Sulphurous acid `(H_(2)SO)` has `Ka_(1) = 1.7 xx 10^(-2) and Ka_(2)=6.4 xx 10^(-8)`. The pH of 0.588 M `H_(2)SO_(3)` is _____ _ (Round off to the Nearest Integer).

Selenious acid (H_(2)SeO_(3)) ,a diprotoc acid has Ka_(1)=10^(-6) and Ka_(2)=10^(-8) respectively. Appoximate pH of 0.01M NaHSeO_(3) is given by :-

Calculate pH of - 10^(-3)M H_(2)SO_(4)

What will be the pH of 1xx10^(-4) M H_(2)SO_(4) solution ?

The solubility of CdSO_4 in water is 8.0 xx 10^(-4) mol L^(-1) . Its solubility in 0.01 M H_2SO_4 solution is ........... Xx 10^(-6)" mol L"^(-1) . (Round off to the Nearest Integer). (Assume that solubility is much less than 0.01M).

The conductivity of a weak acid HA of concentration 0.001 mol L^(-1) is 2.0 xx 10^(-5)S cm^(-1) . If Lambda_(m)^(0)(H A) = 190 S cm^(2) mol^(-1) , the ionization constant (K_(a)) of HA is equal to ________ xx 10^(-6) . (Round off to the Nearest Integer)

Calculate pH solution: 0.1 M CH_(3)COOH (K_a=1.8 xx 10^(-5))