Question

What is the $pH$ of millimolar solution of ammonium hydroxide which is $20\%$ dissociated?
A. 3.699
B. 10.301
C. 4.691
D. 9.301

Answer 1

Hint: Identify the concentration of ammonium hydroxide and degree of dissociation given to find the number of hydroxide ions. Consider the relation between the $pH$ of any solution along with the number of hydrogen or hydroxide ions present. Also recall the relation between $pH$ and $pOH$.

Complete answer:
We will need the following formula to get the answer to this question:
\[\begin{align}
  & pH=-\log [{{H}^{+}}]\text{ and } \\
 & pOH=-\log [O{{H}^{-}}] \\
\end{align}\]
Let us start the solution from the concentration of ammonium hydroxide, that is concentration of $N{{H}_{4}}OH$ solution. It is given that the concentration is a millimolar.
\[[N{{H}_{4}}OH]={{10}^{-3}}M=0.001M\]
We know that the ammonium hydroxide is a weak electrolyte, so it does not dissociate completely. Thus, only a small fraction of the total concentration of the $N{{H}_{4}}OH$ will dissociate into $O{{H}^{-}}$ ions.
Degree of dissociation is given as $20\%$. That is $20\%=\dfrac{20}{100}=0.2$
Then, concentration of $O{{H}^{-}}=\left[ O{{H}^{-}} \right]=$concentration of solution $\times $ degree of dissociation
$[O{{H}^{-}}]=0.001\times 0.2$
\[[O{{H}^{-}}]=2\times {{10}^{-4}}\]
Since, the solution is dilute, we also need to take into account the concentration of ions ${{H}^{+}}$ and $O{{H}^{-}}$ that are contributed by the dissociation of water.
We know that the $pOH$ of water is 7. Thus, the concentration of $O{{H}^{-}}$ ions will be:
\[7=-\log [O{{H}^{-}}]\]
Now, solving for $[O{{H}^{-}}]$ we get:
\[[O{{H}^{-}}]={{10}^{-7}}\]
Adding to the concentration of $O{{H}^{-}}$ ions we get from $N{{H}_{4}}OH$
Total $[O{{H}^{-}}]=(2\times {{10}^{-4}})+({{10}^{-7}})$
Taking ${{10}^{-4}}$ as common
\[[O{{H}^{-}}]=(2+0.001)\times {{10}^{-4}}\]
\[[O{{H}^{-}}]=2.001\times {{10}^{-4}}\]
Now using the formula relating the $pOH$ and $[O{{H}^{-}}]$.
\[pOH=-\log (2.001\times {{10}^{-4}})\]
Now, solving for $pOH$
\[\begin{align}
  & pOH=-\log (2.001)-\log ({{10}^{-4}}) \\
 & pOH=-0.3012+4 \\
 & pOH=3.6988 \\
\end{align}\]
Now, recall the formula that relates $pH$ and $pOH$
\[pH=14-pOH\]
Thus, solving for $pH$ by putting the values, we get:
\[\begin{align}
  & pH=14-3.6988 \\
 & pH=10.3012 \\
\end{align}\]
So, the $pH$ of millimolar solution of ammonium hydroxide which is dissociated $20\%$is 10.3012.
Therefore, the correct option is ‘B. 10.301’ after rounding off.

Note: Remember that the formula that calculates $pOH$ is related to the concentration of the hydroxide ions. Read the question carefully and determine whether the ions formed will be hydroxide ions or hydrogen ions and use the corresponding formula. Also, take note of what value is asked in the question $pH$ or $pOH$ and convert your answer accordingly. In this case, we may get confused and mark the answer as ‘A. 3.699’, please avoid such mistakes.