निम्नलिखित विलयनों की pH ज्ञात कीजिये-
(a) 0.01 M `CH_(3)COOH, K_(a)=1.8 xx 10^(-5)`
(b) 0.01 M `CH_(3)COOH, alpha=2%`
( c) 0.01 M `NH_(4)OH, K_(b)=1.8 xx 10^(-5)`
(d) 0.01 M `NH_(4) OH, alpha =2%`

Explanation

(a) `CH_(3)COOH ltimplies CH_(3)COO^(-) + H^(+)`
दिया हैं- `C=10^(-2) M, K_(a) = 1.8 xx 10^(-5)`
`therefore [H^(+)] C.alpha = C xx sqrt([K_(a)/C])=sqrt(K_(a).C) (therefore K_(a) = Calpha^(2))`
`[H^(+)] = sqrt(10^(-2) xx 1.8 xx 10^(-5)) = 4.24 xx 10^(-4)`
`therefore pH=3.3726`
(b) `CH_(3)COOH ltimplies CH_(3)COO^(-) + H^(+)`
दिया हैं- `C=10^(-2)M , alpha=2/100 = 2xx 10^(-2)`
`therefore [H^(+)]=C alpha= 10^(-2) xx 2 xx 10^(-2) = 2 xx 10^(-4)`
`therefore pH=3.6989`
( c) `NH_(4)OH ltimplies NH_(4)^(+) + OH^(-)`
`C=10^(-2)M, K_(b) = 1.8 xx 10^(-5)`
`therefore [OH^(-)] = Calpha = Csqrt([K_(b))/C] = sqrt(K_(b).C)`
`=sqrt(10^(-2) xx 1.8 xx 10^(-5))`
`=4.24 xx 10^(-4)`
`rArr pOH=3.3726`
`therefore pH=14-pOH=14-3.3726= 10.6274`
(d) `NH_(4)OH ltimplies NH_(4)^(+) +OH^(-)`
`C=10^(-2)M, alpha=2/100 = 2 xx 10^(-2)`
`therefore [OH^(-)] =Calpha= 10^(-2) xx 2 xx 10^(-1) = 2 xx 10^(-3)`
`therefore` pOH = 3.6989
`therefore pH=14-pOH=14-3.6989= 10.3011`