Chapter 9, Problem 10PE

Interpretation Introduction

Interpretation:

The conventional equation, total ionic equation, and net ionic equation for the reaction between solutions of lead $\left(\text{II}\right)$ nitrate and potassium iodide are to be stated.

Concept introduction:

A conventional equation is the balanced chemical reaction equation in the form of molecular formula of the reactants and products with their states in the reaction.

The total ionic equation is the representation of conventional equation by breaking the species which is forming ions in the solutions and writing all the ions on both sides.

The net ionic equation is the resultant of the total ionic equation by cancelling out the same ions present on both sides.

Answer to Problem 10PE

The conventional equation for the reaction between solutions of lead $\left(\text{II}\right)$ nitrate and potassium iodide is shown below.

$\text{2KI}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right)\to {\text{PbI}}_{\text{2}}\left(\text{s}\right)+2{\text{KNO}}_3\left(\text{aq}\right)$

The total ionic equation for the above equation is shown below.

${\text{2K}}^{+}\left(\text{aq}\right)+2{\text{I}}^{-}\left(\text{aq}\right)+{\text{Pb}}^{2+}\left(\text{aq}\right)+2{\text{NO}}_3^{-}\left(\text{aq}\right)\to {\text{PbI}}_{\text{2}}\left(\text{s}\right)+{\text{2K}}^{+}\left(\text{aq}\right)+2{\text{NO}}_3^{-}\left(\text{aq}\right)$

The net ionic equation for the above equation is shown below.

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

Explanation of Solution

The reaction between solutions of lead $\left(\text{II}\right)$ nitrate and potassium iodide, yielding a precipitate of lead $\left(\text{II}\right)$ iodide is given below.

$\text{KI}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right)\to {\text{PbI}}_{\text{2}}\left(\text{s}\right)+{\text{KNO}}_3\left(\text{aq}\right)$

The number of iodine atoms on both sides is not same, therefore, balance $\text{I}$ by multiplying $\text{KI}$ by two on the left-hand side of the equation.

$\text{2KI}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right)\to {\text{PbI}}_{\text{2}}\left(\text{s}\right)+{\text{KNO}}_3\left(\text{aq}\right)$

The number of potassium atom is not same on both sides of the equation, therefore, balance $\text{K}$ by multiplying ${\text{KNO}}_3$ by two on the right-hand side of the equation.

$\text{2KI}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right)\to {\text{PbI}}_{\text{2}}\left(\text{s}\right)+2{\text{KNO}}_3\left(\text{aq}\right)$

The reaction is now completely balanced chemical equation.

Therefore, the net conventional equation is shown below.

$\text{2KI}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right)\to {\text{PbI}}_{\text{2}}\left(\text{s}\right)+2{\text{KNO}}_3\left(\text{aq}\right)$

The total ionic reaction for the above reaction is written by taking the ions involved in the reaction as shown below.

${\text{2K}}^{+}\left(\text{aq}\right)+2{\text{I}}^{-}\left(\text{aq}\right)+{\text{Pb}}^{2+}\left(\text{aq}\right)+2{\text{NO}}_3^{-}\left(\text{aq}\right)\to {\text{PbI}}_{\text{2}}\left(\text{s}\right)+{\text{2K}}^{+}\left(\text{aq}\right)+2{\text{NO}}_3^{-}\left(\text{aq}\right)$

The net ionic reaction is written by cancelling common ions on both sides as shown below.

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

Conclusion

The conventional equation, total ionic equation, and net ionic equation for the reaction between solutions of lead $\left(\text{II}\right)$ nitrate and potassium iodide has been rightfully stated above.