Question

$pH$ of $0.1M$ solutions of $KX,KY$ and $KZ$ are $8,9$ and $10$ respectively. Thus order of acidic strength of $HX,HY$ and $HZ$ is

• A. $HX<HY<HZ$
• B. $HZ<HX<HY$
• C. $HY<HX<HZ$
• D. $HZ<HY<HX$

Answer 1

Answer: (D)

These are the salts of strong base with weak monobasic acid. Solution is basic due to hydrolysis of anion.

$X^{-}+H_{2}O\rightleftharpoons HX+O{H}^{-}$

$pH=7+\frac{p{K}_a}{2}+\frac{\log C}{2}$

$=7+\frac{p{K}_a}{2}+\frac{\log \stackrel{.}{0}1}{2}=7+\frac{p{K}_a}{2}-0.5$

$\therefore p{k}_a=2(pH-6.5)$

$p{k}_a(HX)=2(8-6.5)=3.0,K_a=10^{-3}$

$p{k}_a(HX)=2(9-6.5)=5.0,K_a=10^{-5}$

$p{k}_a(HZ)=2(10-6.5)=7.0,K_a=10^{-7}$

Thus, $HX$ is the strongest acid.

$\therefore HZ<HY<HY$