The ionization constant of propionic 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 in the solution of `0.01N HCI` ?

Explanation

(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)`