In "0.1M" aq `H_(2)S` solution `K_(b)` for `HS^(-)` is `10^(-9)` .If `[S^(2)]` in this solution is `10^(-10)` ,then the dissociation constant for `H_(2)S` `rarr``2H^(+)`+`S^(2-)` will be (approximately)

At 298 K, the [H_(3),O^(+)] of a solution is 2 xx 10^(-9) M. The nature of the solution is

0.1 M H_(2)S has K_(1)=10^(-5) & K_(2)=1.5xx10^(-12) . What will be the concentration of S^(-2) in the solution.

The first ionization constant of H_(2)S is 9.1xx10^(-8) . Calculate the concentration of HS^(Θ) ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant if H_(2)S is 1.2xx10^(-13) , calculate the concentration of S^(2-) under both conditions.

What is [HS^(-)] & [S^(2-)] in solution of 0.01M HCl and 0.1 M H_(2)S ?(Given that for H_(2)S,K_(a_(1))=10^(-7)and K_(a_(2))=10^(-14))

Approximate p.H,0.1M aqueous H_(2)S solution when K_(1) and K_(2) for H_(2)S at 25^(@)C are 1xx10^(-7) and 1.3xx10^(-13) respectively:-

If K_(SP) of Ag_(2)S is 10^(-17) , the solubility of Ag_(2) S in 0.1 M solution of Na_(2)S will be :

If the specific conductance of 1 M H_(2)SO_(4) solution is 26xx10^(-2)S cm^(2) , then the equivalent conductivity would be

In the reaction. H_(2)S+H_(2)O_(2) rarr S+2H_(2)O describes

A drop of solution (0.05 mL) contains 2 xx 10^(6) mol of H^(+) . How long it takes for this H^(+) ot disapper , if the rate constant is 10^(7) mol L^(-1) s^(-1) ?

An aqueous solution contains 0.10 M H_(2)S and 0.20 M HCl. If the equilibrium constants for the formation of HS^(–) from H_(2)S is 1.0xx10^(–7) and that of S^(2-) from HS^(–) ions is 1.2xx10^(–13) then the concentration of S^(2-) ions in aqueous solution is