Question

What is the $\mathrm{pH}$ of millimolar solution of ammonium hydroxide which is $20\%$ dissociated?

Answer 1

Step 1: Find the value of concentration of the ${\mathrm{NH}}_4\mathrm{OH}$ solution

We know that ${\mathrm{NH}}_4\mathrm{OH}$ is present in millimolar concentration.
${\mathrm{NH}}_4\mathrm{OH}=0.001M\dots \dots \left(1\right)$
The weak base ammonium hydroxide, so ${\mathrm{NH}}_4\mathrm{OH}$ will dissociate as ${\mathrm{OH}}^{-}$ions partially.
The dissociation degree of ${\mathrm{NH}}_4\mathrm{OH}$ is $20\%$
degree of dissociation is as follows

$$\begin{gathered} =20\% \\ =\frac{20}{100} \\ =0.2\dots \dots \left(2\right) \end{gathered}$$

Step 2: Find the value of ${\mathrm{OH}}^{-}$
From the value concentration of the above ${\mathrm{OH}}^{-}$ions can be calculated
Concentration of ${\mathrm{OH}}^{-}=$Concentration of solution $\times$degree of dissociation
$\begin{array}{rcl}\left[O{H}^{-}\right]& =& 0.001\times 0.2\\ & =& 2\times {10}^{-4}\dots \dots .\left(3\right)\end{array}$
We know that water has $\mathrm{pOH}=7$

Hence, the concentration of ${\mathrm{OH}}^{-}$ions will be
$$\begin{gathered} \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}pOH=-\log \left[O{H}^{-}\right]\dots \dots ...\left(4\right) \\ \Rightarrow 7=-\log \left[O{H}^{-}\right] \\ \Rightarrow \left[O{H}^{-}\right]={10}^{-7}\dots \dots \dots ..\left(5\right) \end{gathered}$$

To calculate the ${\mathrm{OH}}^{-}$ of total concentration, we should add the equation (3) and (5)
$$\begin{gathered} total\left[O{H}^{-}\right]=\left(2\times {10}^{-4}\right)+{10}^{-7} \\ \Rightarrow total\left[O{H}^{-}\right]={10}^{-4}(2+0.001) \\ \Rightarrow total\left[{\mathrm{OH}}^{-}\right]=2.001\times {10}^{-4} \end{gathered}$$

Step 3: Find the values of $pOH$ and $pH$
Insert ${\mathrm{OH}}^{-}$ in the equation (4)

$\begin{array}{rcl}pOH& =& -\log \left(2.001\times {10}^{-4}\right)\\ & \Rightarrow & pOH=3.6988\end{array}$
The value of $pH$ can be simplified by using the equation

$\begin{array}{rcl}pH& =& 14-pOH\dots ..\left(6\right)\\ & \Rightarrow & \mathrm{pH}=14-3.6988\\ & \Rightarrow & \mathrm{pH}=10.301\end{array}$

Therefore, A millimolar solution of ammonium hydroxide that is $20\%$ dissociated has a $pH$ of $\mathit{10}\mathit{.}\mathit{301}$.