Question

Consider a weak base, $BOH$. Molar concentration of $BOH$ that provides $[OH{]}^{-}$ of $1.5\times {10}^{-3}M$ will be :
Given: $K_b\left(KOH\right)=1.5\times {10}^{-5}$

• A. $1.5\times {10}^{-2}M$
• B. $0.015M$
• C. $1.5\times {10}^{-4}M$
• D. $0.15M$

Answer 1

The correct option is D $0.15M$

The dissociation of BOH may be represented as,
$BOH(aq.)\rightleftharpoons B^{+}(aq.)+O{H}^{-}(aq.)K_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[BOH\right]}$
Since same amount of $\left[B^{+}\right]$ and $\left[O{H}^{-}\right]$, is produced
. $\therefore \left[B^{+}\right]=\left[O{H}^{-}\right]$ at equilibrium

$K_b=\frac{[O{H}^{-}{]}^2}{\left[BOH\right]}1.5\times {10}^{-5}=\frac{(1.5\times {10}^{-3}{)}^2}{\left[BOH\right]}$

$\left[BOH\right]=\frac{(1.5\times {10}^{-3}{)}^2}{1.5\times {10}^{-5}}=0.15M$



Theory:
$K_a$ for weak acids
For the following reaction : $HA\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+A^{-}\left(aq\right)$
$K_a=\frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$
$K_b$ for weak bases
For the following reaction : $MOH\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons M^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
$K_b=\frac{[M+]\left[O{H}^{-}\right]}{\left[MOH\right]}$
Note : More the values of $K_{a}or{K}_b$ more is the strength of acid or base