Problem 132 - Chemistry and Chemical Reactivity

Question

\(A A .000-g\) sample containing \(K C l\) and \(K C 1 O_{4}\) was dis. solved in sufficient water to give \(250.00 \mathrm{mL}\) of solution. A \(50.00-\mathrm{mL}\) portion of the solution required \(41.00 \mathrm{mL}\) of \(0.0750 \mathrm{M} \mathrm{AgNO}_{3}\) in a Mohr titration (page 187 ). Next, a \(25.00-\mathrm{mL}\), portion of the original solution was treated with \(\mathrm{V}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) to reduce the perchlorate ion to chloride, \(8 \mathrm{V}^{3+}(\mathrm{aq})+\mathrm{ClO}_{4}^{-}(\mathrm{aq})+12 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ \mathrm{Cl}^{-}(\mathrm{aq})+8 \mathrm{VO}^{2+}(\mathrm{aq})+8 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) $$ and the resulting solution was tirrated with AgNO, This titration required \(38.12 \mathrm{mL}\) of \(0.0750 \mathrm{M} \mathrm{AgNO}_{3} .\) What is the mass percent of \(\mathrm{KCl}\) and \(\mathrm{KClO}_{4}\) in the mixture?

Step-by-Step Solution

Answer

KCl is 38.5% and KClO₄ is 61.5% of the sample.

Step 1: Calculate moles of AgNO₃ used in first titration

For the first titration, a 50.00 mL portion of the solution required 41.00 mL of 0.0750 M AgNO₃. Firstly, convert 41.00 mL to liters: \( 41.00 \, \text{mL} = 0.04100 \, \text{L} \). The moles of AgNO₃ are: \( \text{Moles of AgNO₃} = 0.04100 \, \text{L} \times 0.0750 \, \text{mol/L} = 0.003075 \text{ mol} \). This corresponds to the moles of Cl⁻ from KCl since Ag⁺ reacts with Cl⁻ in a 1:1 ratio.

Step 2: Calculate concentration of Cl⁻ from KCl in the original solution

The amount of Cl⁻ from KCl in the entire 250.00 mL solution can now be calculated. Since the titration used only 50.00 mL of this solution, we proportion the moles for the entire solution: \( \text{Total moles of Cl⁻ from KCl} = 0.003075 \text{ mol} \times \frac{250.00}{50.00} = 0.015375 \text{ mol}. \)

Step 3: Calculate moles of AgNO₃ used in the second titration

In the second titration, a 25.00 mL portion required 38.12 mL of 0.0750 M AgNO₃. Convert 38.12 mL to liters: \( 38.12 \, \text{mL} = 0.03812 \, \text{L} \). The moles of AgNO₃ are: \( \text{Moles of AgNO₃} = 0.03812 \, \text{L} \times 0.0750 \, \text{mol/L} = 0.002859 \text{ mol}. \)

Step 4: Determine moles of Cl⁻ after reduction

After treatment with V₂(SO₄)₃, the perchlorate was reduced to chloride, and this chloride combined with any original chloride from KCl. For the entire 250.00 mL solution: \( \text{Total moles of Cl⁻} = 0.002859 \text{ mol} \times \frac{250.00}{25.00} = 0.028590 \text{ mol}. \)

Step 5: Calculate moles of ClO₄⁻ reduced to Cl⁻

Subtract the moles of Cl⁻ from KCl (Step 2) from the total moles of Cl⁻ after reduction (Step 4) to get the moles of Cl⁻ contributed by ClO₄⁻ reduction: \( 0.028590 \text{ mol} - 0.015375 \text{ mol} = 0.013215 \text{ mol}. \) This represents the moles of ClO₄⁻ originally present.

Step 6: Calculate mass of KCl and KClO₄

Calculate the mass of KCl from its moles: \( \text{Mass of KCl} = 0.015375 \text{ mol} \times 74.55 \, \text{g/mol} = 1.146 \text{ g}. \) Calculate the mass of KClO₄: \( \text{Mass of KClO₄} = 0.013215 \text{ mol} \times 138.55 \, \text{g/mol} = 1.831 \text{ g}. \)

Step 7: Calculate mass percent of KCl and KClO₄

Calculate mass percentage of KCl: \( \text{Mass \, percent \, of \, KCl} = \left(\frac{1.146 \, \text{g}}{1.146 \, \text{g} + 1.831 \, \text{g}}\right) \times 100\% \approx 38.5 \% \). Calculate mass percentage of KClO₄: \( \text{Mass \, percent \, of \, KClO}_{4} = \left(\frac{1.831 \, \text{g}}{1.146 \, \text{g} + 1.831 \, \text{g}}\right) \times 100\% \approx 61.5 \% \).

Key Concepts

Mohr Method

The Mohr Method is a type of titration used to determine the concentration of chloride ions in a solution. It involves the use of a silver nitrate (\(\text{AgNO}_3 \)) solution as a titrant. When \(\text{AgNO}_3 \) is added to a solution containing chloride ions, a white precipitate of silver chloride (\(\text{AgCl}\)) forms. This process continues until all the chloride ions are reacted, resulting in a permanent white precipitate. This end point can be determined visually or with the help of an indicator.
Some important points about the Mohr Method:

• The reaction that occurs: \[\text{Ag}^+ (\text{aq}) + \text{Cl}^- (\text{aq}) \rightarrow \text{AgCl} (\text{s})\]
• It is essential to carry out this titration in a neutral to slightly alkaline medium to avoid interference from other ions.
• Commonly, potassium chromate (\(\text{K}_2\text{CrO}_4\)) is used as an indicator, signaling the end point by changing color when excess silver ions are present.

Understanding the Mohr Method is crucial as it applies to various fields, such as water quality testing and clinical laboratory analysis.

Perchlorate Reduction

In chemical reactions, perchlorate ions can be reduced to chloride ions, as seen in the given problem. This reduction is carried out using a reducing agent such as \(\text{V}_2(\text{SO}_4)_3\). In the process, perchlorate, \(\text{ClO}_4^-\), undergoes transformation to yield chloride, \(\text{Cl}^-\).The balanced chemical reaction is:\[8\,\text{V}^{3+}(\text{aq}) + \text{ClO}_4^-(\text{aq}) + 12\,\text{H}_2\text{O}(\ell) \rightarrow \text{Cl}^-(\text{aq}) + 8\,\text{VO}^{2+}(\text{aq}) + 8\,\text{H}_3\text{O}^+(\text{aq})\]
This reduction reaction is critical because it allows the transformation of a less reactive species into a more reactive form. In our context, it enables the determination of the amount of perchlorate in the solution by converting it into a detectable ion form (chloride) using the Mohr Method.A deeper understanding involves:

• Recognizing \(\text{V}^{3+}\) as a well-known reducing agent in such reactions.
• Understanding the conditions required for the reduction, such as the presence of acidified water, as \(\text{H}_3\text{O}^+\) ions are necessary.

Mass Percent Calculation

Mass percent is a way of expressing a concentration of a component within a mixture. It is calculated as the mass of the component divided by the total mass of the mixture, multiplied by 100%.In the problem, we calculate the mass percent of \(\text{KCl}\) and \(\text{KClO}_4\) using the formulas:For \(\text{KCl}:\)\[\text{Mass percent of KCl} = \left(\frac{\text{Mass of KCl}}{\text{Total mass of sample}}\right) \times 100\%\]For \(\text{KClO}_4\):\[\text{Mass percent of KClO}_4 = \left(\frac{\text{Mass of KClO}_4}{\text{Total mass of sample}}\right) \times 100\%\]
This calculation is crucial for understanding the composition of chemical mixtures and is widely utilized in laboratories to determine the purity of substances. To find these percentages in our example, you divide the mass of each substance by the total mass of the sample, then multiply by 100% to obtain percentages.

Chemical Solutions

Chemical solutions are homogeneous mixtures composed of solutes dissolved in solvents. Solutions are essential in chemistry because they provide a medium for reactions and allow substances to be delivered effectively for various applications. Key aspects of chemical solutions include:

• **Solvent**: The component that dissolves the solute. Water is a common solvent due to its polarity and ability to dissolve a wide range of substances.
• **Solute**: The substance dissolved in the solvent. It can be solid, liquid, or gas.
• **Concentration**: A measure of how much solute is present within a solution. It can be expressed in molarity, mass percent, and other units.

Understanding chemical solutions is vital for carrying out experiments accurately, ensuring the correct proportions, and interpreting results effectively. These principles guide many laboratory practices, from preparing reactions to analyzing their outcomes.