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, $2.37\times 10^4M$

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

$\text{Moles of phosphate}=\frac{1mg}{94.97g/mol}=\frac{0.001g}{94.97g/mol}=1.053\times 10^{-5}mol$

Now we have to calculate the concentration of phosphate.

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

$\text{Concentration of phosphate}=\frac{1.053\times 10^{-5}mol}{1L}=1.053\times 10^{-5}mol/L$

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: $K_{sp}=\frac{1}{4}$

$\frac{1}{4}=[Fe^{3+}]\times 1.053\times 10^{-5}mol/L$

$[Fe^{3+}]=2.37\times 10^4M$

Thus, the concentration of $Fe^{3+}$ needed is, $2.37\times 10^4M$