Chapter F, Problem G.21E

(a)

Interpretation Introduction

Interpretation:

The concentration of potassium ions in the solution that is prepared has to be calculated.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters.  Unit of molarity is $\text{mol}\cdot {\text{L}}^{-1}$.  The equation for molarity is given as follows;

    $\text{Molarity (M)}=\frac{\text{Amount of solute in moles}}{\text{Volume of solution in liters}}$

From the above equation, the mass of solute can be calculated if the concentration and volume is known.

Answer to Problem G.21E

The concentration of potassium ions is $4.58\times {10}^{-2}\text{M}$.

Explanation of Solution

The solution is said to be prepared by dissolving $\text{0}\text{.500}\text{g}$ of $\text{KCl}$, $\text{0}\text{.500}\text{g}$ of ${\text{K}}_{\text{2}}\text{S}$, and $\text{0}\text{.500}\text{g}$ of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ in $\text{500}\text{.0}\text{mL}$ of water.  The number of moles of potassium ions that is contributed by each component is calculated as follows;

Number of moles of potassium from $\text{0}\text{.500}\text{g}$ of $\text{KCl}$:

Molar mass of $\text{KCl}$ is $74.5\text{g}\cdot {\text{mol}}^{-1}$.  Number of moles of $\text{KCl}$ is calculated as follows;

    $\begin{array}{c}{n}_{\text{(KCl)}}=\frac{0.500\text{g}}{74.5\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0067\text{mol}\end{array}$

One mole of $\text{KCl}$ gives one mole of potassium ions.  Therefore, $\text{0}\text{.0067}\text{mol}$ of $\text{KCl}$ gives $\text{0}\text{.0067}\text{mol}$ of potassium ions.

Number of moles of potassium from $\text{0}\text{.500}\text{g}$ of ${\text{K}}_{\text{2}}\text{S}$:

Molar mass of ${\text{K}}_{\text{2}}\text{S}$ is $110.3\text{g}\cdot {\text{mol}}^{-1}$.  Number of moles of ${\text{K}}_{\text{2}}\text{S}$ is calculated as follows;

    $\begin{array}{c}{n}_{{\text{(K}}_{\text{2}}\text{S)}}=\frac{0.500\text{g}}{110.3\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0045\text{mol}\end{array}$

One mole of ${\text{K}}_{\text{2}}\text{S}$ gives two moles of potassium ions.  Therefore, $\text{0}\text{.0045}\text{mol}$ of ${\text{K}}_{\text{2}}\text{S}$ gives $\text{0}\text{.0090}\text{mol}$ of potassium ions.

Number of moles of potassium from $\text{0}\text{.500}\text{g}$ of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$:

Molar mass of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $212.27\text{g}\cdot {\text{mol}}^{-1}$.  Number of moles of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ is calculated as follows;

    $\begin{array}{c}{n}_{{\text{(K}}_{\text{3}}{\text{PO}}_{\text{4}}\text{)}}=\frac{0.500\text{g}}{212.27\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0024\text{mol}\end{array}$

One mole of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ gives three moles of potassium ions.  Therefore, $\text{0}\text{.0024}\text{mol}$ of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ gives $\text{0}\text{.0072}\text{mol}$ of potassium ions.

Thus, the total potassium ions in the solution is calculated by summing up all the moles of potassium ions.  This is done as follows;

    $\begin{array}{c}\text{Total}\text{No}\text{.}\text{of}{\text{K}}^{\text{+}}\text{ions}=\text{0}\text{.0067}\text{mol}+\text{0}\text{.0090}\text{mol}+\text{0}\text{.0072}\text{mol}\\ =0.0229\text{mol}\end{array}$

The volume of the solution is given as $\text{0}\text{.500}\text{L}$.  Therefore, the concentration of potassium ions is calculated as shown below;

    $\begin{array}{c}\text{Concentration}\text{of}{\text{K}}^{\text{+}}\text{ions}=\frac{\text{0}\text{.0229}\text{mol}}{\text{0}\text{.500}\text{L}}\\ =0.0458\text{M}\\ =4.58\times {10}^{-2}\text{M}\end{array}$

Thus, the concentration of potassium ions in $\text{0}\text{.500}\text{L}$ of solution is $4.58\times {10}^{-2}\text{M}$.

(b)

Interpretation Introduction

Interpretation:

The concentration of sulfide ions in the solution that is prepared has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem G.21E

The concentration of sulfide ions is $9\times {10}^{-3}\text{M}$.

Explanation of Solution

The solution is said to be prepared by dissolving $\text{0}\text{.500}\text{g}$ of $\text{KCl}$, $\text{0}\text{.500}\text{g}$ of ${\text{K}}_{\text{2}}\text{S}$, and $\text{0}\text{.500}\text{g}$ of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ in $\text{500}\text{.0}\text{mL}$ of water.  The number of moles of sulfide ions that is contributed by each component is calculated as follows;

Among the three components, only sulfide ion is contributed by ${\text{K}}_{\text{2}}\text{S}$.  Therefore, the number of moles of sulfide from $\text{0}\text{.500}\text{g}$ of ${\text{K}}_{\text{2}}\text{S}$:

Molar mass of ${\text{K}}_{\text{2}}\text{S}$ is $110.3\text{g}\cdot {\text{mol}}^{-1}$.  Number of moles of ${\text{K}}_{\text{2}}\text{S}$ is calculated as follows;

    $\begin{array}{c}{n}_{{\text{(K}}_{\text{2}}\text{S)}}=\frac{0.500\text{g}}{110.3\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0045\text{mol}\end{array}$

One mole of ${\text{K}}_{\text{2}}\text{S}$ gives one mole of sulfide ions.  Therefore, $\text{0}\text{.0045}\text{mol}$ of ${\text{K}}_{\text{2}}\text{S}$ gives $\text{0}\text{.0045}\text{mol}$ of sulfide ions.

The volume of the solution is given as $\text{0}\text{.500}\text{L}$.  Therefore, the concentration of sulfide ions is calculated as shown below;

    $\begin{array}{c}\text{Concentration}\text{of}{\text{S}}^{\text{2-}}\text{ions}=\frac{\text{0}\text{.0045}\text{mol}}{\text{0}\text{.500}\text{L}}\\ =0.009\text{M}\\ =9\times {10}^{-3}\text{M}\end{array}$

Thus, the concentration of sulfide ions in $\text{0}\text{.500}\text{L}$ of solution is $9\times {10}^{-3}\text{M}$.