Question

At a wastewater treatment plant, FeCl3(s) is added to remove excess phosphate from the effluent. Assume the following reactions occur: FeCl3 ---> Fe3+ + 3Cl- FePO4 ---> Fe3+ + PO4 3␣ The equilibrium constant for the second reaction is Ksp 1⁄4. What concentration of Fe3+ is needed to maintain the phosphate concentration below the limit of 1 mg P/L?

Answer 1

Answer : The concentration of
`Fe^(3+)` needed is,
`2.37* 10^4M`

Explanation :

First we have to calculate the mole of phosphate.

As we are given that, 1 mg P/L that means, 1 mg of phosphate present in 1 L of solution.

`\text{Moles of phosphate}=\frac{\text{Mass of phosphate}}{\text{Molar mass of phosphate}}`

Molar mass of phosphate = 94.97 g/mole

`\text{Moles of phosphate}=(1mg)/(94.97g/mol)=(0.001g)/(94.97g/mol)=1.053* 10^(-5)mol`

Now we have to calculate the concentration of phosphate.

`\text{Concentration of phosphate}=\frac{\text{Moles of phosphate}}{\text{Volume of solution}}`

`\text{Concentration of phosphate}=(1.053* 10^(-5)mol)/(1L)=1.053* 10^(-5)mol/L`

Now we have to calculate the concentration of
`Fe^(3+)`.

The second equilibrium reaction is,

`FePO_4\rightleftharpoons Fe^(3+)+PO_4^(3-)`

The solubility constant expression for this reaction is:

`K_(sp)=[Fe^(3+)][PO_4^(3-)]`

Given:
`K_(sp)=(1)/(4)`

`(1)/(4)=[Fe^(3+)]* 1.053* 10^(-5)mol/L`

`[Fe^(3+)]=2.37* 10^4M`

Thus, the concentration of
`Fe^(3+)` needed is,
`2.37* 10^4M`