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.