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!!

With Ka1 and Ka2 as close as they are, there is some interaction between the two but I will assume you can take the easy way out, assume there is no interaction, and solve the problem this way. The REAL answer won't be that far away (perhaps 3-5%).

..........H2A ==> H^+ + HA^-
I......0.0650.....0......0
C.........-x......x......x
E....0.0-650-x....x......x

Ka1 = (H^+)(HA^-)/(H2A)
Substitute the E line and solve for x = H^+), then convert to pH. You can then evaluate 0.0650-x = (H2A).

For A^2- I would do this
.........HA^- ==> H^+ + A^2-

Ka2 = (H^+)(A^2-)/(HA^-)
(H^+) = (HA^-); therefore,
(A^2-) = Ka2.
Remember you are making the assumption here that Ka1 and Ka2 act independently.

To find the pH of a solution of a diprotic weak acid H2A and the equilibrium concentrations of H2A and A2- in the solution, we can use the principles of equilibrium and the dissociation constant (Ka) values provided.

Step 1: Write the chemical equation for the dissociation of H2A:
H2A ⇌ H+ + HA- (First dissociation, Ka1)
HA- ⇌ H+ + A2- (Second dissociation, Ka2)

Step 2: Set up an ICE table (Initial, Change, Equilibrium) for each dissociation reaction. We will assume that the initial concentration of H2A is 0.0650 M.

For the first dissociation, H2A ⇌ H+ + HA-:
Initial: 0.0650 M 0 M 0 M
Change: -x +x +x
Equilibrium: 0.0650 - x x x

For the second dissociation, HA- ⇌ H+ + A2-:
Initial: x 0 M 0 M
Change: -x +x +x
Equilibrium: 0.0650 - x x x

Step 3: Write the expression for Ka1 using the equilibrium concentrations:
Ka1 = [H+][HA-] / [H2A] = (x)(x) / (0.0650 - x)

Since Ka1 = 2.2 × 10^(-5) M, we can solve the quadratic equation:
2.2 × 10^(-5) = (x)(x) / (0.0650 - x)

Step 4: Solve for x using the quadratic equation. Rearrange the equation to get:
(x)(x) = 2.2 × 10^(-5) (0.0650 - x)

Simplifying the equation will give us:
x^2 + 2.2 × 10^(-5)x - 2.2 × 10^(-5) × 0.0650 = 0

We can solve this quadratic equation using the quadratic formula or by factoring. The positive value of x obtained will give the concentration of H+ ions, and the equilibrium concentrations of H2A and A2- can be calculated using the ICE table.

After calculating x and substituting the values in the equilibrium expressions, we can determine the concentrations of H2A and A2-.

Finally, to find the pH of the solution, use the formula:
pH = -log[H+]

Remember to double-check and verify all calculations.

Note: It is recommended to use a calculator or software that can handle complex calculations involving equilibrium constants and quadratic equations to get accurate results.