The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 1.32 × 10–5 solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Given:

Ka = 1.32 × 10–5


C = 0.005 M


Assuming α to be very small, applying formula directly, we have


α = √


⇒α = √


⇒α = 1.62 × 10-2


Thus, the degree of ionization in 0.05M solution is 1.62 × 10-2


Let the degree of ionization of propanoic acid be x


Ionization of propanoic acid



By applying the formula


Ka =


⇒Ka =


⇒Ka=


As Ka = 1.32 × 10–5


, C = 0.1 M (given)


Hence,


x= 1.32 × 10-3


Thus, the degree of ionization in 0.01M solution is 1.32 × 10-3


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