Well:
pKa = -log Ka, and
Ka = [H+][A-] / [HA]
Therefore Ka = [H+][HCO3-] / [H2CO3]
[H+] = Ka . [H2CO3] / [HCO3-]
does this help?
Write the equation for the dissociation of carbonic acid.
4 answers
not really I am just trying to understand what the chemical equation looks like
H2CO3 + H2O <---> HCO3− + H3O+
carbonic acid is going to be 1M (mol dm-3) 20 ml
How much water in ml should I have to full dissociate 20 ml of 1M carbonic acid? To determine pKa... i was going to find the pH of the solution after it has been dissciated (carbnoic acid is a weak acid, will not fully dissociate).
from the pH... i will find the [H+] and then plug in the formula
Ka = [HCO3−] x [H3O+] / [H2CO3]
This is my last resort i don't have any thing else but done this help at all?
carbonic acid is going to be 1M (mol dm-3) 20 ml
How much water in ml should I have to full dissociate 20 ml of 1M carbonic acid? To determine pKa... i was going to find the pH of the solution after it has been dissciated (carbnoic acid is a weak acid, will not fully dissociate).
from the pH... i will find the [H+] and then plug in the formula
Ka = [HCO3−] x [H3O+] / [H2CO3]
This is my last resort i don't have any thing else but done this help at all?
The dissociation of H2CO3 occurs in two steps.
H2CO3 + H2O ==> H3O^+ + HCO3^-
HCO3^- + H2O ==> H3O^+ + CO3^2-
H2CO3 + H2O ==> H3O^+ + HCO3^-
HCO3^- + H2O ==> H3O^+ + CO3^2-
6 answers