Respuesta :
Answer : The concentration of [tex]Fe^{3+}[/tex] needed is, [tex]2.37\times 10^4M[/tex]
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.
[tex]\text{Moles of phosphate}=\frac{\text{Mass of phosphate}}{\text{Molar mass of phosphate}}[/tex]
Molar mass of phosphate = 94.97 g/mole
[tex]\text{Moles of phosphate}=\frac{1mg}{94.97g/mol}=\frac{0.001g}{94.97g/mol}=1.053\times 10^{-5}mol[/tex]
Now we have to calculate the concentration of phosphate.
[tex]\text{Concentration of phosphate}=\frac{\text{Moles of phosphate}}{\text{Volume of solution}}[/tex]
[tex]\text{Concentration of phosphate}=\frac{1.053\times 10^{-5}mol}{1L}=1.053\times 10^{-5}mol/L[/tex]
Now we have to calculate the concentration of [tex]Fe^{3+}[/tex].
The second equilibrium reaction is,
[tex]FePO_4\rightleftharpoons Fe^{3+}+PO_4^{3-}[/tex]
The solubility constant expression for this reaction is:
[tex]K_{sp}=[Fe^{3+}][PO_4^{3-}][/tex]
Given: [tex]K_{sp}=\frac{1}{4}[/tex]
[tex]\frac{1}{4}=[Fe^{3+}]\times 1.053\times 10^{-5}mol/L[/tex]
[tex][Fe^{3+}]=2.37\times 10^4M[/tex]
Thus, the concentration of [tex]Fe^{3+}[/tex] needed is, [tex]2.37\times 10^4M[/tex]
