Question

A buffer solution is prepared by mixing 1 mole of HA and 1 mole of NaA into 1 L distilled water. Calculate the change in pH when 25 mL of 0.20 M NaOH is added into 500mL of the buffer?

Answer 1

Note that we are not given the value of $p{K_a}$ hence we’ll solve this question without using the $p{K_a}$ . The pH of a solution containing a weak acid (HA) and its conjugate base (NaA) can be given by the Henderson-Hasselbalch equation. This equation is given as:
$pH = p{K_a} + \log \left( {\dfrac{{[{\text{conjugate base]}}}}{{[{\text{weak acid]}}}}} \right)$

Complete Step By Step Answer:
The buffer solution given to us contains a weak acid (HA) and its conjugate base, ${A^ - }$ which come in the solution from the salt NaA. The value of $p{K_a}$ is given as $p{K_a} = - \log ({K_a})$ . ${K_a}$ is the dissociation constant.
The buffer solution is prepared by 1 mol of weak acid and 1 mol of conjugate base. The dissociation of the salt NaA will be in 1:1 ratio. So the no. of moles of salt will be equal to the no. of moles of conjugate base ${A^ - }$ in the solution. The molarity or the concentration of the solution can be given as: $c = \dfrac{n}{V}$
$[HA] = [{A^ - }] = \dfrac{{1\;mol}}{{1L}} = 1M$
The volume of the stock solution taken is 500mL. The no. of moles of HA and ${A^ - }$ will be equal to: $n = c.V$
${n_{HA}} = {n_{{A^ - }}} = 1M \times 500 \times {10^{ - 3}}L = 0.50mol$
Initially when the buffer solution isn’t added in the solution, the pH of the buffer will be equal to the $p{K_a}$ of the acid. (since the concentration of acid and conjugate base is equal). Substituting in the equation we have:
$p{H_1} = p{K_a} + \log \left( {\dfrac{{1M}}{{1M}}} \right) = p{K_a}$
We are adding 25mL of the NaOH solution. The hydroxide ions will react in a 1:1 ratio with the weak acid (neutralization reaction will occur) and a conjugate base will be produced. $H{A_{(aq)}} + O{H^ - }_{(aq)} \to A_{(aq)}^ - + {H_2}{O_{(l)}}$
The no. of moles of hydroxide ions added to the 500mL of the buffer solution will be equal to: ${n_{O{H^ - }}} = 0.20M \times 25 \times {10^{ - 3}}L = 0.0050mol$ of $O{H^ - }$ .
After neutralization reaction, the no. of $O{H^ - }$ left in the solution will be $= 0mol$ (since all the $O{H^ - }$ ions will be utilized)
The no. of moles of HA left after addition of 25ML of NaOH solution $= {n_{HA}} = 0.50mol - 0.0050mol = 0.495mol$
${n_{{A^ - }}} = 0.50mol + 0.0050mol = 0.505mol$
The total volume of the buffer solution after addition of NaOH $= {V_{total}} = 500mL + 25mL = 525mL$
The new concentrations of weak acid and the conjugate base will be equal to:
$[HA] = \dfrac{{0.495mol}}{{525 \times {{10}^{ - 3}}L}} = 0.9429M$
$[{A^ - }] = \dfrac{{0.505mol}}{{525 \times {{10}^{ - 3}}L}} = 0.9619M$
The pH of the solution according to the Henderson-Hasselbalch equation will be: $p{H_2} = p{K_a} + \log \left( {\dfrac{{0.9619M}}{{0.9429M}}} \right) = p{K_a} + 0.00866$
The change in the pH will be the difference between the two pH’s which comes as:
$\Delta pH = p{H_2} - p{H_1}$
$\Delta pH = p{K_a} + 0.00866 - p{K_a} = 0.00866$
This the required change in pH observed.

Note:
If the value of $p{K_a}$ is given to us then the proceeding of the problem would be the same, but the value of $p{K_a}$ will be substituted in the formula of pH and the change in pH is found according. The given buffer is an acidic buffer, since it is a solution of weak acid and conjugate base (salt).