Calculate the (i) hydrolysis constant, (ii) degree of hydrolysis and (iii) pH of 0.05M sodium carbonate solution `pK_(a)` for `HCO_(3)^(-)` is 10.26.

Explanation

(i) Hydrolysis constant:
`h= sqrt((K_(w))/(K_(a)xx c))`
Given `K_(w)=1xx10^(-14)`
`c=0.05M`
`pK_(a)=10.26`
`pK= -log K_(a)`
`K_(3)="antilog of"(-pK_(a))`
`K_(a)="antilog of"(-10.26)`
`K_(a)=5.49xx10^(-11)`
`h= sqrt((1xx10^(-14))/(5.49xx10^(-11)xx0.05))=sqrt(3.642xx10^(-3))`
`h=6.034xx10^(-2)`
(ii) Degree of hydrolysis: `K_(h)=(K_(w))/(K_(a))=(1xx10^(-14))/(5.49xx10^(-11))=1.82xx10^(-4)`
(iii) `pH=7+(pK_(a))/(2)+("log"C)/(2)`
`=7+(10.26)/(2)+(log(0.05))/(2)=7+5.13+((-1.30)/(2))=7+5.13-0.65`
`pH=11.48`