A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product `(K_(sp))`. For the electrolyte, `A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-)`, with solubility S, the solubility product `(K_(sp)) =x^(x)xxy^(y) xx s^(x+y)`. While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its `K_(sp)`, value at a particular temperature, then precipitation occurs.
The solubility of `BaSO_(4)`, in 0.1 M `BaCl_(2)`, solution is `(K_(sp)`, of `BaSO_(4), = 1.5 xx 10^(-9))`

Options

A. ` 1.5 xx 10^(-9) M`
B. `1.5 xx10^(-8) M`
C. ` 2.25 xx 10^(-16)M`
D. ` 2.25 xx 10^(-18) M`

Answer

B

Explanation

` BaSO_4 hArr Ba^(+2) +SO_4^(-2) `
` 0.1 M BaCl_2 =0.1 M Ba^(+2) `
` [SO_4^(-2) ]= S, [Ba^(+2) ]= (0.1 +S) ~~ 0.1M`
` K_(sp) =[Ba^(+2) ] [SO_4^(-2) ], 1.5 xx 10 ^(-9) =(0.1 ) (S) `
`rArr S = 1.5 xx 1 0^(-8) `