This question is about acidic solutions.
The acid dissociation constant, KaK_{\text{a}}Ka, for methanoic acid is given by the expression
Ka=[HCOO−][H+][HCOOH] K_{\text{a}} = \frac{[\text{HCOO}^-][\text{H}^+]}{[\text{HCOOH}]} Ka=[HCOOH][HCOO−][H+]The value of KaK_{\text{a}}Ka, for methanoic acid is 1.78×10−4 mol dm−31.78 \times 10^{-4} \text{ mol dm}^{-3}1.78×10−4 mol dm−3 at 25 ∘C25\ ^\circ\text{C}25 ∘C.
A buffer solution with a pH of 3.603.603.60 was prepared using methanoic acid and sodium methanoate. In the buffer solution, the concentration of methanoate ions was 0.150 mol dm−30.150 \text{ mol dm}^{-3}0.150 mol dm−3.
Calculate the concentration of the methanoic acid in the buffer solution. Give your answer to three significant figures.
In a different buffer solution, the concentration of methanoic acid was 0.420 mol dm−30.420 \text{ mol dm}^{-3}0.420 mol dm−3 and the concentration of methanoate ions was 0.310 mol dm−30.310 \text{ mol dm}^{-3}0.310 mol dm−3.
A 1.20×10−2 mol1.20 \times 10^{-2} \text{ mol}1.20×10−2 mol sample of potassium hydroxide was added to 600 cm3600 \text{ cm}^3600 cm3 of this buffer solution.
Calculate the pH of the buffer solution after the potassium hydroxide was added. Give your answer to two decimal places.
