Chapter 4, Problem 159AP

The police often use a device called a Breatkalyzer to test drivers suspected of being drunk. In one type of device, the breath of a driver suspected of driving under the influence of alcohol is bubbled through an orange solution containing potassium dichromate $\left({\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\right)$ and sulfuric acid $\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)$. The alcohol in the driver's breath reacts with the dichromate ion to produce acetic acid $\left({\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)$. which is colorless, and green chromium(III) sulfate $\left[{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\right]$. The degree of color change from orange to green indicates the alcohol concentration in the breath sample, which is used to estimate blood alcohol concentration. The balanced overall equation for the Breathalyzer reaction is

$3{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(g\right)+2{\text{k}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(aq\right)+8{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to$

$3{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{Cr}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(aq\right)+2{\text{k}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+11{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(a) Classify each of the species in the Breathalyzer reaction as a strong electrolyte, weak electrolyte, or nonelectrolyte, (b) Write the ionic and net ionic equations for the Breathahzer reaction. (c) Determine the oxidation number of each element in the overall equation. (d) One manufacturer of Breathalyzers specifies a potassium dichromate concentration of 0.025 percent weight per volume $\left(\text{0}{\text{.025 g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{per 100 mL of solution}\right)$. Express this concentration in terms of molarity. (e) What volume of 0.014 M stock solution of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ would have to be diluted to 250 mL to make a solution of the specified concentration? (f) Using square bracket notation, express the molarity of each ion in a ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ solution of the specified concentration.

Interpretation Introduction

Interpretation:

The reactants and products in breathalyzer are to be classified as strong, weak, or non-electrolyte, the ionic and the net ionic equation for the reaction are to be represented, and the oxidation state of each element is to be determined in the reaction. The concentration of potassium dichromate in molarity is to be expressed and the volume of stock solution of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ required to prepare $250\text{ mL}$ of the specified solution is to be determined.

Concept introduction:

An electrolyte is a compound which dissociates into its corresponding ions when dissolved in water and conducts electricity. Electrolytes can be strong, weak, or non-electrolyte. The classification of the type of electrolyte is based on the formation of the ions when the electrolyte is dissolved in water.

An ionic reaction always follows the law of conservation of mass, according to which, when a chemical reaction occurs, the mass of ions in products should be equal to the mass of ions in reactants.

Oxidation number is the net charge on an element involved in the formation of a compound in a reaction. It is also known as oxidation state.

The concentration of a solution in terms of molarity is determined as follows:

$M=\left(\frac{Wt}{MW}\right)\left(\frac{1}{V}\right)$

Here, $M$ is the concentration of compound in molarity, $wt$ is the weight of solute, $MW$ is the molecular weight of solute, and $V$ is the volume of solution in liter containing the compound.

Dilution is the process by which a less concentrated solution can be prepared from a more concentrated solution. But, the number of moles of solute remains the same in the original solution and the dilution. So, the concentration or volume of dilution can be determined as follows:

$M_1{V}_1=M_2{V}_2$

Here, $M_1$ is the concentration of solution $1$ in molarity, $V_1$ is the volume of solution $1$

in ml, $M_2$ is the concentration of solution $2$ in molarity, and $V_2$ is the volume of solution 2 in ml.

Answer to Problem 159AP

Solution:

a)

${\text{CH}}_3{\text{CH}}_2\text{OH}$ is non-electrolyte, ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ is strong electrolyte, ${\text{H}}_2{\text{SO}}_4$ is strong electrolyte, ${\text{CH}}_3\text{COOH}$ is weak electrolyte, ${\text{Cr}}_2{\text{SO}}_4$ is strong electrolyte, ${\text{K}}_2{\text{SO}}_4$ is strong electrolyte, and ${\text{H}}_2\text{O}$ is non-electrolyte.

b)

Ionic equation for breathalyzer is as follows:

$\begin{array}{l}3{\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)+4{\text{K}}^{+}\left(aq\right)+2{\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+8{\text{HSO}}_{\text{4}}{}^{-}\left(aq\right)\\ \to 3{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+4{\text{Cr}}^{3+}\left(aq\right)+6{\text{SO}}_4{}^{2-}\left(aq\right)+4{\text{K}}^{+}\left(aq\right)+2{\text{SO}}_4{}^{2-}\left(aq\right)+11{\text{H}}_2\text{O}\left(l\right)\end{array}$

The net ionic equation is as follows:

$\begin{array}{l}3{\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)+2{\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+8{\text{HSO}}_{\text{4}}{}^{-}\left(aq\right)\\ \to 3{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+4{\text{Cr}}^{3+}\left(aq\right)+8{\text{SO}}_4{}^{2-}\left(aq\right)+11{\text{H}}_2\text{O}\left(l\right)\end{array}$

c)

Oxidation states of reactants and products are:

$\begin{array}{l}3{\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)+2{\text{K}}_2{\text{Cr}}_2{\text{O}}_7\left(aq\right)+8{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to \\ 3{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+2{\text{Cr}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(aq\right)+{\text{K}}_2{\text{SO}}_4\left(aq\right)+11{\text{H}}_2\text{O}\left(l\right)\end{array}$

d)

The molar concentration of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ is $8.5\times {10}^{-4}\text{ M}$.

e)

The volume of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ required for dilution is $15\text{ ml}$.

f)

The molarity of ${\text{K}}^{+}=1.7\times {10}^{-3}\text{ M}$, ${\text{Cr}}_2{\text{O}}_7{}^{2-}=8.5\times {10}^{-4}\text{ M}$.

Explanation of Solution

a)

Given information: The reaction for breathalyzer is as follows:

$\begin{array}{l}3{\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)+2{\text{K}}_2{\text{Cr}}_2{\text{O}}_7\left(aq\right)+8{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\\ \to 3{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+2{\text{Cr}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(aq\right)+{\text{K}}_2{\text{SO}}_4\left(aq\right)+11{\text{H}}_2\text{O}\left(l\right)\end{array}$

a) Classify each of the species in the Breathalyzer reaction as a strong electrolyte, weak electrolite or nonelectrolyte.

${\text{CH}}_3{\text{CH}}_2\text{OH}$ is a non-electrolyte as it does not ionizes when dissolved in water. ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$, ${\text{H}}_2{\text{SO}}_4$, ${\text{Cr}}_2{\text{SO}}_4$, and ${\text{K}}_2{\text{SO}}_4$ are strong electrolytes because they dissociate completely into ions when dissolved in water. ${\text{CH}}_3\text{COOH}$ is a weak electrolyte as it dissociates partially into ions that is ${\text{H}}^{+}$ and ${\text{CH}}_3{\text{COO}}^{-}$. ${\text{H}}_2\text{O}$ is a non-electrolyte as pure water does not contain any ions to conduct electricity.

b) The ionic and net ionic equations for the Breathalyzer reaction.

The ionic equation for the breathalyzer reaction is as follows:

$\begin{array}{l}3{\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)+4{\text{K}}^{+}\left(aq\right)+2{\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+8{\text{HSO}}_{\text{4}}{}^{-}\left(aq\right)\\ \to 3{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+4{\text{Cr}}^{3+}\left(aq\right)+6{\text{SO}}_4{}^{2-}\left(aq\right)+4{\text{K}}^{+}\left(aq\right)+2{\text{SO}}_4{}^{2-}\left(aq\right)+11{\text{H}}_2\text{O}\left(l\right)\end{array}$

The net ionic equation for the breathalyzer reaction is as follows:

$\begin{array}{l}3{\text{CH}}_3{\text{CH}}_2\text{OH}\left(g\right)+2{\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+8{\text{HSO}}_{\text{4}}{}^{-}\left(aq\right)\\ \to 3{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+4{\text{Cr}}^{3+}\left(aq\right)+8{\text{SO}}_4{}^{2-}\left(aq\right)+11{\text{H}}_2\text{O}\left(l\right)\end{array}$

c) The oxidation number of each element in the overall equation.

The oxidation numbers of elements in ${\text{CH}}_3{\text{CH}}_2\text{OH}$ are as follows:

$\begin{array}{l}\text{ON of C}=-2\\ \text{ON of H}=+1\times 6=6\\ \text{ON of O}=-2\end{array}$

The oxidation numbers of elements in ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ are as follows:

$\begin{array}{l}\text{ON of K}=+1\times 2=2\\ \text{ON of Cr}=+6\times 2=12\\ \text{ON of O}=-2\times 7=-14\end{array}$

The oxidation numbers of elements in ${\text{H}}_2{\text{SO}}_4$ are as follows:

$\begin{array}{l}\text{ON of H}=1\times 2=2\\ \text{ON of S}=+6\\ \text{ON of O}=-2\times 4=-8\end{array}$

The oxidation numbers of elements in ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ are as follows:

$\begin{array}{l}\text{ON of H}=1\times 4=4\\ \text{ON of C}=0\\ \text{ON of O}=-2\times 2=-4\end{array}$

The oxidation numbers of elements in ${\text{Cr}}_2{\left({\text{SO}}_4\right)}_3$ are as follows:

$\begin{array}{l}\text{ON of Cr}=+3\times 2=6\\ \text{ON of S}=+6\times 3=18\\ \text{ON of O}=-2\times 4\times 3=-24\end{array}$

The oxidation numbers of elements in ${\text{K}}_2{\text{SO}}_4$ are as follows:

$\begin{array}{l}\text{ON of K}=+1\times 2=2\\ \text{ON of S}=+6=6\\ \text{ON of O}=-2\times 4=-8\end{array}$

The oxidation numbers of elements in ${\text{H}}_2\text{O}$ are as follows:

$\begin{array}{l}\text{ON of H}=1\times 2=2\\ \text{ON of O}=-2\end{array}$

d) The concentration in terms of molarity for the reaction

The concentration of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7=0.025\text{ g}/100\text{ ml}$.

The molecular weight of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ is $294.2\text{ g}$. The concentration of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ in molarity is determined as follows:

$M=\left(\frac{wt}{MW}\right)\left(\frac{1}{V}\right)$

Substitute the values in the equation as follows:

$\begin{array}{c}M=\left(\frac{0.025}{294.2}\right)\left(\frac{1000}{100}\right)\\ =\left(8.5\times {10}^{-5}\right)\left(10\right)\\ =8.5\times {10}^{-4}\text{ M}\end{array}$

e) The volume of $0.014\text{ M}$ stock solution of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$, would be have to be diluted to 250 mL to make a solution of specified concentration.

The concentration of stock solution is $0.014\text{ M}$ and the volume of dilution is $250\text{ ml}$.

Consider the two solutions, the stock solution is solution $1$ and the diluted solution is solution $2$. Now, the volume of stock solution required to prepare dilution is as follows:

$0.014\times V_1=8.5\times {10}^{-4}\times 250$

Rearranging the equation to calculate the volume as follows:

$\begin{array}{c}{V}_1=\frac{\left(8.5\times {10}^{-4}\right)\left(250\right)}{0.014}\\ =15\text{ mL}\end{array}$

f) The molarity of each ion in a ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ solution of specified concentration.

Potassium dichromate form the following ions on dissolution:

${\text{K}}_2{\text{Cr}}_2{\text{O}}_7\rightleftharpoons 2{\text{K}}^{+}+{\text{Cr}}_2{\text{O}}_7{}^{2-}$

So, the concentration of these ions is as follows:

$\begin{array}{c}\left[{\text{K}}^{+}\right]=2\times 8.5\times {10}^{-4}\\ =1.7\times {10}^{-3}\end{array}$

$\left[{\text{Cr}}_2{\text{O}}_7{}^{2-}\right]=8.5\times {10}^{-4}\text{ M}$