Calculate the pH of a 1.7 M solution of H₂A (K_a1= 1.0 × 10^-6 and K_a2 is 1.0 × 10^-10). A) 10.00 B) 2.88 C) 11.12 D) 5.77 E) 7.00

Question

Quizplus · Accepted Answer

H₂A can be treated as a weak acid with two ionizable hydrogen atoms. The dissociation reactions are as follows: H₂A ⇌ H+ + _a1- pKa1 = 6.0 _a1- ⇌ H+ + _a2- pKa2 = 10.0 Since H₂A is a weak acid, we can use the equation for weak acids: pH = pKa + log([A-]/[HA]) where [A-] and [HA] are the concentrations of the conjugate base and acid, respectively. At equilibrium, we can assume that x moles of H₂A has dissociated to form x moles of _a1- and x moles of H+. Then, the concentrations can be expressed as: [H₂A] = 1.7 M - x [_a1-] = x [H+] = x Using the first dissociation constant, we get: Ka1 = 10^-pKa1 = [H+][_a1-]/[H₂A] 10^-6.0 = x^2 / (1.7 - x) Since x is very small compared to 1.7, we can approximate (1.7 - x) as 1.7: 10^-6.0 ≈ x^2 / 1.7 x ≈ 5.5 x 10^-4 Now we can use the second dissociation constant to calculate the pH: Ka2 = 10^-pKa2 = [H+][_a2-]/[_a1-] 10^-10.0 = x^2 / (1.7 - x) Since x is very small compared to 1.7, we can approximate (1.7 - x) as 1.7: 10^-10.0 ≈ x^2 / 1.7 x ≈ 7.9 x 10^-6 Now we can calculate the pH: pH = pKa1 + log([A-]/[HA]) pH = 6.0 + log(7.9 x 10^-6 / 5.5 x 10^-4) pH ≈ 2.88 Therefore, the answer is B.

Calculate the PH of a 1