Question

Calculate solubility of $C{a}_3\left(P{O}_4\right)\left(K_p={10}^{-15}\right)$ in presence of $0.1MCaC{l}_2$ soluion.

Answer 1

$\begin{array}{c}{K}_p={\left[3\left(s+0.1\right)\right]}^3\times {\left(2s\right)}^2\\ {10}^{-15}=27{\left(s+0.1\right)}^3\times 4s^2\\ \frac{{10}^{-15}}{108}={\left(s+0.1\right)}^3\times s^2\\ s<<0.1\\ \frac{{10}^{-15}}{108}=s^5\\ s={\left[\frac{{10}^{-15}}{108}\right]}^{1/5}=0.00987\end{array}$