Question

20 mL of 0.2 M $NaOH$ is added to 50 mL of 0.2 M acetic acid to obtain the solution. Calculate the pH of the solution. (${K_a} = 1.8 \times {10^{ - 5}}$)

Answer 1

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 answer:
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.
In the question, we are given that $NaOH$reacts with acetic acid. The chemical reaction can be written as follows:
$NaOH + C{H_3}COOH \to C{H_3}COONa + {H_2}O$
We know that:
$Number{\text{ }}of{\text{ }}millimoles = Molarity\; \times \;Volume{\text{ }}\left( {mL} \right)$

$${Millimoles{\text{ }}of\;NaOH = 0.2 \times 20 = 4} \\\ {Millimoles{\text{ }}of\;C{H_3}COOH = 0.2 \times 50 = 10} \\\ {Millimoles{\text{ }}of\;C{H_3}COONa\;produced\; = 4} \\\ {\therefore \;Millimoles{\text{ }}of\;C{H_3}COOH\;remained\; = 10 - 4 = 6}$$

Now, pH can be expressed as:
$pH = p{K_a} + \log \dfrac{{millimoles{\text{ }}of{\text{ }}salt}}{{millimoles{\text{ }}of{\text{ }}acid}}$
And we know that $p{K_a} = - log{K_a}$
${K_a} = 1.8 \times {10^{ - 5}}$(Given)
Substituting the values in this formula, we will get the value of pH:
$\Rightarrow pH = - \log (1.8 \times {10^{ - 5}}) + \log \dfrac{4}{6} = 4.56$

**Hence, the pH of the solution is 4.56.

Note:**
The pH scale ranges from 0 to 14, and most solutions fall within this range. Any solution lying below 7.0 is considered to be acidic while above 7.0 is alkaline. If the solution has a pH of 7 then it is considered to be neutral. This means that the solution in the question having pH of 4.56 is acidic.