Question

Mixtures of \(\mathrm{CaCl}_{2}\) and \(\mathrm{NaCl}\) are used to melt ice on roads. A dissolved 1.9348 -g sample of such a mixture was analyzed by using excess \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to precipitate the \(\mathrm{Ca}^{2+}\) as \(\mathrm{CaC}_{2} \mathrm{O}_{4} .\) The \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) was dissolved in sulfuric acid, and the resulting \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) was titrated with \(37.68 \mathrm{~mL}\) of \(0.1019 \mathrm{M} \mathrm{KMnO}_{4}\) solution. (a) Write the balanced net ionic equation for the precipitation reaction. (b) Write the balanced net ionic equation for the titration reaction. (See Sample Problem \(4.18 .)\) (c) What is the oxidizing agent? (d) What is the reducing agent? (e) Calculate the mass percent of \(\mathrm{CaCl}_{2}\) in the original sample.

Short answer

The balanced precipitation reaction is \(\text{Ca}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{CaC}_2 \text{O}_4\). The balanced titration reaction is \[ 5 \text{H}_2 \text{C}_2 \text{O}_4 + 2 \text{MnO}_4^- + 6 \text{H}^+ \rightarrow 10 \text{CO}_2 + 2 \text{Mn}^{2+} + 8 \text{H}_2 \text{O} \]. \(\text{MnO}_4^-\) is the oxidizing agent and \(\text{H}_2 \text{C}_2 \text{O}_4\) is the reducing agent. The mass percent of \(\text{CaCl}_2\) in the sample is approximately 55.01%.

Step-by-step solution

Determine the Balanced Net Ionic Equation for the Precipitation Reaction

Identify the reactants and products for the precipitation reaction. For \(\text{Ca}^{2+}\) and \(\text{C}_2\text{O}_4^{2-}\), the net ionic equation is \[ \text{Ca}^{2+} (aq) + \text{C}_2 \text{O}_4^{2-} (aq) \rightarrow \text{CaC}_2 \text{O}_4 (s) \].

Determine the Balanced Net Ionic Equation for the Titration Reaction

Identify the reacting species in the titration: \(\text{H}_2 \text{C}_2 \text{O}_4\) and \(\text{MnO}_4^-\). The balanced net ionic equation for the titration is: \[ 5 \text{H}_2 \text{C}_2 \text{O}_4 (aq) + 2 \text{MnO}_4^- (aq) + 6 \text{H}^+ (aq) \rightarrow 10 \text{CO}_2 (g) + 2 \text{Mn}^{2+} (aq) + 8 \text{H}_2 \text{O} (l) \]

Determine the Oxidizing Agent

The oxidizing agent is the species that gains electrons. Here, \(\text{MnO}_4^-\) is reduced to \(\text{Mn}^{2+}\), making \(\text{MnO}_4^-\) the oxidizing agent.

Determine the Reducing Agent

The reducing agent is the species that loses electrons. \(\text{H}_2 \text{C}_2 \text{O}_4\) is oxidized to \(\text{CO}_2\), making \(\text{H}_2 \text{C}_2 \text{O}_4\) the reducing agent.

Calculate the Moles of \(\text{KMnO}_4\) Used in the Titration

Use the volume and molarity of \(\text{KMnO}_4\) to find moles: \[\text{moles} \text{KMnO}_4 = \text{molarity} \times \text{volume} = 0.1019 \text{ M} \times 0.03768 \text{ L} = 3.838 \times 10^{-3} \text{ mol} \]

Calculate the Moles of \(\text{H}_2 \text{C}_2 \text{O}_4\) Titrated

From the stoichiometry of the titration reaction, 2 moles of \(\text{MnO}_4^-\) react with 5 moles of \(\text{H}_2 \text{C}_2 \text{O}_4\). Thus: \[\text{moles} \text{H}_2 \text{C}_2 \text{O}_4 = \frac{5}{2} \times 3.838 \times 10^{-3} \text{ mol} = 9.595 \times 10^{-3} \text{ mol} \]

Calculate the Mass of \(\text{CaCl}_2\) in the Original Mixture

First, calculate the moles of \(\text{Ca}^{2+}\), which is equal to the moles of \(\text{H}_2 \text{C}_2 \text{O}_4\): \[ \text{moles} \text{CaCl}_2 = \text{moles} \text{Ca}^{2+} = 9.595 \times 10^{-3} \text{ mol} \]. Then calculate the mass of \(\text{CaCl}_2\) using its molar mass (110.98 g/mol): \[ \text{mass of} \text{CaCl}_2 = 9.595 \times 10^{-3} \text{ mol} \times 110.98 \text{ g/mol} = 1.0643 \text{ g} \]

Calculate the Mass Percent of \(\text{CaCl}_2\)

The mass percent is calculated by dividing the mass of \(\text{CaCl}_2\) by the total mass of the sample, then multiplying by 100: \[ \text{Mass percent} = \frac{1.0643 \text{ g}}{1.9348 \text{ g}} \times 100 \approx 55.01\text{%} \]

Key concepts

balanced net ionic equation

An essential aspect of chemical reactions is identifying and writing the balanced net ionic equations. This ensures that the chemical equation accurately represents the actual species involved in the reaction. For precipitation reactions, ions combine to form an insoluble compound. For instance, when calcium ions \(\mathrm{Ca}^{2+}\) react with oxalate ions \(\mathrm{C}_2\mathrm{O}_4^{2-}\), they form calcium oxalate \(\mathrm{CaC}_2\mathrm{O}_4\). The net ionic equation for this process is:

\[ \mathrm{Ca}^{2+} (aq) + \mathrm{C}_2 \mathrm{O}_4^{2-} (aq) \rightarrow \mathrm{CaC}_2 \mathrm{O}_4 (s) \]
Writing balanced net ionic equations allows us to focus only on the ions that participate in the reaction, ignoring the spectator ions. This makes understanding and solving chemical reactions straightforward.

oxidizing agent

In redox reactions, the oxidizing agent is the substance that gains electrons and is reduced in the process. It effectively 'oxidizes' another substance by taking away its electrons. In our example involving the titration of \(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\) with \(\mathrm{KMnO}_4\), the \(\mathrm{MnO}_4^-\) ion is the oxidizing agent. During the reaction, it gets reduced to \(\mathrm{Mn}^{2+}\). The balanced net ionic equation illustrating this is:

\[ 5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 (aq) + 2 \mathrm{MnO}_4^- (aq) + 6 \mathrm{H}^+ (aq) \rightarrow 10 \mathrm{CO}_2 (g) + 2 \mathrm{Mn}^{2+} (aq) + 8 \mathrm{H}_2 \mathrm{O} (l) \]
Throughout this reaction, \(\mathrm{MnO}_4^-\) accepts electrons and is thus reduced, performing the role of the oxidizing agent.

reducing agent

Conversely, the reducing agent is the substance that loses electrons and is oxidized. It 'reduces' another substance by donating electrons to it. In our chemical example, \(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\) (oxalic acid) serves as the reducing agent. It gets oxidized to carbon dioxide \(\mathrm{CO}_2\), losing electrons to the \(\mathrm{MnO}_4^-\) ion:

\[ 5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 (aq) + 2 \mathrm{MnO}_4^- (aq) + 6 \mathrm{H}^+ (aq) \rightarrow 10 \mathrm{CO}_2 (g) + 2 \mathrm{Mn}^{2+} (aq) + 8 \mathrm{H}_2 \mathrm{O} (l) \]
Here, \(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\) gives up electrons, transforming into \(\mathrm{CO}_2\), thus acting as the reducing agent.

molarity calculation

Molarity, a key concept in chemistry, measures the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. Calculating molarity plays a crucial role in titration processes. In our example, molarity is used to determine the number of moles of \(\mathrm{KMnO}_4\) used in the titration. Given the volume (37.68 mL) and molarity (0.1019 M) of \(\mathrm{KMnO}_4\):

\[ \text{Moles of } \mathrm{KMnO}_4 = \text{molarity} \times \text{volume} = 0.1019 \text{ M} \times 0.03768 \text{ L} = 3.838 \times 10^{-3} \text{ mol} \]
This calculated molarity helps determine the amount of oxalic acid \(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\) in the reaction.

mass percent calculation

Mass percent is a way to express the concentration of an element in a compound or a component in a mixture as a percentage of the total mass. In this exercise, we determine the mass percent of \(\mathrm{CaCl}_2\) in the mixture. After calculating the moles of \(\mathrm{CaCl}_2\), we convert it to mass using its molar mass (110.98 g/mol):

\[ \text{Mass of } \mathrm{CaCl}_2 = 9.595 \times 10^{-3} \text{ mol} \times 110.98 \text{ g/mol} = 1.0643 \text{ g} \]
Then, calculate the mass percent of \(\mathrm{CaCl}_2\) by dividing the mass of \(\mathrm{CaCl}_2\) by the total mass of the sample and multiplying by 100:

\[ \text{Mass percent} = \frac{1.0643 \text{ g}}{1.9348 \text{ g}} \times 100 \approx 55.01\text{%} \]
This percentage indicates how much of the original mixture is composed of \(\mathrm{CaCl}_2\).