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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?

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i. Ionization constant of propionic acid `=1.32xx10^(-5)`
Let the degree of ionisation is `alpha`
`alpha=sqrt((K_(a)/C))=sqrt((1.32xx10^(-5))/0.05)=1.63xx10^(-2)`
`[H^(o+)]=Calpha`
`=0.05xx1.63xx10^(-2)=0.0815xx10^(-2)=815xx10^(-6)`
`pH=-log (815xx10^(-6))=-(log 815+ log 10^(-6))`
`=-(2.9112-6)=3.088`
`pH=3.09`
ii. In presence of `0.01 M HCl`
`HCl` is fully dissociates
`{:(HCl,rarr,H^(o+),+,Cl^(Θ)),(0.01,,0,,0),(0,,0.01,,0.01):}`
Total `[H^(o+)]=[H^(o+)]` from `HCl+[H^(o+)]` from fropinoic acid
`CH_(3)CH_(2)COOH rarr CH_(3)CH_(2)COO^(Θ)+H^(o+)`
`{:(1,,0,,0),(C(1-alpha),,Calpha,,Calpha):}`
But `[H^(o+)]=(0.01+Calpha)`
`K_(a)=(Calphaxx(0.01+Calpha))/(C(1-alpha))`
`K_(a)=0.01alpha+Calpha^(2)(1-alpha~~1)=0.01 alpha`
`(Calpha^(2) "is neglected as alpha is very small")`
`alpha=K_(a)/0.01=(1.32xx10^(-5))/0.01=1.32xx10^(-3)`
`:.` In presence of `0.01 M HCl`,
`alpha=1.32xx10^(-3)`
Note: It can be seen that the degree of ionization decreases almost by a factor of `10`.
Ionisation of Propionic acid `=1.63xx10^(-2)`
Ionisation in presence of `0.01 M HCl=1.32xx10^(-3)`
Decreases almost by factor of `10`

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