The answer in the book is 4.1 * 10^..
But now I think their answer is wrong.
Anyway, I'll ask my teacher next week.
Thank you for all the help. :approve:
Thanks for the sites.
What about the equilibrium between HCrO4(-) and CrO4(2-)?
Sorry if these questions are too basic. I'm just having trouble understanding.
:P
Wait...
Does [HCrO4-]=[KHCrO4]=0.0025mol/L ?
Ka = [H+] [CrO4 2-] / [HCrO4-]
= [3.16*10^(-4)]^2 / [0.0025 - 3.16*10^(-4)]
= 4.57*10^(-5) mol/L
Well that's close to the answer in the textbook, but I don't think it's right.
Ok. This is what I know.
1. The concentration of [H+]= 3.16*10^(-4) mol/L
2. KHCrO4 -> H(+) + KCrO4(-)
KHCrO4 -> H(+) + K(+) + CrO4(2-)
3. Add them?
2KHCrO4 -> 2H(+) + K(+) + Cro4(2-) + KCrO4(-)
Is this right so far?
4. Ka
= [H+]^2 * [K(+)] * [CrO4(2-)] * [KCrO4(-)] / [KHCrO4]^2...
A 0.0025 mol/L solution of KHCrO4 has a pH of 3.50.
Calculate the acid dissociation constant (Ka) for the equilibrium between HCrO4(-) and CrO4(2-).
Thanks!:smile: