The solubility product of ${\text{PbI}}_{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}7.2\times {10}^{-9}$ Calculate the maximum mass of NaI which may be added in 500 mL of $0.005\text{\hspace{0.17em}M}-{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$solution without any precipitation of ${\text{PbI}}_{\text{2}}\left(\text{Atomic\hspace{0.17em}weight\hspace{0.17em}of\hspace{0.17em}Na}=23,\text{\hspace{0.17em}I}=127\right)$

Given,
The solubility product of ${\text{PbI}}_{\text{2}},K_{\text{sp}}=7.2\times {10}^{-9}$
Volume,$v=500\text{\hspace{0.17em}mL}$

molarity, M = 0.005 M maximum mass of  , m = ? 
prevent any precipitation of ${\text{PbI}}_{\text{2}}\text{,Q}<K_{\text{sp}}$

or, $\left[{\text{Pb}}^{\text{2+}}\right]{\left[{\text{I}}^{\text{-}}\right]}^{\text{2}}<K_{\text{sp}}$

or, $\left(0.005\right){\left[{\text{I}}^{\text{-}}\right]}^{\text{2}}<7.2\times {10}^{-9}$

or, $\left[{\text{I}}^{-}\right]<1.2\times {10}^{-3}\text{\hspace{0.17em}M}$

Hence, the maximum mass of  , which can be added 
 

$\begin{array}{l}=\left(\frac{1.2\times {10}^{-3}}{1000}\times 500\right)\times 150\text{\hspace{0.17em}g}\\ =0.09\text{\hspace{0.17em}g}\end{array}$