# Question #dbab5

Apr 25, 2017

$\textsf{p H = 9.9}$

#### Explanation:

The ammonium ion ionises:

$\textsf{N {H}_{4}^{+} r i g h t \le f t h a r p \infty n s N {H}_{3} + {H}^{+}}$

For which:

$\textsf{{K}_{a} = \frac{\left[N {H}_{3}\right] \left[{H}^{+}\right]}{\left[N {H}_{4}^{+}\right]} = 5.6 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

When $\textsf{O {H}^{-}}$ ions are added the following takes place:

$\textsf{N {H}_{4}^{+} + O {H}^{-} \rightarrow N {H}_{3} + {H}_{2} O}$

The no. of moles added is given by:

$\textsf{{n}_{O {H}^{-}} = c \times v = 0.75 \times \frac{20.0}{1000} = 0.015}$

This means, from the equation, that the number of moles of $\textsf{N {H}_{4}^{+}}$ consumed = 0.015

$\therefore$ The number of moles remaining is given by:

$\textsf{{n}_{N {H}_{4}^{+}} = 0.025 - 0.015 = 0.01}$

From the equation you can see that the number of moles of $\textsf{N {H}_{3}}$ formed = 0.015.

The total moles of $\textsf{N {H}_{3}}$ is given by:

$\textsf{{n}_{N {H}_{3}} = 0.04 + 0.015 = 0.055}$

The small value of $\textsf{{K}_{a}}$ means we can assume that these are the moles present at equilibrium.

Rearranging the expression for $\textsf{{K}_{a}}$ we get:

$\textsf{\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}}$

Since the total volume is common to both salt and base we can use moles directly:

$\textsf{\left[{H}^{+}\right] = 5.6 \times {10}^{- 10} \times \frac{0.01}{0.055} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left[{H}^{+}\right] = 1.018 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[1.018 \times {10}^{- 10}\right]}$

$\textsf{p H = 9.99}$