In: Chemistry
What is [H3O+] and pH in a 0.10 M solution of HCN at 25
We have , 0.10M HCN solution at 25oc & Ka for HCN = 4.0 * 10-10
HCN + H2O -----------> H3O+ + CN-
Intial concentration 0.1 M 0 M 0 M
Change in concentration -x +X +X
Equilibrium concentration (0.1-x) X X
Ka = [H3O+][CN-] / [HCN]
4.0 * 10-10 = (X)(X) / (0.1-X)
4.0 * 10-10 = (X)(X) / (0.1)
X2 = (0.1) * (4.0 * 10-10)
[H3O+] = X = 6.3 * 10-6 M
[H3O+] = 6.3 * 10-6 M
pH = -log[H3O+]
= -log (6.3 * 10-6)
pH = 5.2