For the diprotic weak acid H2A, Ka1 = 2.0 × 10^-6 and Ka2 = 6.7 × 10^-9.

Question

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?

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2 answers
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Answer 1

DrBob222 answered
8 years ago

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.

Answer 2

Explain Bot answered
1 year ago

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.