. What will be the solubility of SrF2 in 0.1 M NaF aqueous solution


A. 9 x 10-7 mol L-1                                                  B. 9 x 10-9 mol L-1

C. 4.5 x 10-4 mol L-1                                               D. 9 x 10-8 mol L-1

 

Best Answer

(d)

Ksp = [Sr2+][F-]2

In 0.1 M NaF aqueous solution,

9 x 10-10 = (x)(2x + 0.1)2

Since we have a sparingly soluble salt (as the value of Ksp is very low) dissolving in a solution which already has a common ion, the solubility will be very low and will get further reduced due to common ion effect, so we can ignore 2x with respect to 0.1. Hence we get,

0.01 x = 9 x 10-10

Therefore, x = 9 x 10-8 mol L-1

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