Calculate the pH of a 0. 080 mol L^-1 carbonic acid solution, H₂CO₃₍_aq_),that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11. A) 1.10 B) 3.73 C) 6.37 D) 10.25

Question

Quizplus · Accepted Answer

The first step is to write out the equilibrium reactions and dissociation constants for carbonic acid:
H2CO3 ⇌ H+ + HCO3-          Ka1 = 4.3 × 10^-7
HCO3- ⇌ H+ + CO32-          Ka2 = 5.6 × 10^-11

We can set up a table to keep track of the concentrations of each species as we move through the equilibrium reactions:

|                     | H2CO3    | HCO3-    | CO32-    |
|---------------------|---------|----------|----------|
| Initial concentration| 0.080 M |   0      |    0     |
| Change              |  -x     |  +x     |   +x     |
| Equilibrium         | 0.080-x |   x      |    x     |

Using the equilibrium concentrations, we can write out the expression for the first equilibrium reaction, Ka1:
Ka1 = [H+][HCO3-]/[H2CO3]
4.3 × 10^-7 = x^2 / (0.080 - x)

Because x is small relative to 0.080, we can simplify the expression as follows:
4.3 × 10^-7 = x^2 / 0.080

Solving for x, we find that x = 1.140 × 10^-4 M, which represents the concentration of H+ and HCO3- after the first equilibrium step.

Using this result, we can now write out the second equilibrium reaction and the expression for Ka2:
Ka2 = [H+][CO32-]/[HCO3-]
5.6 × 10^-11 = x^2 / (0.080 - 1.140 × 10^-4)

Again, because x is small relative to 0.080, we can simplify the expression as follows:
5.6 × 10^-11 = x^2 / 0.080

Solving for x, we find that x = 7.496 × 10^-7 M, which represents the concentration of H+, CO32-, and the remaining HCO3- after the second equilibrium step.

Finally, we can calculate the pH of the solution as follows:
pH = -log[H+]
pH = -log(1.140 × 10^-4)
pH = 3.73

Therefore, the answer is B.

Calculate the PH of a 0