# 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

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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?

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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).