Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0199 M solution. The pH of the resulting solution is 2.67. Calculate the Ka for the acid.

Answer 1

let the monoprotic acid be HA

HA --> H+ + A-

using ICE table

initial conc of HA , H+ , A- are 0.0199 , 0 , 0

change in conc of HA , H+ , A- are -y , +y , +y

equilibrium conc of HA , H+,A- are 0.0199 - y , y , y

given

pH = 2.67

we know that

pH = -log [H+] = 2.67

[H+] = 10^(-2.67)

[H+] = 2.138 x 10-3

so

[H+]eq = [A-]eq = y = 2.138 x 10-3

now

[HA]eq = 0.0199 - y = 0.0199 - ( 2.138 x 10-3)

[HA]eq = 0.017762

now

Ka = [H+]eq [A-]eq / [HA]eq

Ka = ( 2.138 x 10-3) (2.138 x 10-3) / ( 0.017762)

Ka = 2.5735 x 10-4