Chapter 14, Problem 14.91QE

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for the reaction has to be determined.

Concept Introduction:

There are three kinds of equations used to represent the chemical reaction. These are molecular equation, total ionic equation and net ionic equations.

Net ionic equation is a chemical equation that contains only those ions that participate in the particular reaction. In this, only useful ions are present. These are commonly used reactions in acid-base reactions, redox reactions and double displacement reactions.

Answer to Problem 14.91QE

The net ionic equation for the reaction is ${\text{Ag}}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)\to \text{AgCl}\left(\text{s}\right)$.

Explanation of Solution

The complete reaction between $\text{KCl}$ and ${\text{AgNO}}_{\text{3}}$ is as follows:

  $\text{KCl}\left(\text{aq}\right)+{\text{AgNO}}_3\left(\text{aq}\right)\to {\text{KNO}}_3\left(\text{aq}\right)+\text{AgCl}\left(\text{aq}\right)$

The total ionic equation for the reaction is as follows:

  ${\text{K}}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Ag}}^{+}\left(\text{aq}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)\to {\text{K}}^{+}\left(\text{aq}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)+\text{AgCl}\left(\text{s}\right)$

The common ions on both side of equation get cancelled out and the net ionic equation is as follows:

  ${\text{Ag}}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)\to \text{AgCl}\left(\text{s}\right)$

(b)

Interpretation Introduction

Interpretation:

The mass of insoluble product has to be calculated.

Concept Introduction:

Molarity is defined as number of moles of solute that are dissolved in one litre of the solution. It is represented by $M$.

The formula to calculate the molarity of solution is as follows:

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

Answer to Problem 14.91QE

The mass of insoluble product is $0.0719\text{ g}$.

Explanation of Solution

Rearrange equation (1) to calculate moles of solute.

  $\text{Moles of solute}=\left(\text{Molarity of solution}\right)\left(\text{Volume of solution}\right)$        (2)

Substitute $0.0259M$ for molarity and $20\text{ mL}$ for volume in equation (2) to calculate the moles of ${\text{AgNO}}_{\text{3}}$.

  $\begin{array}{c}{\text{Moles of AgNO}}_{\text{3}}=\left(\frac{\text{0}\text{.0259 mol}}{1\text{ L}}\right)\left(\text{20 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.000518\text{ mol}\end{array}$

Substitute $0.0502M$ for molarity and $10\text{ mL}$ for volume in equation (2) to calculate the moles of $\text{KCl}$.

  $\begin{array}{c}\text{Moles of KCl}=\left(\frac{\text{0}\text{.0502 mol}}{1\text{ L}}\right)\left(\text{10 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.000502\text{ mol}\end{array}$

Since moles of $\text{KCl}$ are less than those of ${\text{AgNO}}_{\text{3}}$ so $\text{KCl}$ is the limiting reagent and number of moles of $\text{AgCl}$ becomes equal to moles of $\text{KCl}$. Therefore number of moles of $\text{AgCl}$ is $0.000502\text{ mol}$.

The formula to calculate moles of $\text{AgCl}$ is as follows:

  $\text{Moles of AgCl}=\frac{\text{Mass of AgCl}}{\text{Molar mass of AgCl}}$        (3)

Rearrange equation (3) to calculate mass of $\text{AgCl}$.

  $\text{Mass of AgCl}=\left(\text{Moles of AgCl}\right)\left(\text{Molar mass of AgCl}\right)$        (4)

Substitute $0.000502\text{ mol}$ for moles of $\text{AgCl}$ and $143.32\text{ g/mol}$ for molar mass of $\text{AgCl}$ in equation (4).

  $\begin{array}{c}\text{Mass of AgCl}=\left(0.000502\text{ mol}\right)\left(\frac{143.32\text{ g}}{1\text{ mol}}\right)\\ =0.0719\text{ g}\end{array}$

Therefore mass of insoluble product is $0.0719\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

The concentrations of anions in equilibrium solution have to be determined.

Concept Introduction:

Solubility product is the equilibrium constant for the reaction that occurs when an ionic compound is dissolved to produce its constituent ions. It is represented by $K_{\text{sp}}$. Consider ${\text{A}}_x{\text{B}}_y$ to be an ionic compound. Its dissociation occurs as follows:

  ${\text{A}}_x{\text{B}}_y\rightleftharpoons x{\text{A}}^{y+}+y{\text{A}}^{x-}$

The expression for its $K_{\text{sp}}$ is as follows:

  $K_{\text{sp}}={\left[{\text{A}}^{y+}\right]}^x{\left[{\text{B}}^{x-}\right]}^y$

Answer to Problem 14.91QE

The concentration of ${\text{Cl}}^{-}$ is $3.4\times {10}^{-7}M$ and that of ${\text{NO}}_3^{-}$ is $0.0173M$.

Explanation of Solution

The dissociation reaction of $\text{AgCl}$ is as follows:

  $\text{AgCl}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

The formula to calculate the solubility product constant of $\text{AgCl}$ is as follows:

  $K_{\text{sp}}=\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]$        (5)

Rearrange equation (5) to calculate $\left[{\text{Cl}}^{-}\right]$.

  $\left[{\text{Cl}}^{-}\right]=\frac{K_{\text{sp}}}{\left[{\text{Ag}}^{+}\right]}$        (6)

Substitute $1.8\times {10}^{-10}$ for $K_{\text{sp}}$ (Refer to Appendix F) and $0.000518M$ for $\left[{\text{Ag}}^{+}\right]$ in equation (6).

  $\begin{array}{c}\left[{\text{Cl}}^{-}\right]=\frac{1.8\times {10}^{-10}}{0.000518}\\ =3.4\times {10}^{-7}M\end{array}$

Therefore the concentration of ${\text{Cl}}^{-}$ is $3.4\times {10}^{-7}M$.

The expression for molarity equation is as follows:

  $M_{\text{conc}}{V}_{\text{conc}}=M_{\text{dil}}{V}_{\text{dil}}$        (7)

Here,

$M_{\text{conc}}$ is the concentration of concentrated solution.

$M_{\text{dil}}$ is the concentration of diluted solution.

$V_{\text{conc}}$ is the volume of concentrated solution.

$V_{\text{dil}}$ is the volume of diluted solution.

Rearrange equation (7) to calculate $M_{\text{dil}}$.

  $M_{\text{dil}}=\frac{M_{\text{conc}}{V}_{\text{conc}}}{V_{\text{dil}}}$        (8)

The total volume after dilution can be calculated as follows:

  $\begin{array}{c}{V}_{\text{dil}}=10\text{ mL}+2\text{0 mL}\\ =\text{30 mL}\end{array}$

Substitute $1.0\times {10}^{-6}M$ for $M_{\text{conc}}$, $20\text{ mL}$ for $V_{\text{conc}}$ and $30\text{ mL}$ for $V_{\text{dil}}$ in equation (8) to calculate the molarity of ${\text{NO}}_3^{-}$.

  $\begin{array}{c}{\text{Molarity of NO}}_3^{-}=\frac{\left(\text{0}\text{.0259 }M\right)\left(20\text{ mL}\right)}{30\text{ mL}}\\ =0.0173M\end{array}$

Therefore the concentration of ${\text{NO}}_3^{-}$ is $0.0173M$.