Question

An acetate buffer solution is prepared by combining 50. mL of a 0.20 M acetic acid, HC2H3O2 (aq), and 50. mL of 0.20 M sodium acetate, NaC2H3O2 (aq). A 5.0 mL sampleof 0.10 M NaOH (aq) is added to the buffer solution. Which of the following is a correct pairing of the acetate species present in greater concentration and of the pH of the solution after the NaOH (aq) is added? (the pKa of acetic acid is 4.7)

A. HC2H3O2 pH < 4.7
B. HC2H3O2 pH >4.7
C. C2H3O2- pH < 4.7
D. C2H3O2- pH > 4.7

Answer 1

After adding NaOH to the acetate buffer, the acetate ion concentration increases and the pH of the solution will be greater than the pKa of acetic acid (4.7). Option D, with C2H3O2- and pH > 4.7, is correct.

Step-by-step explanation:

The question involves calculating the pH of an acetate buffer solution after the addition of sodium hydroxide, NaOH. The buffer consists of acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2). The pKa of acetic acid is 4.7, and equal concentrations of the buffer components are typically ideal. When NaOH is added, it neutralizes some of the acetic acid, converting it to acetate and increasing the pH slightly.

Before the addition of NaOH, the buffer has equal concentrations of acetic acid and acetate, so its pH is close to the pKa of acetic acid, which is 4.7. After adding NaOH, more acetic acid is converted to acetate, but the pH will still be maintained close to the pKa since the buffer action resists changes in pH. Therefore, the correct pairing is that the acetate ion (C2H3O2-) will be present in greater concentration and the pH of the solution will be greater than 4.7 after NaOH is added, making option D correct.

Answer 2

B. CH3COOH pH > 4.7 (4.8)

Step-by-step explanation:

• CH3COOH + NaOH ↔ CH3COONa + H2O
• CH3COONa + NaOH ↔ CH3COONa

∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol

⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol

⇒ C CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)

⇒ C CH3COOH = 0.0905 M

∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol

⇒ C CH3COONa = (0.01 mol + 5 E-4 mol) / (0.105 L )

⇒ C CH3COONa = 0.1 M

∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5

⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])

⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]

⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0

⇒ [H3O+] = 1.5835 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = - Log (1.5835 E-5)

⇒ pH = 4.8004 > 4.7