For the diprotic weak acid H2A, Ka1 = 2.0 × 10^-6 and Ka2 = 6.7 × 10^-9.
What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
Look at k1 vs k2. k2 is about 1000 times weaker than k1 so the pH is essentially determined by k1.
.........H2A ==> H^+ + HA^-
I........0.08....0......0
C........-x......x......x
E......0.08-x....x......x
Substitute the E line into Ka1 expression and solve for x = H^+ and convert to pH. You MAY need to use the quadratic.
(H2A) = 0.08-x = ?
Then k2 = (H^+)(A^2-)/(HA^-) but note from k1 above that (H^+) = (HA^-) which makes (A^2-) = k2.
To find the pH of a 0.0800 M solution of H2A, we need to consider the dissociation of the weak acid and solve for the equilibrium concentrations of H2A and A2–.
Step 1: Write the dissociation reactions for H2A.
The dissociation reactions for H2A can be represented as follows:
H2A ⇌ H+ + HA– (equation 1)
HA– ⇌ H+ + A2– (equation 2)
Step 2: Write the equilibrium expressions for the dissociation reactions.
The equilibrium expressions for equations 1 and 2 are:
Ka1 = [H+][HA–]/[H2A]
Ka2 = [H+][A2–]/[HA–]
Step 3: Use the given values of Ka1 and Ka2 to set up the equations.
Ka1 = 2.0 × 10^-6 = [H+][HA–]/[H2A]
Ka2 = 6.7 × 10^-9 = [H+][A2–]/[HA–]
Step 4: Use the initial concentration of H2A to determine the initial concentrations of H+ and HA–.
Since H2A is a diprotic acid, initially, the concentration of H2A will be equal to the total concentration provided, which is 0.0800 M. Therefore, [H2A]initial = 0.0800 M.
Step 5: Set up an ICE (Initial, Change, Equilibrium) table.
H2A ⇌ H+ + HA– (equation 1)
Initial: [H2A] = 0.0800 M, [H+] = 0 M, [HA–] = 0 M
Change: -x, +x, +x
Equilibrium: [H2A] - x, x, x
HA– ⇌ H+ + A2– (equation 2)
Initial: [HA–] = 0 M, [H+] = 0 M, [A2–] = 0 M
Change: +x, +x, +x
Equilibrium: x, x, x
Step 6: Substitute the equilibrium concentrations into the equilibrium expressions.
For equation 1:
Ka1 = (x)(x)/(0.0800 - x)
2.0 × 10^-6 = x^2/(0.0800 - x)
For equation 2:
Ka2 = (x)(x)/(x)
6.7 × 10^-9 = x^2
Step 7: Solve the equations simultaneously to obtain x, which represents the concentration of H+ in the solution.
From equation 1:
2.0 × 10^-6 = x^2/(0.0800 - x)
Rearranging the equation gives us:
x^2 = (2.0 × 10^-6)(0.0800 - x)
x^2 = 1.6 × 10^-7 - 2.0 × 10^-6x
x^2 + 2.0 × 10^-6x - 1.6 × 10^-7 = 0
Solve this quadratic equation for x, the concentration of H+.
From equation 2:
6.7 × 10^-9 = x^2
Solve this quadratic equation for x, the concentration of H+.
Step 8: Use the concentration of H+ to calculate the pH.
The pH can be calculated using the formula:
pH = -log[H+]
Step 9: Use the value of x, the concentration of H+, to calculate the equilibrium concentrations of H2A and A2–.
[H2A] = 0.0800 - x
[A2–] = x
By following these steps, you can find the pH and the equilibrium concentrations of H2A and A2– in the given solution of H2A.