Q. Calculate the solubility of silver phosphate $\left(\left(Ag\right)_{3} \left(PO\right)_{4}\right)$ in $0.1MAgNO_{3}$ $Ksp$ of $Ag_{3}PO_{4}=1.1\times 10^{- 16}\left.$ :-
Solution:
$\left(Ag\right)_{3}\left(PO\right)_{4}\rightleftharpoons\underset{\left(3 s + 0 . 1\right)}{3 \left(Ag\right)^{+}}+\underset{s}{PO_{4}^{- 3}}$ $Ksp=\left(\right.3s+0.1\left(\left.\right)^{3}\left(s\right)=1.1\times \left(10\right)^{- 16}$
$10^{- 3}\times s=1.1\times 10^{- 16}$
$s=1.1\times 10^{- 13}$
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