Question

Calculate the concentrations for each species present in a 0.15 M aqueous solution of monobasic organic acid with Ka = 2.5 ´ 10-5. Solve using method of the quadratic equation.

Answer 1

Ans-- let us consider a monobasic acid HA

In aqueous solution monobasic acid dissociates as follows

HA(l) H⁺(aq) + A^-(aq)

initiial concentration 0.15 0 0

at equilibrium 0.15-x x x

we know, ka=[H⁺][A^-]/[HA]

   2.5*10^-5=X² / (0.15-X) Since the value of ka 2.510^-5 is very small so x at the denominator can be ignored

2.5*10^-5=X² / 0.15

x2=0.375*10^-5

x= 0.0019

value of concentration can not be negative , so x= 0.0019

at equilibrium , [HA]=0.15-X=0.15-0.0019=0.1481M

[H+]=0.0019M

[A-]=0.0019M