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?
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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.