Chapter 14, Problem 14.98QE

(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 14.98QE

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 ionization reaction of ${\text{K}}_2{\text{SO}}_4$ is as follows:

  ${\text{K}}_2{\text{SO}}_4\left(\text{aq}\right)\to 2{\text{K}}^{+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)$

The total ionic equation after addition of ${\text{Ba}}^{2+}$ ions is as follows:

  ${\text{Ba}}^{2+}\left(\text{aq}\right)+2{\text{K}}^{+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)\to {\text{BaSO}}_{\text{4}}\left(\text{s}\right)+2{\text{K}}^{+}\left(\text{aq}\right)$

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

  ${\text{Ba}}^{2+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)\to {\text{BaSO}}_{\text{4}}\left(\text{s}\right)$

(b)

Interpretation Introduction

Interpretation:

The mass of precipitate 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 14.98QE

The mass of precipitate is $3.407\times {10}^{-4}\text{ g}$.

Explanation of Solution

The concentration of ${\text{Ba}}^{2+}$ ion in $\text{mol/L}$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{Ba}}^{2+}\right]=\left(\frac{\text{10 μg}}{1\text{ L}}\right)\left(\frac{{10}^{-6}\text{ g}}{1\text{ μg}}\right)\left(\frac{1\text{ mol}}{\text{137}\text{.33 g}}\right)\\ =7.28\times {10}^{-5}M\end{array}$

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 $7.28\times {10}^{-5}M$ for molarity and $20\text{ mL}$ for volume in equation (2) to calculate the moles of ${\text{BaSO}}_4$.

  $\begin{array}{c}{\text{Moles of BaSO}}_4=\left(\frac{7.28\times {10}^{-5}\text{ mol}}{1\text{ L}}\right)\left(\text{20 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =1.45\times {10}^{-6}\text{ mol}\end{array}$

Substitute $0.050M$ for molarity and $25\text{ mL}$ for volume in equation (2) to calculate the moles of ${\text{K}}_2{\text{SO}}_4$.

  $\begin{array}{c}{\text{Moles of K}}_2{\text{SO}}_4=\left(\frac{\text{0}\text{.050 mol}}{1\text{ L}}\right)\left(\text{25 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.00125\text{ mol}\end{array}$

Since moles of ${\text{BaSO}}_4$ are less than those of ${\text{K}}_2{\text{SO}}_4$ so ${\text{BaSO}}_4$ is the limiting reagent and number of moles of ${\text{Ba}}^{2+}$ becomes equal to moles of ${\text{BaSO}}_4$. Therefore number of moles of ${\text{Ba}}^{2+}$ is $1.46\times {10}^{-6}\text{ mol}$.

The formula to calculate moles of ${\text{BaSO}}_4$ is as follows:

  ${\text{Moles of BaSO}}_4=\frac{{\text{Mass of BaSO}}_4}{{\text{Molar mass of BaSO}}_4}$        (3)

Rearrange equation (3) to calculate mass of ${\text{BaSO}}_4$.

  ${\text{Mass of BaSO}}_4=\left({\text{Moles of BaSO}}_4\right)\left({\text{Molar mass of BaSO}}_4\right)$        (4)

Substitute $1.46\times {10}^{-6}\text{ mol}$ for moles of ${\text{BaSO}}_4$ and $233.4\text{ g/mol}$ for molar mass of ${\text{BaSO}}_4$ in equation (4).

  $\begin{array}{c}{\text{Mass of BaSO}}_4=\left(1.46\times {10}^{-6}\text{ mol}\right)\left(\frac{233.4\text{ g}}{1\text{ mol}}\right)\\ =3.407\times {10}^{-4}\text{ g}\end{array}$

Therefore mass of precipitate is $3.407\times {10}^{-4}\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

The concentrations of all species 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 14.98QE

The concentration of ${\text{Ba}}^{2+}$ and ${\text{SO}}_4^{2-}$ is $1.05\times {10}^{-5}M$.

Explanation of Solution

The dissociation reaction of ${\text{BaSO}}_4$ is as follows:

  ${\text{BaSO}}_4\left(\text{s}\right)\rightleftharpoons {\text{Ba}}^{2+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)$

The formula to calculate the solubility product constant of ${\text{BaSO}}_4$ is as follows:

  $K_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]\left[{\text{SO}}_4^{2-}\right]$        (5)

Substitute $s$ for $\left[{\text{Ba}}^{2+}\right]$ and $s$ for $\left[{\text{SO}}_4^{2-}\right]$ in equation (5).

  $K_{\text{sp}}=\left(s\right)\left(s\right)$        (6)

Rearrange equation (6) to calculate $s$.

  $s=\sqrt{K_{\text{sp}}}$        (7)

Substitute $1.1\times {10}^{-10}$ for $K_{\text{sp}}$ (Refer to Appendix F) in equation (7).

  $\begin{array}{c}s=\sqrt{1.1\times {10}^{-10}}\\ =1.05\times {10}^{-5}M\end{array}$

Therefore the concentration of ${\text{Ba}}^{2+}$ and ${\text{SO}}_4^{2-}$ is $1.05\times {10}^{-5}M$.