The value of `K_(sp)` for `CaF_(2)` is `1.7 xx 10^(-10)` . If the concentration of NaF is 0.1 M then new solubility of `CaF_(2)` is

The solubility product (K_(sp)) of AgCI is 1.1 xx 10^(-10) and the concentration of Ag^(+) ions in given solution is 1.0 xx 10^(-7) mole per litre. The concentration CI^(-) for precipitation to occur will be

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

The solubility of CaF_(2) " in water at " 298 K is 1.7 xx 10^(-3) grams per 100 mL of the solution. Calculate solubility product of CaF_(2).

The solubility product of MgF_(2) is 7.4xx10^(-11) . Calculate the solubility of MgF_(2) in 0.1 M NaF solution

K_(sp) of AgCl is 1xx10^(-10) . Its solubility in 0.1 M KNO_(3) will be :