The ionization constant of propanoic acid is 1.32xx10^(-5). Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH

27 views

closed
The ionization constant of propanoic acid is 1.32xx10^(-5). Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M on HCl also?

by (65.9k points)
selected

i. Ionization constant of propionic acid =1.32xx10^(-5)
Let the degree of ionisation is alpha
alpha=sqrt((K_(a)/C))=sqrt((1.32xx10^(-5))/0.05)=1.63xx10^(-2)
[H^(o+)]=Calpha
=0.05xx1.63xx10^(-2)=0.0815xx10^(-2)=815xx10^(-6)
pH=-log (815xx10^(-6))=-(log 815+ log 10^(-6))
=-(2.9112-6)=3.088
pH=3.09
ii. In presence of 0.01 M HCl
HCl is fully dissociates
{:(HCl,rarr,H^(o+),+,Cl^(Θ)),(0.01,,0,,0),(0,,0.01,,0.01):}
Total [H^(o+)]=[H^(o+)] from HCl+[H^(o+)] from fropinoic acid
CH_(3)CH_(2)COOH rarr CH_(3)CH_(2)COO^(Θ)+H^(o+)
{:(1,,0,,0),(C(1-alpha),,Calpha,,Calpha):}
But [H^(o+)]=(0.01+Calpha)
K_(a)=(Calphaxx(0.01+Calpha))/(C(1-alpha))
K_(a)=0.01alpha+Calpha^(2)(1-alpha~~1)=0.01 alpha
(Calpha^(2) "is neglected as alpha is very small")
alpha=K_(a)/0.01=(1.32xx10^(-5))/0.01=1.32xx10^(-3)
:. In presence of 0.01 M HCl,
alpha=1.32xx10^(-3)
Note: It can be seen that the degree of ionization decreases almost by a factor of 10.
Ionisation of Propionic acid =1.63xx10^(-2)
Ionisation in presence of 0.01 M HCl=1.32xx10^(-3)
Decreases almost by factor of 10`