# Question #0f450

Oct 28, 2015

${K}_{b} = 1.8 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}$

#### Explanation:

We can firstly find the $p {K}_{a}$ value using this expression for a weak acid:

$p H = \frac{1}{2} \left(p {K}_{a} - \log c\right)$

Where $c$ is the concentration of the acid.

$2.5 = \frac{1}{2} p {K}_{a} - \frac{1}{2} \log \left(0.18\right)$

$2.5 = \frac{1}{2} p {K}_{a} + 0.372$

$\frac{1}{2} p {K}_{a} = 2.128$

$p {K}_{a} = 4.256$

We now use the relationship:

$p {K}_{a} + p {K}_{b} = 14$

$p {K}_{b} = 14 - 4.256 = 9.744$

This means that $- \log {K}_{b} = 9.744$

From which ${K}_{b} = 1.8 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}$