A solution is made by mixing 50 ml of 2 M K2HPO4 and 25 ml of 2.0 M KH2PO4. The solution is diluted to a final volume of 250 ml. What is the pH of the final solution ? (PK = 6.82)
MH SET - 2019, SET Exam - 2019,
Answer: B
The pH is 7.12.
K2HPO4 is the conjugate base and KH2PO4 the acid. Moles of each:
\[ \begin{aligned} \text{base} &= 0.050 \times 2 = 0.10\ \text{mol} \\ \text{acid} &= 0.025 \times 2.0 = 0.05\ \text{mol} \end{aligned} \]
By the Henderson-Hasselbalch equation (dilution to 250 ml changes both equally, so the ratio is unchanged):
\[ \begin{aligned} \text{pH} &= 6.82 + \log \frac{0.10}{0.05} \\ &= 6.82 + 0.30 = 7.12 \end{aligned} \]
6.82 would need equal amounts of acid and base, and 6.52 comes from inverting the ratio.
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