Question #6521
if a weak acid, HA, is 3% dissociated in a 0.25M solution, calculate the Ka and the pH of the solution.
Expert's answer
a=3% (0.03)
C=0.25M
Ka-? pH-?
HA <=>H+ +
A-
a=(K/C) ^ (1/2)
a^ 2= K/C
K=(a^ 2)/C
K= (0.03^ 2)
/0.25=0.0036
[H+]=(K*C) ^ (1/2)=0.03
pH=-log[H+]=1.52
K=0.0036
pH=1.52
C=0.25M
Ka-? pH-?
HA <=>H+ +
A-
a=(K/C) ^ (1/2)
a^ 2= K/C
K=(a^ 2)/C
K= (0.03^ 2)
/0.25=0.0036
[H+]=(K*C) ^ (1/2)=0.03
pH=-log[H+]=1.52
K=0.0036
pH=1.52
