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?

Question

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!!

Answer 1

This is too much.

Answer 2

To find the pH of a solution of the diprotic weak acid H2A with a concentration of 0.0650 M, we need to consider the dissociation of the acid and the equilibrium concentrations of its species.

Step 1: Write the dissociation equations for H2A:
H2A ⇌ H+ + HA- (Dissociation 1)
HA- ⇌ H+ + A2- (Dissociation 2)

Step 2: Find the equilibrium concentrations of each species:
Let x be the concentration of H+ ions formed and the concentration of HA- formed after dissociation.

For Dissociation 1:
[H2A] = 0.0650 M - x
[H+] = x
[HA-] = x

For Dissociation 2:
[HA-] = x
[H+] = x
[A2-] = x

Step 3: Write the equilibrium expressions for each dissociation:
For Dissociation 1:
Ka1 = [H+][HA-] / [H2A]

For Dissociation 2:
Ka2 = [H+][A2-] / [HA-]

Step 4: Set up the equilibrium expressions using the given Ka values:
Ka1 = 2.2 × 10^(-5) M
Ka2 = 7.8 × 10^(-7) M

Step 5: Solve the equilibrium expressions simultaneously to find x, which represents the concentration of H+ ions in the solution.

Using the equilibrium expression for Ka1:
(2.2 × 10^(-5)) = (x)(x) / (0.0650 - x)

Using the equilibrium expression for Ka2:
(7.8 × 10^(-7)) = (x)(x) / x

Step 6: Solve the quadratic equation:
Solving the equations will give us the value of x, which represents the concentration of H+ ions:

For Ka1:
(2.2 × 10^(-5)) = (x)(x) / (0.0650 - x)
Cross-multiply:
2.2 × 10^(-5) × (0.0650 - x) = (x)^2
Simplify:
0.000143 - 2.2 × 10^(-5) × x = x^2

For Ka2:
(7.8 × 10^(-7)) = (x)(x) / x
Cross-multiply:
7.8 × 10^(-7) × x = (x)^2
Simplify:
0.00000078 = x

Step 7: Calculate the pH of the solution:
The pH of a solution is defined as the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log[H+]

Substitute the value of x into the equation to calculate the pH.

Step 8: Calculate the equilibrium concentrations of H2A and A2-:
[H2A] = 0.0650 M - x
[A2-] = x

Substitute the value of x into the corresponding equations to calculate the equilibrium concentrations of H2A and A2-.

By following these steps, you should be able to calculate the pH of the solution and the equilibrium concentrations of H2A and A2-.