Q2/ Calculate the degree of dissociation and pH of a 0.15 moldm-3 solution of weak acid for which Ka = 3×10-5 ?

1 Answer
May 13, 2018

Consider the acid's dissociation equilibrium in solution,

HA rightleftharpoons H^(+) + A^(-)

where,

K_"a" = ([H^+][A^-])/([HA]) approx 3*10^-5

Now, let's construct an ICE table for the same equilibrium,

HA rightleftharpoons H^(+) + A^(-)

puu.shpuu.sh

Hence,

K_"a" = x^2/(0.15 -x) = 3*10^-5

=> x = [H^+] approx 2.11*10^-3

and by extension,

"pH" = -log[H^+] approx 2.68