Question

A solution is prepared by mixing 50.0 mL of 0.10 M Pb(NO₃)2 with 50.0 mL of 1.0 M KCl. Calculate the concentrations of Pb²⁺ and Cl at equilibrium. [Ks for PbCl₂(s) = 1.6 X 10^-5.]

Short answer

Theconcentrations of Pb²⁺ and Cl at equilibrium are$\left[P{b}^{2+}\right]=1\times {10}^{-4} M \&$$\left[C{l}^{-}\right]=0.4002 M$ respectively.

Step-by-step solution

State given information

The solution can be prepared by mixing$50 mL 1 M KCl \& 50 mL of 0.1 M Pb(N{O}_3{)}_2.$

$K_{sp}\left(PbC{l}_2\right(s\left)\right)=1.6\times {10}^{-5}$

Calculate number of moles ofPb2+ and Cl-

The number of moles can be calculated using the formula

$Number of moles=Molarity\times Volume$

On substituting the values

$$\begin{gathered} mol P{b}^{2+}=\left[P{b}^{2+}\right]\times Volume \\ mol P{b}^{2+}=0.1 mol/L\times \frac{50mL\times 1L}{1000 mL} \\ mol P{b}^{2+}=0.0050 mol \\ mol C{l}^{-}=\left[C{l}^{-}\right]\times Volume \\ mol C{l}^{-}=1 mol/L\times \frac{50mL\times 1L}{1000 mL} \\ mol C{l}^{-}=0.05 mol \end{gathered}$$

The volume of solution is $50mL+50mL=100mL(0.1L)$.

Calculate concentration of Pb2+ and Cl-

The molarity can be calculated using the formula

$Molarity=\frac{number of moles}{Volume}$

On substituting, we get

$$\begin{gathered} \left[P{b}^{2+}\right]=\frac{0.005 moles}{0.1 L}=0.5 M \\ \left[C{l}^{-}\right]=\frac{0.05 moles}{0.1 L}=0.5 M \end{gathered}$$

On calculating the equilibrium ionic product, we get

$$\begin{gathered} Q=\left[P{b}^{2+}\right[C{l}^{-}] \\ Q=(0.05)(0.5{)}^2 \\ Q=0.0125 \end{gathered}$$

Calculate the concentration

The formation reaction can be denoted as

$P{b}^{2+}+2C{l}^{-}\to PbC{l}_2\left(s\right)$

The initial concentration can be denoted as

$$\begin{gathered} \left[P{b}^{2+}\right]=0 M \\ \left[C{l}^{-}\right]=0.4 M \end{gathered}$$

The equilibrium concentration can be denoted as

$$\begin{gathered} \left[P{b}^{2+}\right]=x M \\ \left[C{l}^{-}\right]=(0.40+2x) M \\ 1.6\times {10}^{-5}=\left(x\right)(0.40+2x{)}^2 \\ 1.6\times {10}^{-5}=\left(x\right)(0.4{)}^2 \\ x=\frac{1.6\times {10}^{-5}}{0.16} \\ x=1\times {10}^{-4} \\ 0.40+2x=0.40+2\times 1\times {10}^{-4} \\ =0.4002 \end{gathered}$$

The equilibrium concentration of Pb²⁺is 1×10^-4 M.

The equilibrium concentration of Cl^-is0.4002 M.