Question

Suppose you add 0.2489 g of P b C l 2 ( s ) to 50.0 mL of water. In the resulting saturated solution, you find that the concentration of P b 2 + ( a q ) is 0.0159 M and the concentration of C l − ( a q ) is 0.0318 M. What is the value of the equilibrium constant, Ksp, for the dissolution of P b C l 2 ?

Answer 1

`K_(sp)=1.61* 10^(-5)`

Step-by-step explanation:

`PbCl_2` will form its respective ions in the solution as:

Consider the ICE take for
`PbCl_2` as:

PbCl₂ ⇄ Pb²⁺ + 2Cl⁻

At t =equilibrium x 2x

The expression for dissociation constant of
`PbCl_2` is:

Solubility product =
`[Pb^(2+)][Cl^-]^2`

Given that:

`[Pb^(2+)]=0.0159\ M`

`[Cl^-]=0.0318\ M`

So,

Solubility product =
`[Pb^(2+)][Cl^-]^2` =
`0.0159* {0.0318}^2`

`K_(sp)=1.61* 10^(-5)`