Calculate the pH of a 0.080 mol L^-1 aqueous carbonic acid solution,H₂CO_3(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

To calculate the pH of the solution, we need to first write out the dissociation reactions for carbonic acid:
HS1U1B12S1U1B0COS1U1B13(aq) + H2O(l) ⇌ H3O+(aq) + S1U1B1CO3S1U1B0S1U1B13-(aq) 
S1U1B1CO3S1U1B0S1U1B13-(aq) + H2O(l) ⇌ HCO3-(aq) + OH-(aq) 
Using the given dissociation constants, we can calculate the equilibrium concentrations of each species in the solution, and then use the equilibrium concentration of H3O+ to calculate the pH.
Let x be the concentration of H3O+. Then, at equilibrium:
[HS1U1B12S1U1B0COS1U1B13] = 0.080 mol/L
[HCO3-] = (4.3 x 10^-7) mol/L
[CO3^2-] = (5.6 x 10^-11) mol/L
[OH-] = x
We can then use the equilibrium constant expression for each dissociation reaction to set up two equations and solve for x:
Ka1 = (H3O+)(S1U1B1CO3S1U1B0S1U1B13-)/(HS1U1B12S1U1B0COS1U1B13) 
Ka2 = (HCO3-)(OH-)/(S1U1B1CO3S1U1B0S1U1B13-)
Plugging in the known values and solving for x, we get:
x = 1.18 x 10^-4 M
Finally, we can calculate the pH of the solution:
pH = -log[H3O+] = 3.73
Therefore, the best choice is B.

Calculate the PH of a 0