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labwork [276]
3 years ago
5

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?
Chemistry
1 answer:
laiz [17]3 years ago
7 0

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

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