Question

$K_b$ for a monoacidic base whose $0.2M$ solution has a $pH$ of $9.50$ is:

• A. $5\times {10}^{-8}$
• B. $5\times {10}^{-9}$
• C. $6\times {10}^{-6}$
• D. $6\times {10}^{-9}$

Answer 1

The correct option is B $5\times {10}^{-9}$

$BOH\rightleftharpoons B^{+}+O{H}^{-}$
$C00$
$C(1-\alpha )C\alpha C\alpha$

Given that, $pH=9.5$
$\Rightarrow pOH=14-9.5=4.5$
We know,
$pOH=-log\left[O{H}^{-}\right]=4.5$
$\left[O{H}^{-}\right]={10}^{-4.5}$

now, $[OH{]}^{-}=C.\alpha$
(as $\alpha =\sqrt{K_b/C}$ )
so, $[OH{]}^{-}=\sqrt{K_b.C}$

Squaring both sides:
$\left(\right[O{H}^{-}]{)}^2=K_b.C{10}^{-4.5\times 2}=K_b\times 0.2K_b=\frac{{10}^{-9}}{0.2}=5\times {10}^{-9}$