Confusing concentration calculation about vivianite solubility

I'm confused about the results of [PO4] and [Fe(II)] calculated in [a paper](https://aslopubs.onlinelibrary.wiley.com/doi/pdf/10.4319/lo.2003.48.2.0929). The example they posted is shown in the figure below:


[![Fig_example](https://i.stack.imgur.com/BbcLF.png)](https://i.stack.imgur.com/BbcLF.png)


I tried to calculate the phosphate and Fe(II) concentration. The calculation process is below:


$\ce{ Fe3(PO4)2 8H2O <=> 3Fe^2+ + 2PO4^3- + 8H2O}$


Let's set the concentration of $\ce{Fe^2+}$ as 3x mol/L, concentration of $\ce{PO4^3-}$ as 2x mol/L, then:


(3x)^3\*(2x)^2 = 10^-36, which makes x = 2.47$\*$10^-8


So concentration of $\ce{Fe^2+}$ is 3x = 7.42$\*$10^-8 mol/L, concentration of $\ce{PO4^3-}$ is 2x = 4.95$\*$10^-8 mol/L.


However, the paper said that Fe(II) and P concentration is 1.53$\*$10^-5 mol/L and 1.02$\*$10^-5 mol/L, respectively.


Here are my two questions:


(1) Why my calculation is not the same (even in the magnitude) as the paper's results?


(2) Why do they mention "pH = 7", I couldn't find where is this "pH" take part in the solubiity calcculation.


