Question

3. What is the solubility (in M) of SnCl2 ?The Ksp of SnCl2 is 1.59 × 10-5.

A) 2.0 × 10-3

B) 1.1 × 10-4

C) 1.8 × 10-4

D) 7.1 × 10-4

E) 1.6 × 10-5

F) None of the above

4. The concentration of fluoride ions in a saturated solution of Strontium fluoride is ________ M. The solubility product constant of SrF2 is 1.68 × 10-6.

A) 3.8 × 10-4

B) 3.0 × 10-3

C) 1.5 × 10-2

D) 7.5 × 10-3

E) 1.4 × 10-4

F) None of the above

Answer 1

3) Given Ksp=1.59 × 10-5

SnCl2-----> Sn+2 + 2 Cl-

Ksp=[Sn+2][Cl-]2

Let us assume [Sn+2]=x [Cl-]=2x

1.59 × 10-5=(x)(2x)2

x3=(1.59 × 10-5)/4

=> x=(3.975 × 10-6)1/3

x=0.0158=1.58x10-2 M

[Sn+2]=1.58x10-2 M and [Cl-]=2x1.58x10-2 M=3.16x10-2 M

So the solubility of SnCl2=1.58x10-2 M

Option F is correct

4) Given Ksp=1.68 × 10-6

SrF2-----> Sr+2 + 2 F-

Ksp=[Sr+2][F-]2

Let us assume [Sr+2]=x [F-]=2x

1.68 × 10-6=(x)(2x)2

x3=(1.68 × 10-6)/4

=> x=(4.2 × 10-7)1/3

x=7.525x10-3 M

[Sr+2]=7.525x10-3 M and [F-]=2x7.525x10-3 M=1.5x10-2 M

The concentration of fluoride ions=1.5x10-2 M

Option (C) is correct.