Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0129 M solution. The pH of the resulting solution is 2.69. Calculate the Ka for the acid.

Answer 1

from the pH find the [H3O+]

[H3O+] = 10^-pH = 10^-2.69 = 0.00204

construct the ICE table

lets mono protic acid = HA

         HA (aq) + H2O (l) <-----> H3O+ (aq) + A- (l)

I         0.0129                              0                0

C      -x                                         +x             +x

E   0.0129-x                               +x               +x

from the ICE table concentration of H3O+ = x = 0.00204

concentration of H3O+ = [A-] = x = 0.00204

equilibrium concentration of HA = 0.0129 - 0.00204 = 0.01086 M

Ka = [x] [x] / [0.0129 -x]

Ka = [0.0024] [0.0024] / [0.01086]

Ka = 5.3 x 10^-4