Chapter 16, Problem 152E

A chemist combines $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate. (a) How many grams of lead $\left(\text{II}\right)$ iodide will precipitate? (b) What is the final molarity of the potassium ion? (c) What is the final molarity of the lead $\left(\text{II}\right)$ or iodide ion, whichever one is in excess?

Interpretation Introduction

(a)

Interpretation:

The mass of lead $\left(\text{II}\right)$ iodide that would precipitate in the reaction of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is to be calculated.

Concept introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

The relation between number of moles and mass of a substance is given by the expression as shown below.

$\text{n}=\frac{\text{m}}{\text{Molar}\text{mass}}$

Answer to Problem 152E

The mass of lead $\left(\text{II}\right)$ iodide that would precipitate in the reaction of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is $4.45\text{g}$.

Explanation of Solution

The molarity of lead $\left(\text{II}\right)$ nitrate solution is $0.530\text{M}$.

The volume of lead $\left(\text{II}\right)$ nitrate solution is $20.0\text{mL}$.

The conversion of volume in $\text{L}$ is shown below.

$\begin{array}{rll}\text{V}& =& \left(20.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 20.0\times {10}^{-3}\text{L}\end{array}$

The molarity of potassium iodide solution is $0.322\text{M}$.

The volume of potassium iodide solution is $60.0\text{mL}$.

The conversion of volume in $\text{L}$ is shown below.

$\begin{array}{rll}\text{V}& =& \left(60.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 60.0\times {10}^{-3}\text{L}\end{array}$

The molar mass of lead $\left(\text{II}\right)$ iodide is $461.0\text{g}/\text{mol}$.

The molarity of a solution is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ is the number of moles of the solute.

• $\text{V}$ is the volume of the solution.

Rearrange the above equation for the value of $\text{n}$.

$\text{n}=\text{MV}$

Substitute the values of molarity and volume of potassium iodide solution in above expression.

$\begin{array}{rll}\text{n}& =& \left(0.322\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(60.0\times {10}^{-3}\text{L}\right)\\ & =& 19.32\times {10}^{-3}\text{mol}\end{array}$

The number of moles of potassium iodide present in solution is $19.32\times {10}^{-3}\text{mol}$.

Substitute the values of molarity and volume of lead $\left(\text{II}\right)$ nitrate solution in above expression.

$\begin{array}{rll}\text{n}& =& \left(0.530\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(20.0\times {10}^{-3}\text{L}\right)\\ & =& 10.6\times {10}^{-3}\text{mol}\end{array}$

The number of moles of lead $\left(\text{II}\right)$ nitrate present in solution is $10.6\times {10}^{-3}\text{mol}$.

The reaction between potassium iodide and lead $\left(\text{II}\right)$ nitrate is shown below.

$2\text{KI}+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\to {\text{PbI}}_{\text{2}}+2{\text{KNO}}_{\text{3}}$

Two moles of potassium iodide reacts with one mole of lead $\left(\text{II}\right)$ nitrate. The available number of moles of lead $\left(\text{II}\right)$ nitrate is more than half of number of moles of potassium iodide. Therefore, potassium iodide is the limiting reagent of the reaction.

Two moles of potassium iodide produced one mole of lead $\left(\text{II}\right)$ iodide. Therefore, the relation between the number of moles of potassium iodide and lead $\left(\text{II}\right)$ iodide is given by the expression as shown below.

${\text{n}}_{{\text{PbI}}_2}=\frac{{\text{n}}_{\text{KI}}}{2}$ …(1)

Where,

• ${\text{n}}_{\text{KI}}$ is the number of moles of potassium iodide.

• ${\text{n}}_{{\text{PbI}}_2}$ is the number of moles of lead $\left(\text{II}\right)$ iodide.

Substitute the value of ${\text{n}}_{\text{KI}}$ in the equation (1).

$\begin{array}{rll}{\text{n}}_{{\text{PbI}}_2}& =& \frac{19.32\times {10}^{-3}\text{mol}}{2}\\ & =& 9.66\times {10}^{-3}\text{mol}\end{array}$

The relation between number of moles and mass of a substance is given by the expression as shown below.

$\text{m}=\text{n}\times \text{Molar}\text{mass}$

Where,

• $\text{m}$ is the mass of the substance.

• $\text{n}$ is the number of moles of the substance.

Substitute the value of number of moles and molar mass of lead $\left(\text{II}\right)$ iodide in the above equation.

$\begin{array}{rll}\text{m}& =& \left(9.66\times {10}^{-3}\text{mol}\right)\left(461.0\text{g}/\text{mol}\right)\\ & =& 4.45\text{g}\end{array}$

Therefore, the mass of lead $\left(\text{II}\right)$ iodide that would precipitate in the reaction of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is $4.45\text{g}$.

Conclusion

The mass of lead $\left(\text{II}\right)$ iodide that would precipitate in the reaction of potassium iodide with lead $\left(\text{II}\right)$ nitrate is $4.45\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The final molarity of the potassium ion in solution of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is to be calculated.

Concept introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

The relation between number of moles and mass of a substance is given by the expression as shown below.

$\text{n}=\frac{\text{m}}{\text{Molar}\text{mass}}$

Answer to Problem 152E

The final molarity of the potassium ion in solution of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is $0.2415\text{M}$.

Explanation of Solution

The number of moles of potassium iodide present in solution is $19.32\times {10}^{-3}\text{mol}$.

The dissociation of potassium iodide in aqueous solution is shown below.

$\text{KI}\left(\text{aq}\right)\to {\text{K}}^{+}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)$

One mole of $\text{KI}$ gives one mole of potassium ion. Therefore, the number of moles of potassium ions present in solution is $19.32\times {10}^{-3}\text{mol}$.

The total volume of solution of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is shown below.

${\text{V}}_{\text{t}}={\text{V}}_{\text{KI}}+{\text{V}}_{\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$

Where,

• ${\text{V}}_{\text{KI}}$ is the volume of potassium iodide.

• ${\text{V}}_{\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ is the volume of lead $\left(\text{II}\right)$ nitrate.

Substitute the values of ${\text{V}}_{\text{KI}}$ and ${\text{V}}_{\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ in the above equation.

$\begin{array}{rll}{\text{V}}_{\text{t}}& =& 60.0\text{mL}+20.0\text{mL}\\ & =& \left(80.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 80.0\times {10}^{-3}\text{L}\end{array}$

The molarity of a solution is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ is the number of moles of the solute.

• $\text{V}$ is the volume of the solution.

Substitute the values of number of moles of potassium ion and total volume of the solution in the above equation.

$\begin{array}{rll}\text{M}& =& \frac{19.32\times {10}^{-3}\text{mol}}{80.0\times {10}^{-3}\text{L}}\\ & =& \left(0.2415\text{mol}/\text{L}\right)\left(\frac{1\text{M}}{1\text{mol}/\text{L}}\right)\\ & =& 0.2415\text{M}\end{array}$

Therefore, the final molarity of the potassium ion in solution of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is $0.2415\text{M}$.

Conclusion

The final molarity of the potassium ion in solution is $0.2415\text{M}$.

Interpretation Introduction

(c)

Interpretation:

The final molarity of the ion out of lead $\left(\text{II}\right)$ or iodide ion which is present in excess is to be calculated.

Concept introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

The relation between number of moles and mass of a substance is given by the expression as shown below.

$\text{n}=\frac{\text{m}}{\text{Molar}\text{mass}}$

Answer to Problem 152E

The final molarity of the lead $\left(\text{II}\right)$ ion in solution of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is $0.01175\text{M}$.

Explanation of Solution

The reaction between potassium iodide and lead $\left(\text{II}\right)$ nitrate is shown below.

$2\text{KI}+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\to {\text{PbI}}_{\text{2}}+2{\text{KNO}}_{\text{3}}$

Potassium iodide is the limiting reagent. Therefore, all the iodide ion of potassium iodide has been consumed to precipitated lead $\left(\text{II}\right)$ iodide and no iodide ion is present in the solution.

The initial number of moles of lead $\left(\text{II}\right)$ nitrate present in solution is $10.6\times {10}^{-3}\text{mol}$.

The number of moles of lead $\left(\text{II}\right)$ iodide precipitated is $9.66\times {10}^{-3}\text{mol}$.

One mole of lead $\left(\text{II}\right)$ nitrate produces one mole of lead $\left(\text{II}\right)$ iodide. Therefore, the number of moles of lead $\left(\text{II}\right)$ nitrate reacted in the solution is $9.66\times {10}^{-3}\text{mol}$.

The final number of moles of lead $\left(\text{II}\right)$ nitrate in solution is calculated as shown below.

$\begin{array}{rll}{\text{n}}_{\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}& =& 10.6\times {10}^{-3}\text{mol}-9.66\times {10}^{-3}\text{mol}\\ & =& 0.94\times {10}^{-3}\text{mol}\end{array}$

The dissociation of lead $\left(\text{II}\right)$ nitrate in aqueous solution is shown below.

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right)\to {\text{Pb}}^{2+}\left(\text{aq}\right)+2{\text{NO}}_{\text{3}}{}^{-}\left(\text{aq}\right)$

One mole of lead $\left(\text{II}\right)$ nitrate gives one mole of lead ion. Therefore, the number of moles of lead ions present in solution is $0.94\times {10}^{-3}\text{mol}$.

The total volume of solution of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is shown below.

${\text{V}}_{\text{t}}={\text{V}}_{\text{KI}}+{\text{V}}_{\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$

Where,

• ${\text{V}}_{\text{KI}}$ is the volume of potassium iodide.

• ${\text{V}}_{\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ is the volume of lead $\left(\text{II}\right)$ nitrate.

Substitute the values of ${\text{V}}_{\text{KI}}$ and ${\text{V}}_{\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ in the above equation.

$\begin{array}{rll}{\text{V}}_{\text{t}}& =& 60.0\text{mL}+20.0\text{mL}\\ & =& \left(80.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 80.0\times {10}^{-3}\text{L}\end{array}$

The molarity of a solution is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ is the number of moles of the solute.

• $\text{V}$ is the volume of the solution.

Substitute the values of number of moles of lead ion and total volume of the solution in the above equation.

$\begin{array}{rll}\text{M}& =& \frac{0.94\times {10}^{-3}\text{mol}}{80.0\times {10}^{-3}\text{L}}\\ & =& \left(0.01175\text{mol}/\text{L}\right)\left(\frac{1\text{M}}{1\text{mol}/\text{L}}\right)\\ & =& 0.01175\text{M}\end{array}$

Therefore, the final molarity of the lead $\left(\text{II}\right)$ ion in solution of $60.0\text{mL}$ of $0.322\text{M}$ potassium iodide with $20.0\text{mL}$ of $0.530\text{M}$ lead $\left(\text{II}\right)$ nitrate is $0.01175\text{M}$.

Conclusion

The final molarity of the lead $\left(\text{II}\right)$ ion in solution is $0.01175\text{M}$.