Buffer Practice-Key - Practice Worksheet key PDF

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1 Buffer Practice Problems 1.

What would be the pH of a 100.0 mL solution containing 0.24 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.24 M sodium formate (NaCHO2)? pH = 3.74

2.

What would be the pH of a 100.0 mL solution containing 0.15 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.15 M sodium formate (NaCHO2)? pH = 3.74

3.

How do the pH values in Problem 1 and Problem 2 compare? Why is this so? (Henderson-Hasselbalch equation might help you see the “why.”) They are the same. Since the [base] = [acid], the ratio of base to acid is 1. Since log(1) = 0, the pH = pKa. It does not matter how concentrated, if acid concentration equals base concentration, the buffer will have the pH = pKa.

4.

What would be the pH of a 100.0 mL solution containing 0.24 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.15 M sodium formate (NaCHO2)? pH = 3.54

5.

How do the pH values in Problem 2 and Problem 4 compare? Based on the concentration and types of substances in solution, why is this so? Problem 4 is lower in pH. Since [base] < [acid], the pH shifts to a more acidic pH value. Since the acid is more concentrated than the base, we would expect the pH to be more acidic (lower pH) than if the acid and base were equal in concentration (less than pKa).

6.

What would be the pH of a 100.0 mL solution containing 0.15 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.24 M sodium formate (NaCHO2)? pH = 3.95

7.

How do the pH values in Problem 2 and Problem 6 compare? Based on the types of substances in solution, why is this so? Problem 6 is higher in pH. Since [base] > [acid], the pH shifts to a more basic pH value. Since the base is more concentrated than the acid, we would expect the pH to be more basic (higher pH) than if the acid and base were equal in concentration (greater than pKa).

2 8.

What would be the pH of a 100.0 mL solution containing 0.24 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.24 M sodium formate (NaCHO2) if it were contaminated by 0.013 moles of NaOH? HCHO2 + OH—  B 0.024 0.013 C -x -x A 0.024-x 0.013-x 0.011 0

CHO2— + H2O 0.024 --+x --0.024+x --0.037 ---

pH = 4.27

9.

How do the pH values in Problem 1 and Problem 8 compare? Based on the types of substances in solution, why is this so? Has the buffer exceeded its capacity? Problem 8 is higher in pH. Since the contaminant is a base, we should expect the pH to go up. This can also be seen in the neutralization… more conjugate base present than acid. The pH has NOT changed more than 1.0 pH unit, so the buffer has NOT exceeded capacity.

10. What would be the pH of a 100.0 mL solution containing 0.15 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.15 M sodium formate (NaCHO2) if it were contaminated by 0.013 moles of NaOH? HCHO2 + OH—  B 0.015 0.013 C -x -x A 0.015-x 0.013-x 0.002 0

—

CHO2 + H2O 0.015 --+x --0.015+x --0.028 ---

pH = 4.89

11. How do the pH values in Problem 2 and Problem 10 compare? Based on the types of substances in solution, why is this so? Has the buffer exceeded its capacity? Problem 10 is higher in pH. Since the contaminant is a base, we should expect the pH to go up. This can also be seen in the neutralization… more conjugate base present than acid. The pH has changed more than 1.0 pH unit, so the buffer HAS exceeded capacity.

3 12. What would be the pH of a 100.0 mL solution containing 0.24 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.24 M sodium formate (NaCHO2) if it were contaminated by 0.013 moles of HCl? CHO2— + H3O+  B 0.024 0.013 C -x -x A 0.024-x 0.013-x 0.011 0

HCHO2 + H2O 0.024 --+x --0.024+x --0.037 ---

pH = 3.22

13. How do the pH values in Problem 1 and Problem 12 compare? Based on the types of substances in solution, why is this so? Has the buffer exceeded its capacity? Problem 12 is lower in pH. Since the contaminant is an acid, we should expect the pH to go down. This can also be seen in the neutralization… more acid present than conjugate base. The pH has NOT changed more than 1.0 pH unit, so the buffer has NOT exceeded capacity. 14. What would be the pH of a 100.0 mL solution containing 0.15 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.15 M sodium formate (NaCHO2) if it were contaminated by 0.013 moles of HCl? — + CHO2 + H3O  B 0.015 0.013 C -x -x A 0.015-x 0.013-x 0.002 0

HCHO2 + H2O 0.015 --+x --0.015+x --0.028 ---

pH = 2.60

15. How do the pH values in Problem 2 and Problem 14 compare? Based on the types of substances in solution, why is this so? Has the buffer exceeded its capacity? Problem 14 is lower in pH. Since the contaminant is an acid, we should expect the pH to go down. This can also be seen in the neutralization… more acid present than conjugate base. The pH has changed more than 1.0 pH unit, so the buffer HAS exceeded capacity.

4 16. What would be the pH of a 100.0 mL solution containing 0.15 M hydroiodous acid (HIO; Ka = 2.3x10-11) and 0.15 M sodium hypoiodite (NaIO) if it were contaminated by 20.0 mL of 0.20M NaOH? HIO + OH—  IO— + H2O B 0.015 0.004 0.015 --C -x -x +x --A 0.015-x 0.004-x 0.015+x --0.011 0 0.019 --pH = 10.88 17. What would be the pH of a 100.0 mL solution containing 0.24 M hydroiodous acid (HIO; Ka = 2.3x10-11) and 0.24 M sodium hypoiodite (NaIO) if it were contaminated by 20.0 mL of 0.20M HCl? IO— + H3O+  HIO + H2O B 0.024 0.004 0.024 --C -x -x +x --A 0.024-x 0.004-x 0.024+x --0.020 0 0.028 --pH = 10.49 18. What would be the pH of a 100.0 mL solution containing 0.50 M acetic acid (HC2H3O2; Ka = 1.8x10-5) if 50.0 mL of 0.60 M NaOH were added? HC2H3O2 + OH—  C2H3O2 — + H2O B 0.050 0.030 0 --C -x -x +x --A 0.050-x 0.030-x x --0.020 0 0.030 --(Forms a buffer… weak acid and conj. base still there) pH = 4.92 19. What would be the pH of a 100.0 mL solution containing 0.10 M pyridine (C5H5N; Kb = 1.7x10-9) if 50.0 mL of 0.12 M HCl were added? C5H5N + H3O+  HC5H5N + H2O B 0.010 0.006 0 --C -x -x +x --A 0.010-x 0.006-x x --0.004 0 0.006 --(Forms a buffer… weak acid and conj. base still there) pH = 5.05 20. What would be the pH of a 100.0 mL solution containing 0.15 M sodium bicarbonate (NaHCO3) and 0.20 M sodium carbonate (Na2CO3)? (For H2CO3, Ka1 = 4.3x10-7 and Ka2 = 5.6x10-11) Buffer equilibrium involves the HCO3— and the CO3 2—. pH = 10.38...