0.1 M NaOH is titrated with 0.1 M , 20 ml HA till the end point, K a ( HA )=6 × 10-6 and degree of dissociation of HA is small as compared to unity. What is the pH of the resulting solution at the end point?

Question

0.1 M NaOH is titrated with 0.1 M , 20 ml HA till the end point, K a ( HA )=6 × 10-6 and degree of dissociation of HA is small as compared to unity. What is the pH of the resulting solution at the end point? (A) 6.23 (B) 9.22 (C) 7.21 (D) 8.95

Tardigrade Admin · Accepted Answer

underset2 NaOH + underset2 HA arrow NaA + H 2 O At end point ≡ 0.1 × 20=2 ∵ 20 mL NaOH is required for the complete neutralization of HA NaA is a salt of strong base and weak acid Thus, will undergo hydrolysis and solution will becomes basic C =[ NaA ]=(2/20+20)=0.05 M And pK a =- log (6 × 10-6)=5.2 pH at the end point =7+(1/2)( pK a + log C ) 7+(1/2)(5.2+ log 0.05)=8.95

Q. 0.1MNaOH is titrated with 0.1M,20mlHA till the end point, Ka​(HA)=6×10−6 and degree of dissociation of HA is small as compared to unity. What is the pH of the resulting solution at the end point?

6.23

9.22

7.21

8.95

Q. $0.1 M N a O H$ is titrated with $0.1 M, 20 m l H A$ till the end point, $K_{a} (H A) = 6 \times 1 0^{- 6}$ and degree of dissociation of $H A$ is small as compared to unity. What is the $p H$ of the resulting solution at the end point?