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?

Explanation

Assuming `alpha` to be very small, applying the formula directly, we have
`alpha=sqrt(K_(a)//c)=sqrt((1.32xx10^(-5))//0.05)=1.62xx10^(-2)`
`CH_(3)CH_(2)CO OH hArrCH_(3)CH_(2)CO O^(-)+ H^(+)`
In presence of HCl, equilibrium will shift in the backward direction, i.e., concentration of `CH_(3)CH_(2)CO OH` will increase, i.e., amount dissociated will be less. If c is the initial concentration and x is the amount now dissociated, then at equilibrium `[CH_(3)CH_(2)CO OH]=c-x, [CH_(3)CH_(2)CO O^(-)]=x, [H^(+)]=0.01 + x `
`:. K_(a) = (x(0.01xx x))/(c-x)~=(x(0.01))/(c) or (x)/(c) =(K_(a))/(0.01)=(1.32xx10^(-5))/(10^(-2))=1.32xx10^(-3)`. But `(x)/(c) ` means `alpha`.
Hence, `alpha = 1.32xx10^(-3)`.