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1. A buffer solution is formed by adding 0.500 mol of sodium acetate and 0.500 mol of acetic acid to

1.00 L H2O. What is the pH of the solution at equilibrium? (Ka = 1.80 x 10-5)

(A) 5.05 (B) 4.74 (C) 4.44 (D) 2.38 (E) None of these

Mixing a weak acid and a salt that contains the anion of a weak acid in roughly equal amounts makes a buffer. That is what’s done here. If they are in exactly the same amount, the Henderson-Hasselbach equation reduces to pH = pKa. pH = -log(1.80 x 10-5) = 4.74. The correct choice was B.

2. The amount (in grams) of sodium acetate (MW = 82.0) to be added to 500.0 mL of 0.200 molar acetic acid (Ka = 1.80 x 10-5) in order to make a buffer with pH = 5.000 is

(A) 69 (B) 0.180 (C) 14.9 (D) 29.5 (E) None of these

Mixing a weak acid and a salt that contains the anion of the weak acid to make a buffer. Use the Henderson-Hasselbach equation. pH = pKa + log[C2H3O2-]/[HC2H3O2]. Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles. 5 = -log(1.80 x 10-5) + log(x)/0.1. Solve for x = 0.18 moles. Multiply by the molar mass and you get 14.9 grams. The correct choice was C.

3. What volumes of 0.500 M HNO2 (Ka = 4.0 X 10-4) and 0.500 M NaNO2 must be mixed to prepare 1.0 L of a solution buffered at pH 3.55?

(A) 500 mL of each solution

(B) 703 mL 0.500 M HNO2, 297 mL 0.500 M NaNO2

(C) 413 mL 0.500 M HNO2, 587 mL 0.500 M NaNO2

(D) 297 mL 0.500 M HNO2, 703 mL 0.500 M NaNO2

(E) 587 mL 0.500 M HNO2, 413 mL 0.500 M NaNO2

Mixing a weak acid and a salt that contains the anion of the weak acid to make a buffer. Use the Henderson-Hasselbach equation. pH = pKa + log[NO2-]/[HNO2]. Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles. 3.55 = -log(4.0 X 10-4) + log[NO2-]/[HNO2]. Solve for [NO2-]/[HNO2] = 1.42. Set up the ratio of moles according to the solution concentrations. (.5V1)/(.5V2) = 1.42 and the other constraint is that the total volume sums to 1 liter: V1 + V2 = 1. Substitute and solve. V1 = 0.587 L V2 = 0.413 L. The correct choice was C.

4. Which of the following would not make a good buffering system?

(A) SO42- and H2SO4

(B) HCO3- and H2CO3

(C) NH3 and NH4+

(D) CH3COO- and CH3COOH

There are two ways to make a buffer. Either mix roughly equal amounts of a weak acid and a salt that contains the anion of the weak acid or mix half as much strong base with a weak acid. There are no strong bases among the choices so pick the weak acid/acid anion pair. Choices B, C, and D are all weak acid/acid anion pairs and would make a buffer. Choice A is not a weak acid/acid anion (conjugate base) pair. The correct choice was A.

5. If 1.00 L of 1.00 M CH3COOH is mixed with 0.25 mole of solid NaOH (assume no volume change), what will be the pH of the resulting solution? (Ka = 1.8 x 10-5).

(A) 4. 14 (B) 4.26 (C) 4.74 (D) 5.35

Mixing a weak acid with a strong base is one way to make a buffer. Use the Henderson-Hasselbach equation. pH = pKa + log[CH3COO-]/[ CH3COOH]. Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles. pH = -log(1.8 x 10-5) + log(.25)/(.75). Solve for pH = 4.26. The correct choice was B.

6. Which solution would show the least change in pH upon addition of 3.0 mL of 1.0 M KOH? (Assume equal volumes of each solution are used. Ka for HC2H3O2 = 1.8 x 10-5)

(A) A solution that is 0.50 M acetic acid and 0.50 M sodium acetate.

(B) A solution that is 0.10 M acetic acid and 0.10 M sodium acetate.

(C) A solution that is 1.0 M acetic acid.

(D) A solution that is 0.50 M sodium acetate.

This question is asking two things: (1) which is a buffer and (2) if a buffer, which has the greatest buffer capacity. Choices A and B are both buffers. The greater concentration of weak acid and acid anion in choice A indicates that it has a greater buffer capacity than choice B. The correct choice was A.

7. What volume of 0.100 M NaOH must be added to 1.00 L of 0.0500 M HA

(Ka = 4.0 x 10-8) to achieve a pH of 8.00?

(A) 1.00 L (B) 5.00 L (C) 2.00 L (D) 4.00 L (E) None of these

Mixing a weak acid with a strong base is one way to make a buffer. Use the Henderson-Hasselbach equation. pH = pKa + log[A-]/[HA]. Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles. 8.00 = -log(4.0 x 10-8) + log(x)/(.05 – x). Solve for x = .2 moles of strong base. (.1 M)(V) = .2 moles. The correct choice was C, 2 liters.

8. Determine the pH of a solution in which 1.00 mol H2CO3 (Ka = 4.2 x 10-7) and 1.00 mole NaHCO3 are dissolved in enough water to form 1.00 L of solution.

This is a buffer. The weak acid is carbonic acid and the conjugate base is bicarbonate

Use the Henderson-Hasselbach equation. pH = pKa + log[HCO3-]/[H2CO3].

Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles.

pH = -log(4.2 x 10-7) + log(1). pH = pKa = 6.4

9. How many moles of NaHCO3 should be added to one liter of 0.100 M H2CO3 (Ka = 4.2 x 10-7) to prepare a buffer with pH = 7.00?

Mixing a weak acid and a salt that contains the anion of the weak acid to make a buffer.

The weak acid is carbonic acid and the conjugate base is bicarbonate

Use the Henderson-Hasselbach equation. pH = pKa + log[HCO3-]/[H2CO3].

Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles.

pH = -log(4.2 x 10-7) + log[HCO3-]/[ H2CO3].

7.00 = 6.4 + log[HCO3-]/(0.1). Solve and you get .4 moles of sodium bicarbonate should be added.

10. Determine the pH of 0.01 M NH3 (Kb= 1.8 x 10-5) when an equal volume of 0.05 M NH4Cl is added.

Mixing a weak acid and a salt that contains the anion of the weak acid to make a buffer.

The weak acid is ammonium and the conjugate base is ammonia.

KaKb = Kw = 1.0 x 10-14 Ka = (1.0 x 10-14)/(1.8 x 10-5) = 5.6 x 10-10

Use the Henderson-Hasselbach equation. pH = pKa + log[NH3]/[NH4+].

Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles.

pH = -log(5.6 x 10-10) + log(0.01)/(0.05) = 8.6

11. Determine the pH of an HCN/KCN buffer containing 0.10 mole HCN and 0.07 mole KCN before and after the addition of 0.001 mol HCl to one liter of buffer. (Ka = 4.79 x 10-10)

Mixing a weak acid and a salt that contains the anion of the weak acid to make a buffer.

The weak acid is cyanic acid and the conjugate base is the cyanide anion.

Before the addition of hydrocholoric acid.

Use the Henderson-Hasselbach equation. pH = pKa + log[CN-]/[HCN].

pH = -log(4.79 x 10-10) + log(0.07)/(0.10) = 9.16

After the addition of hydrocholoric acid. Cyanide anion is converted to cyanic acid.

pH = -log(4.79 x 10-10) + log(0.07 - 0.001)/(0.10 + 0.001) = 9.15

12. If 25 mL of 0.2 M NaOH is added to 50 mL of 0.1 M CH3COOH (Ka = 1.8 x 10-5), what is the pH?

Mixing a weak acid with a strong base is one way to make a buffer.

The moles of acid anion will be equal to the amount of strong base added and the moles of weak acid will be equal to what you start with minus the amount of strong base added. Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles.

Moles CH3COO- = (.025 L)(0.2 M) = 0.005 moles

Moles CH3COOH = (.050 L)(0.1 M) = 0.005 moles

Use the Henderson-Hasselbach equation. pH = pKa + log[CH3COO-]/[ CH3COOH].

The log term is zero since the ratio = 1.

pH = pKa = 4.74

13. Calculate the mass of (NH4)2SO4 that must be added to 1 L of 0.50 M NH3 in order to prepare a buffer with a pH of 8.65. (Ka of NH4+ = 5.6 x 10-10).

Mixing a weak acid and a salt that contains the anion of the weak acid to make a buffer.

The weak acid is ammonium and the conjugate base is ammonia.

Use the Henderson-Hasselbach equation. pH = pKa + log[NH3]/[NH4+].

Point out that the H-H equation is volume independent and therefore the calculation can be based on a ratio of moles.

8.65 = -log(5.6 x 10-10) + log(0.50)/(x)

8.65 = 9.25 + log(0.50 moles)/(x)

(0.50 moles)/(x) = .25 and x = 2.0 moles. Since you get 2 moles of ammonium cation per mole of ammonium sulfate, you need 1 mole of ammonium sulfate or 132 grams.

14. Calculate the pH of the solution prepared when 0.250 mol HClO4 is added to 2.00 L of a solution that is 0.500 M CH3CO2H and 0.400 M CH3CO2Na. (Ka of CH3CO2H = 1.8 x 10-5)

Buffer that has a strong acid added to it

Initial moles CH3COO- = (2.00 L)(0.400 M) = 0.800 moles

Initial moles CH3COOH = (2.00 L)(0.500 M) = 1.00 moles

Moles CH3COO- after addition of strong acid = 0.800 moles – 0.250 moles = 0.550 moles

Moles CH3COOH after addition of strong acid = 1.00 moles + 0.250 moles = 1.25 moles

Use the Henderson-Hasselbach equation. pH = pKa + log[CH3COO-]/[ CH3COOH].

The H-H equation is volume independent and therefore the calculation can be based on a ratio of moles

pH = -log(1.8 x 10-5) + log(0.550)/(1.25) = 4.39

15. A buffer is prepared in which the ratio of HCO3-/CO32- is 5.00.

(A) What is the pH of this buffer? (Ka for HCO3- = 4.80 x 10-11)

Use the Henderson-Hasselbach equation. pH = pKa + log[CO32-]/[ HCO3-].

pH = -log(4.8 x 10-11) + log(1/5) = 9.62

(B) Enough strong acid is added to make the pH of the buffer 9.30. What is the ratio of HCO3-/CO3-2 at this point?

Use the Henderson-Hasselbach equation. pH = pKa + log[CO32-]/[ HCO3-].

9.30 = -log(4.8 x 10-11) + log(x)

9.30 = 10.3 + log(x) x = .1 and therefore the ratio is 10.

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