Question

If the equilibrium constant of the reaction of weak acid HA with strong base is ${10}^9,$ then pH of 0.1 M NaA is

• A. 5
• B. 9
• C. 7
• D. 8

Answer 1

Answer: (B)

$HA+O{H}^{-}{H}_{2}O+A^{-}$ $K=\frac{[A^{-}]}{[HA]O{H}^{-}]}$ $HA{H}^{+}+A^{-}$ $K_a=\frac{[H^{+}][A^{-}]}{[HA]}$ $\frac{K_a}{K}=[H^{+}][O{H}^{-}]=K_w$ Or $K_a=K_{2}K={10}^{-14}\times {10}^9={10}^{-5}$ $p{K}_a=5$ $A^{-}$ solution is alkaline due to hydrolysis $\therefore$ $pH=7+\frac{p{k}_a}{2}+\frac{\log C}{2}=2+\frac{5}{2}+\frac{\log 0.1}{2}=9$