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

Answer:

B. CH3COOH pH > 4.7 (4.8)

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