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

The ionization constant of propanoic acid is 1.32xx10^(-5). Calculate

1.	The ionization constant of propanoic acid is 1.32xx10^(-5). Calculate the degree of ionization if its solution is 0.05 M. What will be its degree of ionization if the solution is 0.01 M in HCl also?
Answer» Solution :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)`.