Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0175 M solution. The pH of the resulting solution is 2.35. Calculate the Ka for the acid.

Answer 1

Solution :-

pH = 2.35

lets calculate the [H3O+] using the pH

[H3O+] = antilog [-pH]

            = antilog [-2.35]

            = 0.00447 M

HA + H2O ----- > H3O+ + A-

0.0175 M            0              0

-x                         +x            +x

0.0175-x           0.00447      0.00447

0.0175 - 0.00447 = 0.01303

ka= [H3O+] [A-] / [HA]

ka = [0.00447][0.00447]/[0.01303]

Ka= 1.53*10^-3

so the Ka of the acid is 1.53*10^-3