For the diprotic weak acid H2A, Ka1 = 2.2 × 10-5 M and Ka2 = 7.8 × 10-7 M. What is the pH of a 0.0650 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Thank you for the help!!

To find the pH of a 0.0650 M solution of H2A, we can use the dissociation constants (Ka1 and Ka2) to determine the concentrations of H2A, H+, and A2- at equilibrium.

Step 1: Set up the dissociation reactions for the acid H2A:

H2A ⇌ H+ + HA-
HA- ⇌ H+ + A2-

Step 2: Write the equilibrium expressions for Ka1 and Ka2:

Ka1 = [H+][HA-]/[H2A]
Ka2 = [H+][A2-]/[HA-]

Step 3: Define the variables:

Let [H2A] = X
Let [HA-] = Y
Let [A2-] = Z
Let [H+] = H

Step 4: Set up the initial and equilibrium concentrations table:

H2A ⇌ H+ + HA- Initial: X 0 0 Change: -H +H +H Equilibrium: X - H H H

HA- ⇌ H+ + A2- Initial: 0 Y 0 Change: +H -H +H Equilibrium: H Y - H H

Step 5: Use the equilibrium concentrations to rewrite the equilibrium expressions:

Ka1 = (H)(Y - H)/(X - H)
Ka2 = (H)(H)/(Y - H)

Step 6: Substitute the known values into the equations:

(2.2 × 10^-5) = (H)(Y - H)/(0.0650 - H)
(7.8 × 10^-7) = (H)(H)/(Y - H)

Step 7: Solve the equations to find H:

1. Solve the first equation for Y - H:

(0.0650 - H)(2.2 × 10^-5) = (H)(Y - H) --> 0.00143 - 2.2 × 10^-5H = HY - H^2

Simplifying: HY - H^2 = 0.00143 - 2.2 × 10^-5H

2. Solve the second equation for Y:

(7.8 × 10^-7) = (H)(H)/Y

Simplifying: Y = (H)(H)/(7.8 × 10^-7)

3. Substitute Y from the second equation into the first equation:

(H^2)/(7.8 × 10^-7) - H^2 = 0.00143 - 2.2 × 10^-5H

Simplifying: H^2 - (H^2)/(7.8 × 10^-7) = 0.00143 - 2.2 × 10^-5H

Cross-multiplying and rearranging: (7.8 × 10^-7)(0.00143 - 2.2 × 10^-5H) = H^2

Expanding and simplifying: 0.0011114 - 1.716 × 10^-8H = H^2

Rearranging: H^2 + 1.716 × 10^-8H - 0.0011114 = 0

Solving the quadratic equation using the quadratic formula, we find that H ≈ 0.0301 M.

Step 8: Calculate the equilibrium concentrations of H2A and A2-:

[H2A] = X - H ≈ 0.0650 M - 0.0301 M ≈ 0.0349 M
[A2-] = H ≈ 0.0301 M

Step 9: Calculate the pH of the solution:

pH = -log[H+]
= -log(0.0301)
≈ 1.52

Therefore, the pH of the 0.0650 M solution of H2A is approximately 1.52, and the equilibrium concentrations of H2A and A2- are approximately 0.0349 M and 0.0301 M, respectively.

Still to much.