Question

The dissociation constant of a base $MOH$ is $5.5\times {10}^{-6}$, then calculate the dissociation constant of its conjugate acid.

• A. $1.81\times {10}^{-9}$
• B. $1.81\times {10}^{-8}$
• C. $2.81\times {10}^{-7}$
• D. $2.81\times {10}^{-8}$

Answer 1

The correct option is A $1.81\times {10}^{-9}$

$MOH\left(aq\right)\rightleftharpoons M^{+}\left(aq\right)+O{H}^{-}\left(aq\right)BaseConjugateAcid{K}_b=\frac{\left[M^{+}\right].\left[O{H}^{-}\right]}{\left[MOH\right]}{M}^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons MOH\left(aq\right)+H^{+}\left(aq\right)K_a=\frac{\left[MOH\right]\left[H^{+}\right]}{\left[M^{+}\right]}$
We see that,
$Kw=\left[H^{+}\right]\left[O{H}^{-}\right]=K_a\times K_b$
$\Rightarrow Ka=\frac{K_w}{K_b}$

putting the values,
$\Rightarrow Ka=\frac{{10}^{-14}}{5.5\times {10}^{-6}}=1.81\times {10}^{-9}$