Question

What mass of ${Pb}^{2+}$ ion is left in solution when $50.0mL$ of $0.20M$ $Pb{\left({NO}_3\right)}_2$ is added to $50mL$ of $1.5M$ $NaCl$?

[Given, $K_{sp}$ for $Pb{Cl}_2$ is $1.7\times {10}^{-4}$]

• A. $6mg$
• B. $12mg$
• C. $18mg$
• D. $36mg$

Answer 1

The correct option is B $12mg$

$P{b}^{2+}+2C{l}^{-}\to PbC{l}_2$

moles of $P{b}^{2+}=50\times {10}^{-3}\times 0.2=10\times {10}^{-3}$ moles

moles of $C{l}^{-}=1.5\times 50=75$ milimoles $=75\times {10}^{-3}$ moles

$\begin{array}{cc}\text{moles of}C{l}^{-}\text{left}& =(75-20)\times {10}^{-3}\\ & =55\times {10}^{-3}\text{moles}\\ & =55\text{millimoles}\end{array}$

$\left[C{l}^{-}\right]=\frac{no.\text{of}moles}{\text{Total volume}}=\frac{55}{50+50}=0.55M$

$k_{sp}=\left[P{b}^{2+}\right]{\left[C{l}^{-}\right]}^2$

$1.7\times {10}^{-4}=\left[{Pb}^{2+}\right](0.55{)}^2$

$\left[P{b}^{2+}\right]=\frac{1.7\times {10}^{-4}}{(0.55{)}^2}$

$\left[P{b}^{2+}\right]=5.6\times {10}^{-4}M$

$\begin{array}{cc}\text{milimoles of}\left[P{b}^{2+}\right]=& 5.6\times {10}^{-4}\times 100\\ =& 5.6\times {10}^{-2}\text{milimoles}\end{array}$

$\begin{array}{cc}\mathrm{mass}\text{of}P{b}^{2+}& =5\cdot 6\times {10}^{-2}\times {10}^{-3}\times 208\\ & =12mg\end{array}$