Chapter 19, Problem 79AP

A piece of magnesium metal weighing 1.56 g is placed in 100.0 mL of $\text{0}\text{.100 }M{\text{ AgNO}}_{\text{3}}\text{at 25°C}\text{. Calculate }\left[{\text{Mg}}^{\text{2-}}\right]\text{ and }\left[{\text{Ag}}^{\text{-}}\right]$ in solution at equilibrium. What is the mass of the magnesium left? The volume remains constant.

Interpretation Introduction

Interpretation:

The concentrations of ${\text{Mg}}^{2+}{\text{ and Ag}}^{+}$ and the mass of the magnesium remaining are to be calculated with given mass of magnesium and volume and concentration of ${\text{AgNO}}_{\text{3}}$.

Concept introduction:

The expression to calculate the electrode potential of cell $\left(E_{c\text{ell}}^o\right)$ is:

$E_{c\text{ell}}^o=E_{\text{cathode}}^o-E_{\text{anode}}^o$

Here, $E_{\text{cathode}}^o$ is the standard reduction potential of the cathode and $E_{\text{anode}}^o$ is the standard reduction potential of the anode.

The number of moles of a compound can be calculated as:

$\text{moles=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}$

The number of moles of a compound can be calculated as:

$\text{moles}\text{=}\text{concentration}\text{×}\text{volume}$

The relationship between liters and milliliters can be expressed as: $\text{1mL = 0}\text{.001 L}$

To convert milliliters to liters, the conversion factor is $\frac{0.001\text{ L}}{1\text{ mL}}$

The relation between equilibrium constant and standard reduction potential of a cell is given as:

$\Delta E^{\circ }=\frac{RT}{nF}\ln K$

Here, $n$ is the number of electrons transferred in a reaction, $R$ is the gas constant, $T$ is the temperature, $K$ is the equilibrium constant, and $F$ is the Faraday’s constant.

the equilibrium constant is calculated as:

$K=e^{\frac{n{E}_{\text{cel}l}^o}{0.0257}}$

For a general reaction: $a\text{A}\text{+}b\text{B}\rightleftarrows c\text{C + }d\text{D}$

The general formula for writing equilibrium expression for the reaction is given as:

$K_C=\frac{{\left[\text{C}\right]}^c{}_{eqm}{\left[\text{D}\right]}^d{}_{eqm}}{{\left[\text{A}\right]}^a{}_{eqm}{\left[\text{B}\right]}^b{}_{eqm}}$

Here, $K_C$ is equilibrium constant, $C$ in $K_C$ stands for the concentration. $\left[\text{A}\right]$, $\left[\text{B}\right]$, $\left[\text{C}\right]$, and $\left[\text{D}\right]$ are the equilibrium concentration of reactants A and B and product C and D.

A and B are reactants, C and D are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

Answer to Problem 79AP

Solution: $0.0500\text{ M;}7\times {10}^{-55}\text{ M};1.44\text{ g}$

Explanation of Solution

Given information: The mass of magnesium is $1.56\text{ g}$.

Volume of ${\text{AgNO}}_{\text{3}}$ is $100.0\text{ mL}$.

Concentration of ${\text{AgNO}}_{\text{3}}$ is $0.100\text{ M}$.

The mass of $1\text{ mole}$ of magnesium is equal to its molar mass, i.e., $\text{24}\text{.31 g}$.

$\text{1 mol Mg}=24.31\text{ g}\left[\text{molar mass of Mg}=24.31\text{ g/mol }\right]$

Thus, the conversion factor becomes $\left(\frac{24.31\text{ g Mg}}{\text{1 mol Mg}}\right)$.

Invert the conversion factor and multiply it with $1.56\text{ g}$ of $\text{Mg}$ in order to calculate the moles of $\text{Mg}$ as:

$\begin{array}{c}\text{Moles of Mg}=\left(\frac{\text{1 mol Mg}}{24.31\cancel{\text{g Mg}}}\right)\times 1.56\cancel{\text{g Mg}}\\ =0.0642\text{ mol Mg}\end{array}$

Molarity is the number of moles of solute dissolved in $1\text{}L$ of solution. Thus, the given molarity, i.e., $0.100\text{ M}$ suggests that $0.100\text{ mol}$ is present in $1L$ of solution. Thus, the given molarity can be mathematically expressed as: $\frac{0.100\text{ mol}}{1\text{ L}}$.

Multiply the conversion factor with volume to calculate the moles of ${\text{Ag}}^{+}$ as:

$\begin{array}{c}{\text{Moles of Ag}}^{+}=\frac{0.100\text{ mol}}{1\cancel{\text{L}}}\times 100.0\cancel{\text{mL}}\times \frac{0.001\cancel{\text{L}}}{1\cancel{\text{mL}}}\\ =0.0100{\text{ mol Ag}}^{+}\end{array}$

Write the balanced chemical equation between ${\text{Ag}}^{+}$ and $\text{Mg}$ as:

${\text{2Ag}}^{+}\left(aq\right)+\text{Mg}\left(s\right)\to 2\text{Ag}\left(s\right)+{\text{Mg}}^{2+}\left(aq\right)$

According to the above reaction, $1\text{ mol Mg}$ gives $2{\text{ mol Ag}}^{+}$. So, the conversion factor becomes:

$\left(\frac{1\text{ mol Mg}}{2{\text{ mol Ag}}^{+}}\right)$.

Multiply the conversion factor with the calculated moles of ${\text{Ag}}^{+}$ to calculate the amount of $\text{Mg}$ as:

$\begin{array}{c}\text{Amount of Mg}=\left(\frac{1\text{ mol Mg}}{2\cancel{{\text{mol Ag}}^{+}}}\right)\times 0.0100\cancel{{\text{mol Ag}}^{+}}\\ =0.00500\text{ mol Mg}\end{array}$

Calculate the remaining moles of $\text{Mg}$ as:

$\begin{array}{c}\text{Moles of Mg}=\left(0.0642-0.00500\right)\text{mol}\\ =0.0592\text{ mol}\end{array}$

The mass of $1\text{ mole}$ of magnesium is equal to its molar mass, i.e., $\text{24}\text{.31 g}$.

$\text{1 mol Mg}=24.31\text{ g}\left[\text{molar mass of Mg}=24.31\text{ g/mol }\right]$

Thus, the conversion factor becomes $\left(\frac{24.31\text{ g Mg}}{\text{1 mol Mg}}\right)$.

Multiply the conversion factor with $0.0592\text{ mol}$ of $\text{Mg}$ in order to calculate the remaining mass of $\text{Mg}$ as:

$\begin{array}{c}\text{Mass of Mg}=\left(\frac{24.31\text{ g Mg}}{\text{1 }\cancel{\text{mol Mg}}}\right)\times 0.0592\cancel{\text{mol Mg}}\\ =1.44\text{ g Mg}\end{array}$

According to the above reaction, the mole ratio of $\text{Mg}$ and ${\text{Mg}}^{2+}$ is $1:1$. Calculate the concentration of ${\text{Mg}}^{2+}$ as:

$\begin{array}{c}\left[{\text{Mg}}^{2+}\right]=\frac{0.00500\text{ mol}}{0.100\text{ L}}\\ =0.0500\text{ M}\end{array}$

The formula to calculate the standard electrode potential $\left(E^o\right)$ is:

$E^o=E_{\text{cathode}}^o-E_{\text{anode}}^o$

From the standard reduction potential table, the reduction potential of ${\text{Mg}}^{2+}$ is $-2.37\text{ V}$ and that of ${\text{Ag}}^{+}$ is $0.80\text{ V}$.

From the reaction, it is clear that the reduction of ${\text{Ag}}^{+}$ takes place at the cathode and the oxidation of $\text{Mg}$ takes place at the anode.

Substitute $0.80\text{ V}$ for $E_{\text{cathode}}^o$ and $-2.37\text{ V}$ for $E_{\text{anode}}^o$ in the above equation as:

$\begin{array}{c}{E}^o=0.80\text{ V}-\left(-2.37\text{ V}\right)\\ =3.17\text{ V}\end{array}$

Now, calculate the equilibrium constant as follows:

$K=e^{\frac{n{E}_{\text{cel}l}^o}{0.0257}}$

According to the reaction, the number of electrons transferred is 2.

Substitute 2 for $n$ and $3.17\text{ V}$ for $E_{\text{cell}}^o$ in the above expression as:

$\begin{array}{c}K=e^{\frac{2\times 3.17\text{ V}}{0.0257}}\\ =1\times {10}^{107}\end{array}$

Let $x$ be the amount of ${\text{Mg}}^{2+}$ that is needed to reach the equilibrium. As the value of the equilibrium is very large, the reaction will move in the forward direction, which means that it will be product favored. So, the concentration of ${\text{Ag}}^{+}$ initially will be zero. Make an ICE table as:

$\begin{array}{l}{\text{ 2Ag}}^{+}\left(aq\right)+\text{Mg}\left(s\right)\to 2\text{Ag}\left(s\right)+{\text{Mg}}^{2+}\left(aq\right)\\ \text{Initial}\left(\text{M}\right)\text{ 0}\text{.0000 0}\text{.0500}\\ \text{Change}\left(\text{M}\right)\text{ 2}x-x\\ \text{Equilibrium}\left(\text{M}\right)\text{ 2}x\left(0.0500-x\right)\end{array}$

According to the above, the expression of $K$ will be as follows:

$K=\frac{\left[{\text{Mg}}^{2+}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}$

Substitute $1\times {10}^{107}$ for $K$, $\left(0.0500-x\right)$ for $\left[{\text{Mg}}^{2+}\right]$ and $2x$ for ${\text{Ag}}^{+}$ in the above expression as:

$\begin{array}{c}1\times {10}^{107}=\frac{\left(0.0500-x\right)}{{\left(2x\right)}^2}\\ 1\times {10}^{107}=\frac{\left(0.0500-x\right)}{{\left(2x\right)}^2}\\ {\left(2x\right)}^2=\frac{\left(0.0500-x\right)}{1\times {10}^{107}}\end{array}$

As the value of $K$ is very large, the term $\left(0.0500-x\right)$ will be approximately equal to 0.0500.

$\begin{array}{c}{\left(2x\right)}^2=\frac{0.0500}{1\times {10}^{107}}\\ =0.0500\times {10}^{-107}\\ =5.00\times {10}^{-109}\\ =50.0\times {10}^{-110}\end{array}$

On solving,

$\begin{array}{c}2x=\sqrt{50.0\times {10}^{-110}}\\ =7\times {10}^{-55}\end{array}$

So, the concentration of ${\text{Ag}}^{+}$ will be equal to $2x$, that is, $7\times {10}^{-55}\text{ M}$.

The concentration of ${\text{Mg}}^{2+}$ at equilibrium is $\left(0.0500-x\right)$.

The value of $2x$ is $7\times {10}^{-55}\text{ M}$. From this, calculate the value of $x$ as:

$\begin{array}{c}x=\frac{7\times {10}^{-55}\text{ M}}{2}\\ =3.5\times {10}^{-55}\text{ M}\end{array}$

As the value of $x$ is very small, it can be neglected. So, the concentration of ${\text{Mg}}^{2+}$ will be equal to $0.0500\text{ M}$.

Conclusion

The concentrations of ${\text{Mg}}^{2+}{\text{ and Ag}}^{+}$ are $0.0500\text{ M \& }7\times {10}^{-55}\text{ M}$, respectively. The mass of the magnesium remaining is $1.44\text{ g}$.