(I). Calculation of pH of 0.01M chloroactic acid solution
According to available data, `K_(a) =1.35 xx 10^(-3) : C = 0.1 M`
`[H^(+)] =(K_(a) xx C)^(1//2) =(1.35 xx 10^(-3) xx 10^(-1))^(1//2) =1.16 xx 10^(-2)M`
`pH =- log [H^(+)] =- log (1.16 xx 10^(-2))`
`= 2- log 1.16 =2 -0.06 =1.94`
(II). Calculation of pH of 0.01 M solution salt of acid
`K_(h) =(K_(w))/(K_(a)) =(1xx 10^(-14))/(1.35 xx 10^(-3)) =7.4 xx 10^(-12)`
Degree of hydrolysis `(h) = sqrt((K_(h))/(C)) = ((7.4 xx 10^(-12))/(0.1))^(1//2) =8.6 xx 10^(-6)`
the sodium salt of the acid upon hydrolysis will form a basic solution
`[OH^(+)] = Ch = 0.1xx 8.6 xx 10^(-6) =8.6 xx 10^(-7)`
`[H^(+)] =(K_(w))/[[OH^(+)]] =(1.0xx 10^(-14))/(8.6 xx 10^(-7)) =1.16 xx 10^(-8)`
`=[8 - log 1.16] =8 - 0.064 =7.94.`
