-‐7

2 [Ba2+] ≅ [F-‐] sub into Ksp Ksp = [Ba2+][F-‐]2 = [Ba2+](2[Ba2+])2 [Ba2+] = (Ksp/4)1/3 = 7.5e-‐3M

25

Problem Set 7 (starts with question 3) 2] What is the pH after 100.00 mL of 0.100 M HClO4 is added to 50.0 mL of 0.100 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 Kf’ for CaY

2-‐ is related to Kf through which of the relationships? a) Kf’ = Kf at pH =0 b) Kf’ = αy4-‐Kf c) Kf = αy4-‐Kf’ d) Kf’ = 1 / Kf e) Kf’ = Kf

2 5] In reference to EDTA titrations the symbol, αy4-‐, indicates which of the following?

a) The fraction of metal chelated by EDTA b) The concentration of EDTA in the Y4-‐ form. c) The fraction of EDTA in the Y4-‐ form. d) The analytical concentration of metal. e) The fraction of EDTA not in the Y4-‐ form.

6] A spontaneous electrochemical cell would have which of the following? a) Ecell = 0 b) Ecell > 0 c) Ecell < 0 d) Ecell ≤ 0 e) Ecell ≥ 0

7] If A + e-‐ = B has E0 = 0.775 V then the E0 for 2A + 2e-‐ = 2B is __________________.

26

8] The standard cell potential for the following is Fe(s)/Fe2+(aq)//Sn2+(aq)/Sn(s) Fe2+ + 2e-‐ = Fe(s) E0 = -‐0.44 V Sn2+ + 2e-‐ = Sn(s) E0 = -‐0.141 V 9] The E0 for the following is FeCO3(s) + 2e

-‐ = Fe(s) + CO32-‐ E0 = ?

Fe2+ + 2e-‐ = Fe(s) E0 = -‐0.44 V Ksp {FeCO3(s)} = 2.1e-‐11 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 Y4-‐ fraction d) basic pH’s to minimize Y4-‐ fraction e) acidic pH’s to maximize Y4-‐ fraction

11] Which is true of the equivalence point for the redox titration of Fe2+ with Ce4+?

a) only [Fe2+] = [Fe3+] b) [Fe2+] = [Ce3+] and [Ce4+] = [Fe3+] c) [Fe2+] = [Ce4+] and [Fe3+] = [Ce3+] d) [Fe2+] = [Fe3+] and [Ce4+] = [Ce3+] e) [Fe2+] = 0

12] A pH electrode was found to have a potential of 0.241 V in a pH 4.01 buffer solution. A sample solution was found to have a potential of 0.252 V. What is the pH of that sample? E = const – 0.0592 pH 13] KMnO4 can be standardized with which of the following?

a) H2O2 b) CH4 c) H2O d) NaC2O4 e) C2H4

14] Calculate pCa if 20.0 mL of 0.050 M of EDTA is added to 15.0 mL of 0.050 M Ca2+ at pH 9.0.

27

15] Calculate the cell potential when 25.0 mL of 0.010 M Ce4+ is added to 15.0 mL of 0.010 M Fe2+. Ce4+ + e-‐ = Ce3+ E0 = 1.70 V Fe3+ + e-‐ = Fe2+ E0 = 0.767 V 16] How many grams of K2C2O4 (MW 166.22) must be added 25.0 mL of 0.500 M HCl to give a pH of 4.500 when this solution is diluted to 500.0 mL.

Ka1 = 5.60e-‐2 Ka2 = 5.42e-‐5

Answers to Problem Set 7 2] [H+] = 5.00 mmol / 150.0 mL = 3.33e-‐2 pH = 1.48 3] its transition pH which should overlap the steepest portion of the titration curve. 4] Kf’ = αy4-‐Kf 5] The fraction of EDTA in the Y4-‐ form. 6] Ecell > 0 7] 0.775 V 8] E = -‐0.141 –(-‐0.44) = 0.30 V 9] E = -‐0.44 – (0.0592/2) log 1/Ksp = -‐0.756 V 10] basic pH’s to maximize Y4-‐ fraction 11] [Fe2+] = [Ce4+] and [Fe3+] = [Ce3+] 12] 3.82 13] NaC2O4

28

14] mol EDTA = 20.0 mL * 0.050 M = 1.0 mmol mol Ca2+ = 15.0 mL * 0.050 M = 0.75 mmol excess EDTA region where, [CaY2-‐] = 0.75 mmol / 35.0 mL = 2.1e-‐2 M

[EDTA] = 0.25 mmol / 35.0 mL = 7.1e-‐3 M

Kf = [CaY2-‐] / [Ca2+]*[Y4-‐]

[Y4-‐] = αY4-‐ [EDTA] Kf *αY4-‐ = Kf’ = [CaY

2-‐] / [Ca2+]*[EDTA] Kf = 4.9e10 Kf’ = 5.4e-‐2*4.9e10 = 2.6e9

2.6e9 = 2.1e-‐2 M / [Ca2+]*7.1e-‐3 M [Ca2+] = 1.1e-‐9 M pCa = 8.94 15] Mol Ce4+ = 25.0 mL * 0.010 M = 0.25 mmol

Mol Fe2+ = 15.0 mL * 0.010 M = 0.15 mmol Excess Ce4+ region Ce4+ + Fe2+ = Fe3+ + Ce3+ 0.25 0.15 0 0 -‐0.15 -‐0.15 +0.15 +0.15 0.10 0 0.15 0.15 E = 1.70 – (0.0592) log (0.15/0.10) = 1.69 V 16] pKa1 = 1.25 pKa2 = 4.27

Therefore only Ka2 is important. HC2O4

-‐ = H+ + C2O42-‐

[C2O4

2-‐]i = x / 0.500 L [H+]i = 25.0*0.500 mmol / 500.0 mL = 0.025 M [H+]f = 10

-‐4.500 = 3.16e-‐5 M x = mol C2O4

2-‐

29

H+ + C2O42-‐ = HC2O4

-‐ 0.025 x / 0.500L 0 -‐y -‐y +y 3.16e-‐5 0.025 – y = 3.16e-‐5 y = 0.025 M Ka = [H

+][C2O42-‐] / [HC2O4

-‐] 5.42e-‐5 = [3.16e-‐5][C2O4

2-‐] / [0.025] [C2O4

2-‐] = 4.29e-‐2 M 4.29e-‐2 M = x / 0.500L – 3.16e-‐5 x = 2.15e-‐2 mol g K2C2O4 = 2.15e-‐2 mol * 166.22 g/mol = 3.57 g

30

Problem Set 8 Problems 1-2 involve the following titration: 1] 75.0-‐mL of 0.100 M CH3COOH(aq) (pKa = 4.757) is titrated with 0.0500 M NaOH(aq). What is the volume of NaOH solution required to reach the equivalence point? (5 points) a) 150-‐mL b) 75.0-‐mL c) 50.0-‐mL d) 100-‐mL e) 125-‐mL

2] What is the pH of the equivalence point in problem 1? (5 points) a) 7.000 b) 6.569 c) 10.368 d) 7.889 e) 8.640

3] What is the final pH if solutions of 200.0-‐mL of 0.0500 M NaOH(aq) and 75.0-‐mL of 0.100 M CH3COOH(aq) are added together? (5 points) a) 10.554 b) 2.345 c) 7.000 d) 11.959 e) 14.000

4] 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] 50.0-‐mL of 0.0500 M oxalic acid (H2C2O4, pka1 = 1.252, pKa2 = 4.266) is titrated with 50.0-‐mL of 0.0500 M NaOH. What is the pH of that titrated solution? (5 points) a) 7.00 b) 1.252 c) 2.759 d) 4.266 e) 8.667

6] What is Kf’ for SrEDTA

2-‐ at pH 11? a) 0.85 b) 108.73 c) 0.85×8.73 d) 0.85×108.73 e) 1.55e-‐14

31

7] The formal concentration of EDTA is 1.00 mM. What is the concentration of the Y4-‐ form at pH 4? a) 1.00 mM b) 3.8×10-‐9 × 1.00 mM c) 3.8×10-‐9 mM d) 0.85 mM e) 0.85×108.73

Problems 8-‐11 involve the following titration. 8] A solution of 50.0-‐mL of 1.00×10-‐3 M NiCl2(aq)is titrated with 1.00×10

-‐3 M EDTA in a solution of 0.100 M NH3 at pH 11.00. What is pNi if 25.0-‐mL of the titrant solution is added? Note that αNi2+ = 1.34×10

-‐4 at 0.100 M NH3. (5 points) a) 7.350 b) 8.442 c) 5.311 d) 10.673 e) 11.995

9] What is Kf’’ for the NiEDTA

2-‐ complex in 0.100 NH3 at pH 11? (5 points) a) 1.34×10-‐4(0.85)18.62 b) 1.34×10-‐4 × 1018.62

c) (0.85)1018.62 d) 1.34×10-‐4(0.85)1018.62

10] What is [NiEDTA2-‐] if 75.0-‐mL of titrant is added to the NiCl2 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 75.0-‐mL of 1.00×10-‐3 M EDTA titrant is added to the 50.0-‐mL of 1.00×10-‐3 M NiCl2 solution in 0.1M NH3? Assume equilibrium conditions. (5 points) a) [Ni2+] = [EDTA]

b) [NiEDTA2-‐] > [EDTA]

c) [NiEDTA2-‐] = [EDTA] d) [Ni2+] > [EDTA]

12] An electrochemical cell will discharge spontaneously if (5 points) a) Ecell < 0 b) Ecell > 0 c) Ecell = 0 d) does not depend on Ecell 13] The reductions take place at which electrode? (5 points) a) anode b) toade c) cathode d) alkaline e) amphiprotic

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 E0cell for the reaction below? (5 points)

F2 + 2Fe2+ = 2F-‐ + 2Fe3+ F2 + 2e

-‐ = 2F-‐ E0red = 2.890 V Fe3+ + e-‐ = Fe2+ E0red = 0.771 V

a) -‐2.119 V b) -‐1.348 V c) 1.348 V d) 0.655 V e) 2.119 V

16] What is E0cell for the reaction below? (5 points)

Hg2SO4(s) + 2e-‐ = 2Hg(l) + SO42-‐

Hg2

2+ + 2e-‐ = 2Hg(l) E0red = 0.796 V Hg2SO4(s) = Hg2

2+ + SO42-‐ Ksp = 7.4e-‐7

a) sp

cell KE 1log

20592.0796.00 −=

b) spcell KE log20592.0796.00 −=

c) sp

cell KE 1log0592.0796.00 −=

d) sp

cell KE 1log796.00 −=

17] What are the final concentrations of each ion when 25.0-‐mL of 0.0500 M Ce4+ is mixed with 15.0-‐mL of 0.0500 M Cu+? (5 points) Ce4+ + e = Ce3+ E0red = 1.44 V Cu2+ + e = Cu+ E0red = 0.161 V 18] What is the potential of the final solution when 25.0-‐mL of 0.0500 M Ce4+ is mixed with 15.0-‐mL of 0.0500 M Cu+? (5 points)

33

Answers to Problem Set 8 1 & 2] eq. pt. mol CH3COOH = mol OH-‐

mol CH3COOH = 75.0-‐mL*0.100 M = 7.50 mmol vol NaOH = 7.50 mmol/0.0500 M = 150-‐mL total volume = 150-‐mL+75.0-‐mL = 225-‐mL [CH3COO

-‐] = 7.50 mmol/225-‐mL = 3.33e-‐2 M CH3COO

-‐ + H2O = CH3COOH + OH-‐ 3.33e-‐2 -‐-‐ 0 0 -‐ x -‐-‐ + x + x Kb = Kw/Ka = 5.71e-‐10 = x

2/(3.33e-‐2-‐x) ≈ x2/(3.33e-‐2) x = 4.36e-‐6 pOH = 5.360 pH =8.640

3] mol CH3COOH = 75.0-‐mL*0.100 M = 7.50 mmol

mol OH-‐ = 200.0-‐mL*0.0500 M = 10.0 mmmol excess OH-‐ = 10.0 – 7.50 mmol = 2.50 mmol [OH-‐] = 2.50 mmol/275-‐mL = 9.09e-‐3 pOH = 2.041 pH = 11.959

4] a] The equivalence point is where mol acid = mol base and the end point is where indicator changes color. 5] mol of OH-‐ = 50.0-‐mL*0.0500 M = 2.50 mmol mol of H2C2O4 = 50.0-‐mL*0.0500 M = 2.50 mmol Therefore we have only HC2O4

-‐ which is an amphiprotic species. pH is pH = ½(pka1 + pKa2) = ½(4.266 + 1.252) = 2.759 6] Kf’ = αy4-‐Kf = 0.85*5.4e8 = 4.6e8 7] [Y4-‐] = 3.8e-‐9*1.00e-‐3 M = 3.8e-‐12 M b and c are correct.

34

8] Initial mol Ni2+ = 50.0-‐mL*1.00e-‐3 M = 0.0500 mmol Added mol EDTA = 25.0-‐mL*1.00e-‐3 M = 0.0250 mmol Excess Ni2+ = 0.0500 – 0.0250 mmol = 0.0250 mmol CNi2+ = 0.0250 mmol / 75.0-‐mL = 3.33e-‐4 M Free [Ni2+] = αNi2+ CNi2+ = 1.34e-‐4*3.33e-‐4 = 4.47e-‐8 M pNi = 7.350

9] Kf’’ = αNi2+αY4-‐*Kf = 1.34e-‐4*0.85*10

18.62 = 4.7e14 10&11]Initial mol Ni2+ = 50.0-‐mL*1.00e-‐3 M = 0.0500 mmol

Added mol EDTA = 75.0-‐mL*1.00e-‐3 M = 0.0750 mmol [NiEDTA] = 0.0500 mmol / 125.0-‐mL = 4.00e-‐4 M

Excess EDTA = 0.0250 mmol / 125.0-‐mL = 2.00e-‐4 M

Kf’’ = [NiEDTA]/CNi*[EDTA] = 4.00e-‐4/CNi*2.00e-‐4 = 4.7e14

CNi = 4.3e-‐15 [Ni2+] = 1.34e-‐4*4.3e-‐14 = 5.8e-‐18 M pNi = 17.24 Therefore [NiEDTA2-‐] > [EDTA] 12] b) Ecell > 0 13] cathode 14] d) to provide a stable potential in which the cathode reaction can be compared. 15] Ecell = Ecath – Eanod = 2.890 – 0.771 = 2.119 V 16] E = 0.796 – 0.0592/2 log 1/[Hg2

2+] Ksp = 7.4e-‐7 = [Hg2

2+][SO42-‐]

[Hg22+] = 7.4e-‐7/[SO4

2-‐] E = 0.796 – 0.0592/2 log [SO4

2-‐]/7.4e-‐7 = 0.615 V

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17&18] Reaction: Ce4+ + Cu+ = Ce3+ + Cu2+ Initial mol Ce4+ = 25.0-‐mL*0.0500 M = 1.25 mmol Initial mol Cu+ = 15.0-‐mL*0.0500 M = 0.75 mmol More Ce4+ than Cu+ therefore Mol Ce3+ = 0.75 mmol [Ce3+] = 0.75 mmol / 40.0-‐mL = 1.9e-‐2 M Mol Ce4+ = 0.50 mmol [Ce4+] = 0.50 mmol / 40.0-‐mL = 1.3e-‐2 M Mol Cu+ = 0.00 mmol [Cu+] = 0.00 Mol Cu2+ = 0.75 mmol [Cu2+] = 0.75 mmol / 40.0-‐mL = 1.9e-‐2 M E = 1.44 – 0.0592 log 1.9e-‐2 / 1.3e-‐2 = 1.43 V

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-‐ + H2O Ý HA + OH

-‐ predominates. d) It is impossible to know since it could be acidic or basic depending on the Ka of

the acid. e) Strongly basic since excess OH-‐ is present.

2] A sample solution of 50.00 mL 0.0500 M oxalic acid (H2C2O4) is titrated with 50.00 mL of 0.1000 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 [H2C2O4] = [HC2O4

-‐] 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 50.00 mL 0.0500 M oxalic acid (H2C2O4) is titrated with 35.00 mL of 0.1000 M of NaOH. Which of the following is true after the two solutions are mixed?

a) There is 2.50 mmol of H2C2O4 present. b) There is 1.50 mmol of OH-‐ present. c) There is 2.50 mmol of HC2O4

-‐ present. d) There is 1.00 mmol of H2C2O4 present. e) There is 1.00 mmol of C2O4

2-‐ present 4] What is the fraction of EDTA in the Y4-‐ form at pH 7.00? 5] What is the conditional formation constant of CaEDTA2-‐ 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]

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a) β

α+

=1

][Mm

b) ][1][L

Mm β

α+

=

c) ][1

1Lm β

α+

=

d) 2][][1

1LLm ββ

α++

=

e) β

α+

=11

m

7] What is the pH of a titration solution that consists of 0.100 M CH3COOH (Ka = 1.75e-‐5) and 0.050 M NaOH? 8] What is E0 for the following half reaction if E0 for Zn2+ + 2e-‐ Ý Zn(s) is -‐0.762 V? ½ Zn2+ + e-‐ Ý ½ Zn(s) 9] What is E0cell for the following reaction? 2Na(s) + 2H+ Ý 2Na+ + H2(g) Na+ + e-‐ Ý Na(s) E0 = -‐ 2.7143 V 2H+ + 2e-‐ Ý H2(g) E0 = 0.0000 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 0.355 V in standardize 1.00e-‐3 M solution. The response of this electrode in an unknown solution of F-‐ is 0.407 V. What is [F-‐] for that unknown solution? 12] 18.00 mL of 0.125 M Sn4+ is titrated with 0.100 M Ti2+ in the following reaction: Sn4+ + 2Ti2+ = Sn2+ + 2Ti3+ What is the added volume of titrant required to reach the equivalence point?

38

13] Which of the following is true at the equivalence point of problem 12?

a) [Sn4+] = [Ti2+] & [Sn2+] = [Ti3+] b) [Sn4+] = [Ti2+] = [Sn2+] = [Ti3+] c) 4[Sn4+] = 2[Ti2+] & 2[Sn2+] = 3[Ti3+] d) 2[Sn4+] = [Ti2+] & 2[Sn2+] = [Ti3+] e) [Sn4+] = 2[Ti2+] & [Sn2+] = 2[Ti3+]

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 25.00 mL of 0.500 M NaOH and 25.00 mL of 0.250 M H3AsO4 (arsenic acid). H3AsO4 = H2AsO4

-‐ + H+ Ka1 = 5.8e-‐3 H2AsO4

-‐ = HAsO42-‐+ H+ Ka2 = 1.10e-‐7

HAsO42-‐ = AsO4

3-‐+ H+ Ka3 = 3.2e-‐12 17] Which of the following species is the strongest reducing agent? A+ + e-‐ = A E0 = 0.75 V B + e-‐ = B-‐ E0 = 0.25 V D2+ + e-‐ = D+ E0 = -‐0.50 V

a) A+ b) B-‐ c) B d) D2+ e) D+

39

18] Calculate the standard state cell potential for the following Cu(s)/Cu2+(aq)//K+(aq)/K(s) 19] What is the standard state reduction potential for the following reaction?

AgBr(s) + e-‐ = Ag(s) + Br-‐ Ag+ + e-‐ = Ag(s) E0 = 0.799 V AgBr(s) = Ag+ + Br-‐ Ksp = 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-‐ = A2-‐ + H20 2.50 1.00 0 -‐1.00 -‐1.00 +1.00 1.50 0 1.00 1.00 mmol A2-‐ 4] b 5] a: Kf’ = 0.36*10

10.69 = 1.8e-‐10 6] c 7] e: pH = pKa 8] e 9] d: E0cell = 0.0000 – (-‐2.7143) V 10] c: E = E0 – 0.0592 log 1/[H+] = 0.0000 – 0.0592 log 1/[1.00e-‐5] = -‐0.296 V 11] c: E = const – 0.0592 log [F-‐] 0.355 = const – 0.0592 log [1.00e-‐3] const = 0.177 0.407 = 0.177 – 0.0592 log [F-‐] [F-‐] = 1.30e-‐4 M 12] a: 18.00 mL*0.125 M Sn4+*(2mol Ti2+/mol Sn4+)*1/0.100 M Ti2+ = 45.0 mL 13] d 14] c 15] d 16] b: 2nd eq. pt where all H3AsO4 is titrated to HAsO4

2-‐ which is intermediate of H2AsO4-‐

and AsO43-‐. These two equilibria become important:

H2AsO4

-‐ = HAsO42-‐+ H+ Ka2 = 1.10e-‐7

HAsO42-‐ = AsO4

3-‐+ H+ Ka3 = 3.2e-‐12 Therefore pH = ½(pKa2 + pKa3) = ½ (6.9586 + 11.495) = 9.23

40

17] e 18] a: Ecell = Ecath – Eanod = -‐2.936 – 0.339 = -‐3.275 V 19] Start with: E = E0(Ag+/Ag) – 0.0592 log 1/[Ag+] Realize that Ksp = [Ag

+] [Br-‐] [Ag+] = Ksp / [Br-‐] sub into Nernst Eqn above

E = E0(Ag+/Ag) – 0.0592 log [Br-‐]/Ksp let [Br-‐] = 1 for standard state conditions E0 = 0.799 – 0.0592 log 1/5.00e-‐13 = 0.0708 V

problemset1( - university of idaho 253/homewor… · · 2013-05-29answers(to(problem(set(1(! 1]...

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