Question

A certain weak acid has $K_a=1\times {10}^{-4}.K_{eq}$ for its reaction with NaOH is

• A. $1\times {10}^{-4}$
• B. $1\times {10}^{-10}$
• C. $1\times {10}^{-10}$
• D. $1\times {10}^{-14}$

Answer 1

The correct option is C $1\times {10}^{-10}$

NaOH being a strong base is complete ionized, the reaction of acid actually take place with $O{H}^{-}$.
$HA+O{H}^{-}\rightleftharpoons A^{-}+H_{2}O$
Which is the addition of
$HA+H_{2}O\rightleftharpoons H_3{O}^{+}+A^{-};k_{a}and{H}_3{O}^{+}+O{H}^{-}\rightleftharpoons 2H_{2}O;\frac{1}{k_w}$
$\Rightarrow k=\frac{k_a}{k_w}={10}^{10}$