Question

4 ml of HCl solution of pH = 2 is mixed with 6 ml of NaOH solution of pH = 12. what would be the final pH of solution ? log 2 = 0.3

• A. 10.3
• B. 11.3
• C. 11
• D. 4.3

Answer 1

Answer: (B)

11.3

Answer 2

pH = 2, [HCl] = 10^–2 (M) & pOH = 2

or [NaOH] = 10^–2 (M)

4 ml of 10^–2 (M) HCl ŗ 4 × 10^–5 moles HCl.

6 ml of 10^–2 (M) NaOH ŗ 6 × 10^–5 moles NaOH

After mixing excess moles of OH^– = 2 × 10^–5

[OH^–] = $\frac{2 \times 10^{- 5}}{10}$× 10³ = 2 × 10^–3

or pOH = 3 – log 2 = 3 – 0.3 = 2.7

or pH = 11.3