Question

What is the $pOH$ of $0.1MKB$ (salt of weak acid and strong base) at ${25}^{\circ }C$?

(Given: $p{K}_b$ of $B^{-}=7$)

Answer 1

Since, $p{K}_b=7$, $K_b=1\times {10}^{-7}$
The base dissociation equilibrium is as shown below.
$B^{-}+H_{2}O\rightleftharpoons BH+O{H}^{-}$.
Let the hydroxide ion concentration be $x$ M.

	$B^{-}$	$BH$	$O{H}^{-}$
Initial concentration (M)	$0.1$	0	0
Equilibrium concentration (M)	$0.1-x$	$x$	$x$

The expression for the equilibrium constant is $K=\frac{\left[BH\right]\left[O{H}^{-}\right]}{\left[B^{-}\right]}$
Substitue 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 $4$.