Question

You made $\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{mL}$ of a lead nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only role="math" localid="1657263505731" $\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{mL}$ left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{mL}$ of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of $\mathbf{3}\mathbf{.}\mathbf{407}\mathbf{g}$.What was the concentration of the original lead nitrate solution?

Short answer

The concentration of original lead nitrate solution is$4.88\mathrm{M}$

Step-by-step solution

Introduction

Mole is a SI unit for measuring the amount of a material. The mole is the amount of a material that contains approximately the same number of atoms as $\mathbf{12}\mathbf{.}\mathbf{00}\mathbf{g}$ of $\mathbf{Carbon}\mathbf{-}\mathbf{12}$.

The number of moles of a substance is defined as the ratio of the compound's given mass to its molar or molecular mass.

The mathematical formula is as follows:

$\mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{compound}}{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{compound}}$

In the case of dilution, the number of moles of solute before and after dilution is the same. Thus, the connection between molarity and volume may be represented as follows:

${\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2$

Where,

${\mathrm{M}}_1$Initial molarity

${\mathrm{V}}_1$Initial volume

${\mathrm{M}}_2$Final molarity

${\mathrm{V}}_2$Final volume

Explanation

Volume of solution $=100.0\mathrm{mL}$

In the next lab session,

The volume of solution left$=80.0\mathrm{mL}$

The mass of the solid obtained after the reaction$=3.407\mathrm{g}$

The balanced reaction between lead nitrate and sodium chloride is:

$\mathbf{2}\mathbf{NaCl}\mathbf{+}\mathbf{Pb}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\to }{\mathbf{PbCl}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{NaNO}}_{\mathbf{3}}$

Calculating the number of moles

The molar mass of ${\mathrm{PbCl}}_2=278.1\mathrm{g}/\mathrm{mole}$

Mass of ${\mathrm{PbCl}}_2=3.407\mathrm{g}$

The number of moles can be calculated by:

$\begin{array}{c}{\mathrm{PbCl}}_2=\frac{3.407\mathrm{g}}{278.1\mathrm{g}/\mathrm{mole}}\\ =\mathbf{0}\mathbf{.}\mathbf{0122}\mathbf{mole}\end{array}$

The ratio of ${\mathrm{PbCl}}_2$and$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ is 1:1, so the number of moles of lead(II) nitrate is also $\mathbf{0}\mathbf{.}\mathbf{0122}\mathbf{mole}$.

Calculating the molarity

The molarity can be calculated as:

$\mathrm{Molarity}=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}}{\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{L}}$

On converting the volume of solution into liters:

$\begin{array}{c}\mathrm{Volume}\mathrm{in}\mathrm{L}=2.00\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}\\ =0.002\mathrm{L}\end{array}$

On calculating:

$\begin{array}{c}\mathrm{Molarity}\mathrm{of}\mathrm{lead}\mathrm{nitrate}=\frac{0.0122\mathrm{mole}}{0.002\mathrm{L}}\\ =6.10\mathrm{M}\end{array}$

Calculating the concentration of solution

The formula is:

${\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2$

The values are:

$\begin{array}{l}{\mathrm{M}}_1=6.10\text{\hspace{0.17em}}\mathrm{M}\\ {\mathrm{V}}_1=80\mathrm{mL}\\ {\mathrm{M}}_2=\mathrm{Molarity}\mathrm{of}\mathrm{original}\mathrm{solution}\\ {\mathrm{V}}_2=100\mathrm{mL}\end{array}$

On substituting,

$\begin{array}{l}6.10\mathrm{M}\times 80\text{}\mathrm{mL}={\mathrm{M}}_2\times 100\mathrm{mL}\\ {\mathrm{M}}_2=\frac{6.10\mathrm{M}\times 80\mathrm{mL}}{100\text{\hspace{0.17em}}\mathrm{mL}}\\ {\mathrm{M}}_2=\mathbf{4}\mathbf{.}\mathbf{88}\mathbf{M}\end{array}$

Interpretation

Thus, the concentration of the initial lead nitrate solution was estimated using the dilution rate.