Question

Above what Fe2+ concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 8.36 ?

The

Answer 1

We know that, Ksp of Fe(OH)₂=$4.87\times10^{-17}$

pOH= 14-8.36=5.64

$[OH^{-1}]=-log(5.64)=2.291\times10^{-6}$

$[Fe^{2+}][OH^{-1}]=4.87\times10^{-17}$

$[Fe^{2+}][2.291\times10^{-6}]=4.87\times10^{-17}$

$[Fe^{2+}]=2.12 \times 10^{-11}$

Above this it will make buffer