Question

When 100 ml of 0.4 M CH3COOH are mixed with 100 ml of 0.2 M NaOH, the [H3O+] in the solution is approximately : [Ka(CH3COOH) = 1.8 × 10–5]

• A. 1.8 × 10–6 M
• B. 1.8 × 10–5 M
• C. 9 × 10–6 M
• D. 9 × 10–5 M.

Answer 1

Answer: (B)

1.8 × 10–5 M

Answer 2

CH3COOH + NaOH $\longrightarrow$ CH3COONa + H2O

time t = 0 40 mmole 20 mmole

time t = t 20 mmole – 20 mmole

pH = pKa+ log $\left( \frac{20}{20} \right) \Rightarrow$ pH = pKa

$\Rightarrow$ [H+] = Ka = 1.8 × 10–5 M