`2H_(2)O hArr H_(3)O^(o+) + overset(Theta)OH,K_(w) = 10^(-14)` at `25^(@)C`, hence `K_(a)` is

2H_(2)OhArrH_(3)O^(+)+OH^(-) K_(w)=1xx10^(-14) at 25^(@)C . Hence, K_(a) is:

Calculate the dissociation constant of NH_(4)OH at 298k , if DeltaH^(Theta) and DeltaS^(Theta) for the given changes are as follows:- NH_(3) + H^(o+) hArr overset(o+)NH_(4) , DeltaH^(Theta) =- 52.2 KJ mol^(-1), DeltaS^(Theta) = 1.67 J K^(-1)mol^(-1) H_(2)O hArr H^(o+) + overset(Theta)OH, DeltaH^(Theta) = 56.6 kJ mol^(-1) . DeltaS^(Theta) =- 76.53 JK^(-1)mol^(-1)

Calculate the dissociation constant of NH_(4)OH at 298k , if DeltaH^(Theta) and DeltaS^(Theta) for the given changes are as follows:- NH_(3) + H^(o+) hArr overset(o+)NH_(4) , DeltaH^(Theta) =- 52.2 KJ mol^(-1), DeltaS^(Theta) = 1.67 J K^(-1)mol^(-1) H_(2)O hArr H^(o+) + overset(Theta)OH, DeltaH^(Theta) = 56.6 kJ mol^(-1) . DeltaS^(Theta) =- 76.53 JK^(-1)mol^(-1)

The K_(w) for 2H_(2)O hArr H_(3)O^(+)_OH^(-) changes from 10^(-14) at 25^(@)C to 9.62xx10^(-14) at 60^(@)C . What is pH of water at 60^(@)C ? What happens to its neutrality ?

Calculate the [overset(Theta)OH] of [NH_(2)C_(2)H_(4)NH_(3)]^(o+) and [H_(3)N-C_(2)H_(4)NH_(3)]^(2+) in 0.15M ethylene diamine (aq) if NH_(2)C_(2)H_(4)NH_(2)+H_(2)O hArr NH_(2)C_(2)H_(4)overset(o+)NH_(3)+overset(Theta)OH (K_(1) = 8.5 xx 10^(-5)) NH_(2)C_(2)H_(4)overset(o+)NH_(3) +H_(2)O hArr [NH_(3)C_(2)H_(4)NH_(3)]^(2+) + overset(Theta)OH (K_(2) = 2.7 xx 10^(-8))

Calculate [overset(Theta)OH] of [NH_(2)C_(2)H_(4)NH_(3)]^(o+) and [H_(3)N-C_(2)H_(4)NH_(3)]^(2+) in 0.15M ethylene diamine (aq) if NH_(2)C_(2)H_(4)NH_(2)+H_(2)O hArr NH_(2)C_(2)H_(4)overset(o+)NH_(3)+overset(Theta)OH (K_(1) = 8.5 xx 10^(-5)) NH_(2)C_(2)H_(4)overset(o+)NH_(3) +H_(2)O hArr [NH_(3)C_(2)H_(4)NH_(3)]^(2+) + overset(Theta)OH (K_(2) = 2.7 xx 10^(-8))

In atmosphere, SO_(2) and NO are oxidised to SO_(3) and NO_(2) , respectively,w hcih react with water to given H_(2)SO_(4) and HNO_(3) . The resultant solution is called acid rain. SO_(2) dissolves in water to form diprotic acid. SO_(2)(g) +H_(2)O(l) hArr HSO_(3)^(Theta) + H^(o+), K_(a_(1)) = 10^(-2) . HSO_(3)^(Theta) hArr SO_(3)^(2-) + H^(o+), K_(a_(2)) = 10^(-7) and for equilibrium, SO_(2)(aq) + H_(2)O (l) hArr SO_(3)^(2-)(aq) +2H^(o+)(aq) K_(a) = K_(a_(1)) xx K_(a_(2)) = 10^(-9) at 300K . The pH of 0.01M aqueous solutioon of sodium sulphite (Na_(2)SO_(3))