Asked by Lily
A 0.942 M sample of carbonic acid, H2CO3, has a measured hydronium ion concentration of 6.36 ´ 10–4 M. Calculate the acid-ionization constant of carbonic acid. The equilibrium equation is:
H2CO3(aq) + H2O(l) <--><--> H3O+(aq) + HCO-3(aq)
H2CO3(aq) + H2O(l) <--><--> H3O+(aq) + HCO-3(aq)
Answers
Answered by
DrBob222
Ka = (H3O^+)(HCO3^-)(H2CO3)
Set up an ICE chart, substitute into Ka expression, and solve for Ka.
Post your work if you get stuck.
Set up an ICE chart, substitute into Ka expression, and solve for Ka.
Post your work if you get stuck.
Answered by
Lily
what is an ICE chart?
Answered by
DrBob222
ICE.
I = initial
C = change
E = equilibrium.
H2CO3 + H2O ==> H3O^+ + HCO3^-
Ka = (H3O^+)(HCO3^-)/(H2CO3).
initial:
H2CO3 = 0.942 M
(H3O^+) = 0
(HCO3^-) = 0
equilibrium:
(H3O^+) = 6.36 x 10^-4 M
(HCO3^-) = 6.36 x 10^-4 M
(H2CO3) = 0.942-6.36 x 10^-4 = 0.942
This solution assumes Ka is the only ionization constant for H2CO3; in reality, H2CO3 has a k1 and a k2 but we are ignoring k2. The error is small.
I = initial
C = change
E = equilibrium.
H2CO3 + H2O ==> H3O^+ + HCO3^-
Ka = (H3O^+)(HCO3^-)/(H2CO3).
initial:
H2CO3 = 0.942 M
(H3O^+) = 0
(HCO3^-) = 0
equilibrium:
(H3O^+) = 6.36 x 10^-4 M
(HCO3^-) = 6.36 x 10^-4 M
(H2CO3) = 0.942-6.36 x 10^-4 = 0.942
This solution assumes Ka is the only ionization constant for H2CO3; in reality, H2CO3 has a k1 and a k2 but we are ignoring k2. The error is small.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.