Question

One litre of solution contains 1M $HOCl$ [$Ka = {10^{ - 8}}$] and 1M $NaOH$. What is the pH of the solution?

Answer 1

Hint- To calculate the pH of any solution, this formula is used i.e. $- log{\text{ }}\left[ {{H^ + }} \right]$. So the very first thing we need is the concentration of ${H^ + }$ ions in the solution. And also remember that when acid and base are mixed together in a solution, then there is always a formation of ${H_2}O$ according to the availability of ions present.

Complete step by step solution
For a generalised chemical reaction taking place in a solution:
$aA + bB \rightleftharpoons cC + dD\;$
The equilibrium constant can be expressed as follows:
$K = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
where [A], [B], [C] and [D] refer to the molar concentration of species A, B, C, D respectively at equilibrium. The coefficients like a, b, c, and d in the generalised chemical equation become exponents as seen in the above expression.
$HOCl$ is a weak acid which dissociate partially and its dissociation constant is given in question (i.e. $Ka = {10^{ - 8}}$). $HOCl$ dissociates according to the following chemical equation:
$HOCl \to \;\;{H^ + } + {\text{ }}OC{l^ - }$
$1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0\; \;\;\;\;\;\;\;\; 0\;\;\;\;\;$ initial concentration on chemical species
$(1-\alpha) \;\;\;\;\;\;\; \alpha \;\;\;\;\;\;\;\;\; \alpha \;\;\;\;\;$ final concentration of chemical species
Equilibrium constant for this reaction is $K = {10^{ - 8}} = \dfrac{{{\alpha ^2}}}{{1 - \alpha }}$
α is the degree of dissociation of this acid and it is a very small quantity compared to 1 so we neglect it from the denominator.
So now ${\alpha ^2} = {10^{ - 8}}$from above equation. And now, we have to calculate α by taking the square root of both sides. Thus, $\alpha {\text{ }} = {\text{ }}{10^{ - 4}}$
This means that the total concentration of ${H^ + }$ ions in this solution is also${10^{ - 4}}M$.
Now we can calculate the number of $O{H^ - }$ from the basic solution. As we know that $NaOH$ is a strong base so it dissociates completely and provides equal concentration of $O{H^ - }$ ions from its solutions. So there is 1 mole of OH- in the solution.
$NaOH\xrightarrow{{complete{\text{ }}dissociation}}N{a^ + } + O{H^ - }$

Now calculate the number of total $O{H^ - }$ ions after neutralization in solution mixture and it is equal to $\left[ {O{H^ - }} \right] - \left[ {{H^ + }} \right]$.
As we know the values of both, substitute these values to get the total number of $O{H^ - }$ions.
$1M{\text{ }} - {\text{ }}{10^{ - 4}}M{\text{ }} = {\text{ }}0.9999M{\text{ }} = {\text{ }}\left[ {O{H^ - }} \right]$
Now the volume of solution is becoming double so that means the concentration will get half.
So now, $\left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}0.49995M$
Finally we will find out the pH of solution.
First we will calculate $pOH$ and then convert it to $pH$ by using this given relation.
$pH{\text{ }} + {\text{ }}pOH{\text{ }} = {\text{ }}14$
The formula of $pOH$using concentration of $O{H^ - }$ions is written below:.

pOH{\text{ }} = {\text{ }} - log{\text{ }}\left[ {O{H^ - }} \right] \\\ \begin{array}{*{20}{l}} {pOH{\text{ }} = {\text{ }} - log{\text{ }}\left[ {0.49995} \right]} \\\ {pOH{\text{ }} = {\text{ }}0.30107} \end{array} \\\

As we know that:
$pH = {\text{ }}14{\text{ }}-{\text{ }}pOH$
Substituting the values of $pOH$, get the value of $pH$
$pH = 14-0.30107 = 13.698$

**Hence, the pH of the resulting solution is 13.698.

Note: **
The pH scale ranges from 0 to 14, and most solutions fall within this range. Any solution lying below 7 is considered to be acidic while above 7 is alkaline. If the solution has a pH of 7 then it is considered to be neutral.