Question

The concentration of pb2+ in a solution saturated with pbbr2(s) is 2.14 ✕ 10-2 m. calculate ksp for pbbr2.

Answer 1

Answer : The value of $K_{sp}$ is, $3.92\times 10^{-5}$

Explanation :

The balanced equilibrium reaction will be,

$PbBr_2\rightleftharpoons Pb^{2+}+2Br^{1-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=4(s)^3$

Now put the value of solubility in this expression, we get the value of $K_{sp}$

$K_{sp}=4(2.14\times 10^{-2})^3$

$K_{sp}=3.92\times 10^{-5}$

Therefore, the value of $K_{sp}$ is, $3.92\times 10^{-5}$

Answer 2

Concentration = 2.14 âś• 10-2 m 
For [Br-], there are 2 ions so 2 x 2.14 x 10^-2 =4.28 x 10^-2  
Ksp = [Pb][Br]^2 = 2.14 âś• 10-2 x (4.28 x 10^-2 )^2 = 39.20 x 10^-6 
Ksp = 3.92 x 10^-5