Question

If equilibrium constant for the reaction,
$XO ^{-}( aq )+ H _{2} O (1) \rightleftharpoons HXO ( aq )+ OH ^{-}( aq )$
is $0.36 \times 10^{-6}$ then find the value of dissociation constant $\left( K _{ a }\right)$ for $HX O$ ?

• A. $0.36 \times 10^{-8}$
• B. $2.8 \times 10^{-8}$
• C. $2.8 \times 10^{-10}$
• D. $0.36 \times 10^{-6}$

Answer 1

Answer: (B)

$HXO \rightleftharpoons H ^{+}+ XO ^{-}$
$K _{ a }=\frac{\left[ H ^{+}\right][ XO ]}{[ HXO ]}$
$XO ^{-}+ H _{2} O \rightleftharpoons HXO + OH ^{-}$
$K _{ eq }=\frac{[ HXO ]\left[ OH ^{-}\right]}{\left[ XO ^{-}\right]}$
$K _{ a } \cdot K _{ eq }=\left[ H ^{+}\right]\left[ OH ^{-}\right]$
$= K _{ w }=1 \times 10^{-4}$
$K _{ a }=\frac{ K _{ w }}{ K _{ eq }}=\frac{1.0 \times 10^{-14}}{0.36 \times 10^{-6}}$
$=2.8 \times 10^{-8}$