The Ka of propionic acid is `1.34 xx 10^(-5)`. What is the pH of a solution containing 0.5 M propionic acid and 0.5 M sodium proportionate ? What happens to the pH of the solution when volume is doubled by adding water ?

Explanation

`Ka" of propionic acid "=1.34xx10^(-5)`
`therefore" "pK_(a)=-logK_(a)=-log(1.34xx10^(-5))`
`=4.87`
By Herderson - Hasselbalch equation
`pH=pK_(a)+log.(["salt"])/(["acid"])`
`=4.87+log""(0.5)/(0.5)`
`pH=4.87`
87
Alternative solution : The dissociation equilibrium of propionic acid will be
`K_(a)=([C_(2)H_(5)COO^(-)][H^(+)])/([C_(2)H_(5)COOH])=(0.5xx[H^(+)])/(0.5)`
`=[H^(+)]`
`therefore" "pH=-log[H^(-)]`
`=-logK_(a)=-log(1.34xx10^(-5))`
`therefore pH=4.87`