Question

The equilibrium constant for the reaction given below is $3.6\times {10}^{-7}$.

$O{Cl}^{-}+H_{2}O\left(l\right)\rightleftharpoons HOCl(aq.)+{OH}^{-}(aq.)$

What is $K_a$ for $HOCl$?

• A. $2.8\times {10}^{-8}$
• B. $3.6\times {10}^{-7}$
• C. $6\times {10}^{-4}$
• D. $2.8\times {10}^{-6}$

Answer 1

The correct option is A $2.8\times {10}^{-8}$

$K$ for the given reaction:

$K_{eq}=\frac{\left[HOCl\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]\left[OC{l}^{-}\right]}$

or $K_b=\frac{\left[HOCl\right]\left[O{H}^{-}\right]}{\left[OC{l}^{-}\right]}=3.6\times {10}^{-7}$

Also, $K_a\times k_b=k_w$

$\therefore 3.6\times {10}^{-7}\times K_a={10}^{-14}$

$K_a\approx 2.8\times {10}^{-8}$