Chapter 11, Problem 71QP

(a)

Interpretation Introduction

Interpretation:

The volume of $1.0\text{M KI}$ required for complete precipitation of lead ions for the given solution is to be calculated.

Explanation of Solution

Given Information: $100\text{mL}$ of $0.15\text{M}{\text{Pb}}^{2+}$ ion solution and $1.0\text{M KI}$ solution are given.

The given ionic reaction given is:

${\text{Pb}}^{2+}(aq)+2{\text{I}}^{-}(aq)\to {\text{PbI}}_2(s)$

The molarity of ${\text{Pb}}^{2+}$ ions is $0.15\text{M}$ . This means that $0.15\text{mole}$ of ${\text{Pb}}^{2+}$ ions are present per litre of the solution. The volume of $0.15\text{M}{\text{Pb}}^{2+}$ solution is $100\text{mL}$ . This must be converted into litres. The relationship between molarity, moles of solute and volume of solution is:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution in litre}}$ …..(1)

So, moles of solute can be calculated by rearranging the equation (1).

$\text{Moles of solute}=\text{Molarity}\times \text{Volume of solution in litre}$ …..(2)

Moles of $0.15\text{M}{\text{Pb}}^{2+}$ present in the solution, using equation (2) is

$\begin{array}{c}{\text{mol Pb}}^{2+}=100\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\times \frac{0.15\text{mol}}{1\text{L}}\\ =0.015\text{mol}\end{array}$

The mole ratio of the reaction is: $1\text{mol}$ of ${\text{Pb}}^{2+}$ reacts with $2\text{mol}$ iodide ions to give $1\text{mol}$ of ${\text{PbI}}_2$ solid. So, the moles of $\text{KI}$ present are:

$\begin{array}{c}\text{mol KI}=0.015{\text{mol Pb2}}^{\text{+}}\text{ion}\times \frac{2\text{mol KI}}{1{\text{mol Pb}}^{\text{2+}}\text{ion}}\\ =0.03\text{mol}\end{array}$

Volume of $\text{KI}$ required is

$\begin{array}{c}\text{KI}\text{volume}=0.03\text{mol}\times \frac{1\text{L KI solution}}{1.\text{0 mol}}\\ =0.03\text{L}\end{array}$

So, $0.03\text{L}$ of KI is required to completely precipitate $0.15\text{M}{\text{Pb}}^{2+}$ ion in $100\text{mL}$ solution. Since the value is very small, the volume must be converted to unit of millilitres.

$\begin{array}{c}\text{KI}\text{volume in millilitre}=0.03\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =30.0\text{mL}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The mass of ${\text{PbI}}_2$ precipitated in the above reaction is to be calculated.

Explanation of Solution

Given Information: $100\text{mL}$ of $0.15\text{M}{\text{Pb}}^{2+}$ ions and $1.0\text{M KI}$ solution .

The given ionic reaction is:

${\text{Pb}}^{2+}(aq)+2{\text{I}}^{-}(aq)\to {\text{PbI}}_2(s)$

To calculate the mass of ${\text{PbI}}_2$ , mol of ${\text{PbI}}_2$ must be calculated from the mole ratio of the balanced reaction.

$\begin{array}{c}{\text{Moles PbI}}_2=0.015{\text{mol Pb}}^{\text{2+}}\text{ion}\times \frac{1{\text{mol PbI}}_2}{1{\text{mol Pb}}^{\text{2+}}\text{ion}}\\ =0.015{\text{mol PbI}}_2\end{array}$

The molar mass of ${\text{PbI}}_2$ is $461.01{\text{g mol}}^{-1}$ . So, the mass of ${\text{PbI}}_2$ precipitate is:

$\begin{array}{c}{\text{Mass of PbI}}_2=0.015\text{mol}\times \text{461}\text{.01}{\text{g mol}}^{-1}\\ =\text{6}\text{.91}\text{g}\end{array}$

So, $6.91\text{g}$ of ${\text{PbI}}_2$ will precipitate in the reaction.