Question

Calculate $pOH$ of $0.1M(aq.)$ solution of weak base $BOH(K_b={10}^{-7})$ at ${25}^{\circ }C$.

Answer 1

The base dissociation equilibrium is as shown below.
$BOH\rightleftharpoons B^{+}+O{H}^{-}$.

Let the hydroxide ion concentration be x M.

	$BOH$	$B^{+}$	$O{H}^{-}$
Initial concentration (M)	$0.1$	$x$	$x$
Equilibrium concentration (M)	$0.1-x$	$x$	$x$

The expression for the equilibrium constant is $K=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[BOH\right]}$

Substitute values in the above expression.

$1\times {10}^{-7}=\frac{x\times x}{0.1-x}$

Since the value of the equilibrium constant is very small, the value of x is also very small.
Hence, $0.1-x\approx =0.1$

The equilibrium constant expression becomes $\frac{x^2}{0.1}=1\times {10}^{-7}$
Hence, $x=1\times {10}^{-4}$.

Hence, the $pOH$ of the solution is $pOH=-log\left[O{H}^{-}\right]=-log1\times {10}^{-4}=4$.
Hence, the $pOH$ of the solution is $4$.