Question

A diprotic acid, H, A, has acid dissociation constants of K₁1 = 3.78 x 104 and K 4.21 x 10-11. Calculate the pH and molar

concentrations of H, A, HA, and A2 at equilibrium for each of the solutions.
A 0.177 M solution of H₂A.
pH = 2.09
[HA]-
pH=
A 0.177 M solution of NaHA.
Incorrect
0.00817
[HA-] =
pH=
6.90
A 0.177 M solution of Na, A.
0.177
13.8
Incorrect
[HA]-
4.19 x10-5
Incorrect
M
M
M
(H₂A) =
(A) 4.21 x10-11
0.169
(H₂A)- 5.9 x10-5
[A-]= 5.96 x10-5
[H₂A] -
[A-]-
1.11 x10-7
Incorrect
0.176958
Incorrect
M
M
M
M
M
M

Answer 1

To calculate the pH and molar concentrations at equilibrium for each solution, we can use the given acid dissociation constants (K₁ and K₂) and the initial concentrations of the diprotic acid (H₂A) and its conjugate bases.

1. **For a 0.177 M solution of H₂A:**
- \( \text{pH} = 2.09 \)
- [HA] = \(1.11 \times 10^{-7}\) M
- [H₂A] = \(0.176958\) M
- [A²-] = \(4.21 \times 10^{-11}\) M

2. **For a 0.177 M solution of NaHA:**
- \( \text{pH} = 6.90 \) (Incorrectly stated as "0.00817" in the provided information)
- [HA⁻] = \(5.9 \times 10^{-5}\) M
- [H₂A] = \(0.00817\) M
- [A²-] = \(4.21 \times 10^{-11}\) M

3. **For a 0.177 M solution of Na₂A:**
- [HA] = \(4.19 \times 10^{-5}\) M (Incorrectly stated as "13.8" in the provided information)
- [H₂A] = \(5.96 \times 10^{-5}\) M
- [A⁻] = \(0.177\) M
- [A²-] = \(4.21 \times 10^{-11}\) M