Answer: pH ≈ 3.186
Explanation:
[tex]p^{H} &=p^{k a}+\left[\frac{A^{-}}{H A}\right] \\3.26 &=p^{k a}+ \log \frac{0.37}{0.34} \\3.26 &=p^{k a}+\log\ 1.089 \\3.26 &=p^{k e}+0.037 \\p^{k a} &=3.26-0.1105 \\p^{k a} &=3.1495[/tex]
[tex]$0.34 \frac{\mathrm{mol}}{2} \times 0.71 \mathrm{~L}=0.2414 \ {\mathrm{mol} \rightarrow HA \\\\[/tex]
[tex]$0.37 \frac{\mathrm{mol}}{2} \times 0.71 \mathrm{~L}=0.2627 \ {\mathrm{mol} \rightarrow A^{-} \\\\[/tex]
[tex]$0.2414-0.032 \mathrm{mol}=0.2094 \mathrm{\ mol} \ HA[/tex]
[tex]$0.2627+0.032 \mathrm{mol}=0.2947 \mathrm{\ mol} \ A^{-}[/tex]
[tex]\begin{aligned}&p H=3.1495+\log\frac{0.2627}{0.2414} \\&p H=3.1495+0.0367 \\&p H=3.186\end{aligned}[/tex]
Therefore, the pH after 0.032 mol of HCl are added to 0.71 L of the solution is approximately 3.186
