Given a diprotic acid, H2A, with two ionization constants of Ka1 = 1.63× 10–4 an

Question

Given a diprotic acid, H2A, with two ionization constants of Ka1 = 1.63× 10–4 and Ka2 = 2.93× 10–12, calculate the pH and molar concentrations of H2A, HA–, and A2– for each of the solutions below.

Given a diprotic acid, H2A, with two ionization constants of Kat-1.63x 10-4 and Ka2 = 2.93x 10-2 calculate the pH and molar concentrations of H2A, HAT, and Afor each of the solutions below (a) a 0.195 M solution of H2A [H,A] pH HA Number Number Number (b) a 0.195 M solution of NaHA [HA] pH HA Number Number Number (c) a 0.195 M solution of Na2A [H,A HA- Number Number Number Number

Explanation / Answer

The reactions are

H2A HA- + H+;................ Ka1 = 1.63*10^-4
HA A(^2-) + H+;................Ka2 = 2.93*10^-12

Because Ka2 is very low, very little HA- will dissociate, so that reaction is ignored and the first reaction treated as monoprotic

If x moles of H2A dissociate, then the remaining concentration of H2A is
[H2A] = 0.195 - x
and the concentration of the products is
[HA-] = x
[H+] = x

1.63*10^-4 = [HA-]*[H+]/[H2A] = x²/(0.195 - x)

Ka1 is small, only a small amount of H2A dissociates so x << 0.195 and we can use the approximation 0.195 - x 0.195; then x² = [1.63*10^-4*0.195] = 0.00563

[H+] = x = 0.00563
pH = 2.25

[HA-] also = x
[HA-] = 0.00563 M
[H2A] = 0.195 - x = 0.195 - 0.00563 M = 0.1894 M

Now consider the second reaction

Ka2 = [H+]*[A(2^-)]/[HA-]

from above we know the [H+] = [HA-] so Ka2 = [A-]

[A(^2-)] = 2.93*10^-12 M

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.