Question

The pH of solution obtained on mixing 50 mL of 0.1 M formic acid with 150 mL of 0.02 M NaOH is (K_a of HCOOH = 1.8 × 10^–4 ; log 1.8 = 0.255)

• A. 3.92
• B. 2.8
• C. 3.79
• D. 1.4

Answer 1

Answer: (A)

3.92

Answer 2

HCOOH + NaOH $\rightleftharpoons$HCOONa + H₂O

t = 0 50 × 0.1=5 150 × 0.02=3 0 0

at end 2 0 3

pH = pK_a + log $\frac { 3 } { 2 }$

= – log 1.8 × 10^–4 + 0.48 – 0.30

= 3.745 + 0.48 – 0.30

$\approx$ 3.92