`K_(sp)` of salt `A_(3)B_(2)`, for solubility `x mol L^(-1)`, is

Match the salt in Column I with its relation for K_(sp) in Column II in terms of solubility s in mol L^(-1).

Solubility of salt A_(2)B_(3) is 1 xx 10^(-4) , its solubility product is

Solubility of salt A_(2)B_(3) is 1 xx 10^(-4) , its solubility product is

A salt MX has K_(sp)=3.2xx10^(-11) . Value of K_(sp) must another salt, KX_(2) , have, if the molar solubility of the two salts is to be identical?

What will be solubility of A_(2)X_(3) if its solubility product (K_(sp)) is equal to 1.08 xx 10^(-23) ?

What will be solubility of A_(2)X_(3) if its solubility product (K_(sp)) is equal to 1.08 xx 10^(-23) ?

K_(sp) of a sparingly soluble salt AB_(2) is 4xx10^(-12) mol^(3) L^(-3) . The solubility of the salt is

A_(3),B_(2) is a sparingly soluble salt of molar mass M (g mol^(-1) and solubility x g L^(-1) The solubility product satisfies K_(sp)=a(x)/(M)^(5) . The value of a is__________(Integer answer)

The K_(sp) of CaF_2 at 298 K is 1.7xx10^(-10) .Calculate its solubility in mol L^(-1)

For the sparingly soluble salts, an equilibrium is established between the undissolved solid unit and ions of the dissolved salt. For a solution of the salt like A_x B_y . A_x B_y =xA^(y+)+yB^(x-) K_(sp)=[A^(y+)]^x [B^(x-)]^y For precipitation to occur , the ionic product must be greater than K_(sp) . K_(sp) for A_2B_3 salt is 1.1xx10^(-23) . Its solubility is