Answer : The concentration of
needed is, 
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.

Molar mass of phosphate = 94.97 g/mole

Now we have to calculate the concentration of phosphate.


Now we have to calculate the concentration of
.
The second equilibrium reaction is,

The solubility constant expression for this reaction is:
![K_{sp}=[Fe^{3+}][PO_4^{3-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B3%2B%7D%5D%5BPO_4%5E%7B3-%7D%5D)
Given: 
![\frac{1}{4}=[Fe^{3+}]\times 1.053\times 10^{-5}mol/L](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%3D%5BFe%5E%7B3%2B%7D%5D%5Ctimes%201.053%5Ctimes%2010%5E%7B-5%7Dmol%2FL)
![[Fe^{3+}]=2.37\times 10^4M](https://tex.z-dn.net/?f=%5BFe%5E%7B3%2B%7D%5D%3D2.37%5Ctimes%2010%5E4M)
Thus, the concentration of
needed is, 