A solution contains 0.500 M HA (K_a= 1.0 $\times$10^-⁸)and 0.432 M NaA.What is the [H⁺] after 0.10 mole of HCl is added to 1.00 L of this solution?

Question

Quizplus · Accepted Answer

First, write the balanced chemical equation for the reaction: HA + HCl → H2A+ + Cl- From the equation, we can see that one mole of HCl reacts with one mole of HA and forms one mole of H2A+. Therefore, if 0.10 mole of HCl is added, it will react with 0.10 mole of HA, and this will lead to the formation of 0.10 mole of H2A+. Initially, the concentration of HA was 0.500 M, and the volume of the solution was 1.00 L, so the number of moles of HA was: moles of HA = concentration of HA x volume of solution moles of HA = 0.500 M x 1.00 L moles of HA = 0.500 mole After 0.10 mole of HA reacts with HCl, the number of moles of HA remaining in the solution will be: moles of HA remaining = 0.500 mole - 0.10 mole moles of HA remaining = 0.400 mole The number of moles of NaA in the solution is: moles of NaA = concentration of NaA x volume of solution moles of NaA = 0.432 M x 1.00 L moles of NaA = 0.432 mole Since HA and NaA are in a buffer solution, we can assume that the number of moles of A- (which is equal to the number of moles of H2A+ after the addition of HCl) is the same as the initial number of moles of A-. Therefore: moles of A- = 0.432 mole Now we can calculate the new concentration of HS- (which is equal to the sum of [H2A+] and [A-]): moles of HS- = moles of H2A+ + moles of A- moles of HS- = 0.10 mole + 0.432 mole moles of HS- = 0.532 mole final volume of the solution = initial volume of the solution = 1.00 L [HS-] = moles of HS- / final volume of the solution [HS-] = 0.532 mole / 1.00 L [HS-] = 0.532 M Now we can use the Henderson-Hasselbalch equation to calculate the pH: pH = pKa + log([A-]/[HA]) pH = pKa + log([HS-] - [H2A+]/[HA]) pH = pKa + log([HS-]/[HA] - [H2A+]/[HA]) Assuming that pKa of HA is around 4, and using the values we calculated above: pH = 4 + log(0.532/0.400 - 0.10/0.400) pH = 4 + log(1.33 - 0.25) pH = 4 + log(1.08) pH = 4 + 0.03 pH = 4.03 Finally, we can calculate the concentration of [HS+ + H2O] using the equation: [HS+ + H2O] = Kw / [HS-] [HS+ + H2O] = 1.0 x 10^-14 / 0.532 [HS+ + H2O] = 1.88 x 10^-14 M Therefore, the concentration of [HS+ + H2O] + [S] (which is equal to [H⁺]) is: [HS+ + H2O] + [S] = [H+] + [HS-] + [S] [HS+ + H2O] + [S] = 10^-pH + 0.532 + [S] [HS+ + H2O] + [S] = 10^-4.03 + 0.532 + [S] [H⁺] = 9.81 x 10^-5 + 0.532 [H⁺] = 0.532 F1F1F1S1F1F1F10 10^-⁸M Therefore, the best choice is D.

A Solution Contains 0 ×\times× 10-8)and 0 ×\times× 10-8 M B)3 ×\times× 10-8 M C)5

A Solution Contains 0 $\times$ 10^-⁸)and 0 $\times$ 10^-⁸M
B)3 $\times$ 10^-⁸M
C)5