Question

g [McQuarrie 22-7] The value of Ksp for PbCrO4(s) in equilibrium with water at 25◦C is 2.8 · 10−13M2 . Write the chemical equation that represents the solubility equilibrium for PbCrO4(s) and calculate its solubility in grams per liter in water at 25◦C.

Answer 1

The solubility equilibrium reaction will be:

`PbCrO_4\rightleftharpoons Pb^(2+)+CrO_4^(2-)M^2`

The solubility in grams per liter in water at
`25^oC` is,
`1.7* 10^(-4)g/L`

Explanation :

The solubility equilibrium reaction will be:

`PbCrO_4\rightleftharpoons Pb^(2+)+CrO_4^(2-)M^2`

Let the solubility be 's'.

The expression for solubility constant for this reaction will be,

`K_(sp)=[Pb^(2+)][CrO_4^(2-)]`

`K_(sp)=(s)* (s)`

`K_(sp)=s^2`

Given:

Solubility constant =
`K_(sp)` =
`2.8* 10^(-13)`

Now put all the given values in the above expression, we get:

`K_(sp)=s^2`

`2.8* 10^(-13)=s^2`

`s=5.3* 10^(-7)M=5.3* 10^(-7)mol/L`

Now we have to convert the solubility in gram per liter.

Solubility =
`(5.3* 10^(-7)mol/L)* \text{Molar mass of }PbCrO_4`

`\text{Molar mass of }PbCrO_4=323.2g/mol`

Solubility =
`(5.3* 10^(-7)mol/L)* 323.2g/mol`

Solubility =
`1.7* 10^(-4)g/L`

Therefore, the solubility in grams per liter in water at
`25^oC` is,
`1.7* 10^(-4)g/L`