📖 generic · CBSE Class 11 English medium · CHEMISTRY · Page 26question

 of hydrogen · Part 21

Chapter 6: Equilibrium · CHEMISTRY

by Q sp (section . . ) when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions K sp = Q sp but otherwise it gives the direction of the processes of precipitation or dissolution.

The solubility product constants of a number of common salts at 298K are given in Table . . Table . The Solubility Product Constants, K sp of Some Common Ionic Salts at 298K.

Problem . Calculate the solubility of A X in pure water, assuming that neither kind of ion reacts with water. The solubility product of A X , K sp = . × – .

A X → 2A + + 3X – K sp = [A + ] [X – ] = . × – If S = solubility of A X , then [A + ] = 2S; [X – ] = 3S therefore, K sp = (2S) (3S) = 108S = . × – thus, S = × – S = . × – mol/L.

Problem . The values of K sp of two sparingly soluble salts Ni(OH) and AgCN are . × – and × – respectively. Which salt is more soluble?

Explain. AgCN Ag + + CN – K sp = [Ag + ][CN – ] = × – Ni(OH) Ni + + 2OH – K sp = [Ni + ][OH – ] = × – Let [Ag + ] = S , then [CN - ] = S Let [Ni + ] = S , then [OH – ] = 2S S = × – , S = . × – (S )(2S ) = × – , S = . × – Ni(OH) is more soluble than AgCN.

. . Common Ion Effect on Solubility of Ionic Salts It is expected from Le

Related topics

Have a question about this topic?

Get an AI answer grounded in your actual textbook — with the exact page reference.

Ask AI about this topic →