Answer to Question #186712 in Chemistry for Jowheecelle

Question

Calculate the pH of the resulting solution when the original solution is mixed with 35.0 mL of 0.975 M NaOH solution

Answer 1

An erlenmeyer flask contains 50.0 mL of an aqueous solution of 0.425 M acetic acid (CH₃COOH). Given that acetic acid has a K_a = 1.75×10^-5. Calculate the pH of the resulting solution when the original solution is mixed with 35.0 mL of 0.975 M NaOH solution.

Solution:

V_solution = 50.0 mL + 35.0 mL = 85.0 mL

C(CH₃COOH) = 0.425 M × (50.0 mL / 85.0 mL) = 0.25 M

C(NaOH) = 0.975 M × (35.0 mL / 85.0 mL) = 0.40 M

The balanced chemical equation:

CH₃COOH + NaOH → CH₃COONa + H₂O

ICE Table:

NaOH >> CH₃COONa

So, pH is determined by the concentration of excess NaOH

NaOH → Na⁺ + OH^-

[OH^-] = C(NaOH) = 0.15 M (according to the ICE table)

pOH = -log[OH^-] = -log(0.15) = 0.82

pH = 14 - pOH = 14 - 0.82 = 13.18

pH = 13.18

Answer: pH = 13.18