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

28 views

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

by (73.7k points)
selected

(I). Caculation of alpha for propionic acid
According to Ostwald Dilution Law,
alpha +((K_(a))/(C))^(1/2) = ((1.32 xx 10^(-5))/(0.05))^(1/2) =(2.64 xx 10^(-4))^(1//2) = 1.62 xx 10^(-2)
(II). Calculation of pH of the solution
[H^(+)] =(K_(a) xx C)^(1//2) =(1.32 xx 10^(-5)xx 5xx 10^(-2))^(1//2)
=(6.6 xx 10^(-7))^(1//2) = (66 xx 10^(-8))^(1//2) =8.124 xx 10^(-4)
pH =- log (8.124 xx 10^(-4)) =- (log 8.124 -4 log 10)
=(4- log 8.124) = (4-0.909) = 3.09
(III). Calculation of alpha for propionic acid in 0.01 M HCI solution.
CH_(3)CH_(2)COOH hArr CH_(3)CH_(2)COO^(-) + H^(+)
In the pressure of HCI. the ionisation of CH_(3)CH_(2)COOH will decrease. If C is the initial concentration of acid adn x is the amount dissociated at equilibrium
[CH_(3)CH_(2)COOH] =C- x , [CH_(3)CH_(2)COO^(-)] = x , [H^(+)] = 0.01 +x
K_(a) =[[CH_(3) CH_(2)COO^(-)][H^(+)]]/[[CH_(3)CH_(2)COOH]] =((x)xx(0.01 +x))/((C -X)) = (x(0.01))/(C)
" or "" " (x)/(C) =(K_(a))/(0.01) =(1.32 xx 10^(-5))/(10^5) =(1.32 xx 10^(-5))/(10^(-2)) =1.32 xx 10^(-3) , alpha =(x)/(C) =1.32 xx 10^(-3)