# The pH of a 5.95\times 10^-1 M solution of a weak base is 11.54. What is the pK_b for this base to two decimal places?

Nov 29, 2017

$\textsf{p {K}_{b} = 4.69}$

#### Explanation:

From an ICE table you can use this expression which applies to a weak base:

$\textsf{p O H = \frac{1}{2} \left[p {K}_{b} - \log b\right]}$

b is the concentration of the base.

$\textsf{p O H + p H = 14}$

$\therefore$$\textsf{p O H = 14 - 11.54 = 2.46}$

$\therefore$$\textsf{2.46 = \frac{1}{2} \left[p {K}_{b} - \log \left(0.595\right)\right]}$

$\textsf{2.46 = \frac{1}{2} p {K}_{b} + 0.1127}$

$\textsf{\frac{1}{2} p {K}_{b} = 2.3473}$

$\textsf{p {K}_{b} = 4.69}$