Chapter 11, Problem 8PP

Interpretation Introduction

Interpretation:

The volume of lead nitrate required to react with given potassium chromate solution and the mass of lead chromate formed are to be determined.

Concept Introduction:

Titration is a method to determine the concentration of a substance in solution by allowing it to react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances comes to an end. In precipitation reactions, on reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

Answer to Problem 8PP

Solution:

The volume of lead nitrate required to react with given potassium chromate solution is $114.286\text{ mL}$ and mass of lead chromate formed is $3.878\text{ g}$ .

Explanation of Solution

Given Information:

The molarity of lead nitrate solution is $0.105\text{ M}$ and the molarity of potassium chromate is $0.120\text{ M}$ . The volume of potassium chromate is $100\text{ ml}$ .

The given chemical reaction is,

$\text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)+{\text{ K}}_2{\text{CrO}}_4\left(aq\right)\to {\text{ PbCrO}}_4\left(s\right)+{\text{ 2KNO}}_3\left(aq\right)$

Thus, one mole of lead nitrate reacts with one mole of potassium chromate to form a mole of lead chromate.

The normality $\left(N\right)$ of a solution is defined as the ratio of the molarity $\left(M\right)$ of the solution to the oxidation state ions of solute. The oxidation state of both the solution is $2$ . Therefore, the formula becomes:

$N=\frac{M}{2}$ …… (1)

The product of volume of lead nitrate solution $\left(V_1\right)$ and the normality of lead nitrate solution $\left(N_1\right)$ is equal to the product of volume of potassium chromate solution $\left(V_2\right)$ and normality of potassium chromate solution $\left(N_2\right)$ .

$V_1{N}_1=V_2{N}_2$ …… (2)

$N_1$ and $N_2$ can be calculated by substituting $M_1$ as $0.105\text{ M}$ and $M_2$ as $0.120\text{ M}$ in equation (1) as follows:

$\begin{array}{c}{N}_1=\frac{0.105\text{ M}}{2}\\ =0.0525\text{ N}\\ N_2=\frac{0.120}{2}\\ =0.06\text{ N}\end{array}$

The volume of lead nitrate solution is determined by using equation (2). Substitute $V_2$ as $\text{100 mL}$ , $N_1$ as $\text{0}\text{.0525 N}$ and $N_2$ as $0.06\text{ N}$ in equation (2) as follows:

$\begin{array}{c}{V}_1\times \text{ 0}\text{.0525 N}=\text{100 mL}\times \text{ 0}\text{.06 N}\\ V_1=\frac{\text{100 mL}\times \text{0}\text{.06 N}}{\text{0}\text{.0525 N}}\\ =\text{114}\text{.286 mL}\end{array}$

Thus, the volume of lead nitrate required is $114.286\text{ mL}$ .

Convert volume units from milliliters to litres as follows:

$\begin{array}{c}1\text{ L}=1000\text{ mL}\\ \text{1}\text{mL}=\frac{1\text{ L}}{1000\text{mL}}\end{array}$

Convert $114.286\text{ mL}$ volume from milliliters to litres as follows:

$\begin{array}{c}\text{114}\text{.286}\text{mL}=\frac{1\text{ L}}{1000\text{mL}}\times \text{114}\text{.286}\text{mL}\\ =0.1143\text{ L}\end{array}$

The molarity of the solution $\left(\text{M}\right)$ is defined as the ratio of number of moles $\left(n\right)$ to the volume of the solution $\left(V\right)$ in litres. The number of moles is equal to the ratio of mass of solute $\left(m\right)$ to the molar mass $\left(MM\right)$ of the solution.

$\text{M}=\frac{n}{V}=\frac{m}{MM\times V}$ …… (3)

Substitute $\text{M}$ as $0.105{\text{ mol L}}^{-1}$ , $MM$ as $331.2{\text{g mol}}^{-1}$ and $V$ as $0.1143\text{L}$ in equation (3) as follows:

$\begin{array}{c}0.105{\text{ mol L}}^{-1}=\frac{m}{331.2{\text{ g mol}}^{-1}\times 0.1143\text{L}}\\ m=0.105{\text{mol L}}^{-1}\times 0.1143\text{L}\times 331.2{\text{ g mol}}^{-1}\\ =3.974\text{ g}\end{array}$

From the equation, it can be summarized that one mole of lead nitrate produces one mole of lead chromate.

$\begin{array}{c}\text{1 mole of lead nitrate = one mole lead chromate}\\ \text{331}\text{.2 g of lead nitrate = 323}\text{.19 g lead chromate}\\ \text{1 g of lead nitrate = }\frac{\text{323}\text{.19 g lead chromate}}{331.2\text{g lead nitrate}}\end{array}$

For $\text{3}\text{.974 g}$ of lead nitrate, the lead chromate produced is,

$\begin{array}{c}\text{3}\text{.974 g lead nitrate produces}=\frac{\text{323}\text{.19 g lead chromate}}{\text{331}\text{.2}\text{g lead nitrate }}\times \text{3}\text{.974 g lead nitrate}\\ =\text{3}\text{.878 g lead chromate}\end{array}$

Thus, the mass of lead chromate formed is $\text{3}\text{.878 g}$ .

Conclusion

The volume of lead nitrate required to react with given potassium chromate solution is $114.286\text{ mL}$ and the mass of lead chromate formed is $\text{3}\text{.878 g}$ .