`K_(sp)` of `SrF_(2)` (`s`) in water is `3.2xx10^(-11)`. The solubility `SrF_(2)` (`s`) in `0.1 (M) NaCl` solution is

The solubility product of a soluble salt A_(x)B_(y) is given by : K_(sp)=[A^(y+)]^(z)[B^(x-)]^(y) . As soon as prodcut of concentration of A^(y+) and B_(x-) . Becomes more than its K_(sp) , the salt starts precipitation. It may practically be noticed that AgCI is more soluble in water and its solubility decreases dramatically in 0.1m NaCI or 0.1 m AgNO_(3) solution. It can be concluded that in presence of a common ion the solubility of salt decreases. K_(sp) of SrF_(2) in water is 8xx10^(-10) . The solubility of SrF_(2) in 0.1MNaF aqueous solution is :

The solubility of Agcl (K_sp=1.2xx10^(-10)) in a 0.10 M NaCl solution is :

The K_(sp) of M(OH)_(2)" is "5 times 10^(-10)M^(3) . The molar solubility of M(OH)_(2) in a 0.1 M NaOH solution is

The solubility product of SrF_(2) in water is 8xx10^(-10) . Calculate its solubility in 0.1M NaF aqueous solution.

The solubility product of SrF_(2) in water is 8xx10^(-10) . Calculate its solubility in 0.1M NaF aqueous solution.

The solubility product of SrF_(2) in water is 8xx10^(-10) Calculate its solubility in 0.1 M of aqueous NaF solution. If its solubility is expressed as y xx 10^(-8) then what is the value of 'y' ?

The solubility product of SrF_(2) in water is 8 xx 10^(-10) Calculate its solubility in 0.1 M of aqueous NaF solution. If its solubility is expressed as y xx 10^(-8) then what is the value of 'y' ?

The K_(sp) of Ag_(2)CrO_(4) at 298 K is 1 times 10^(-12)M^(3) The solubility of Ag_(2)CrO_(4) in a 0.1 M AgNO_(3) solution at 298 K is