Question

A mixture was known to contain both \(\mathrm{KNO}_{3}\) and \(\mathrm{K}_{2} \mathrm{SO}_{3}\). To \(0.486 \mathrm{~g}\) of the mixture, dissolved in enough water to give \(50.00 \mathrm{~mL}\) of solution, was added \(50.00 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}\) (an excess of \(\mathrm{HCl}\) ). The reaction mixture was heated to drive off all of the \(\mathrm{SO}_{2}\), and then \(25.00 \mathrm{~mL}\) of the reaction mixture was titrated with \(0.100 \mathrm{MKOH}\) The titration required \(13.11 \mathrm{~mL}\) of the \(\mathrm{KOH}\) solution to reach an end point. What was the percentage by mass of \(\mathrm{K}_{2} \mathrm{SO}_{3}\) in the original mixture of \(\mathrm{KNO}_{3}\) and \(\mathrm{K}_{2} \mathrm{SO}_{3}\) ?

Short answer

The percentage by mass of K2SO3 in the original mixture is 21.33%.

Step-by-step solution

Write down the balanced equations for the reactions

For the reactions between K2SO3 and HCl, then between HCl and the SO2 that's formed, we have the following balanced equations:1. K2SO3 + 2HCl -> 2KCl + H2O + SO22. SO2 + 2HCl -> 2Cl2 + H2ONote that in step 2, the SO2 is driven off by heating. No further reaction with HCl is required.

Calculate the moles of KOH used in the titration

Use the volume and concentration of KOH to find the moles of KOH. Molarity (M) is moles per liter (L), so multiply the volume (in liters) by the molarity:(0.100 M) * (13.11 mL * 0.001 L/mL) = 0.001311 moles

Relate the moles of KOH back to the moles of K2SO3

The reaction between KOH and HCl that didn't react with K2SO3 is:HCl + KOH -> KCl + H2OFor every mole of KOH, there is 1 mole of HCl that didn't react with K2SO3. But every mole of K2SO3 would've produced 2 moles of HCl, so we divide the moles of KOH by 2 to get the moles of K2SO3:(0.001311 moles KOH) / 2 = 0.0006555 moles K2SO3

Calculate the mass of K2SO3

Multiply the moles of K2SO3 by its molar mass to find the mass of K2SO3 in the original mixture:0.0006555 moles * 158.26 g/mol (molar mass of K2SO3) = 0.1037 g

Calculate the percentage by mass of K2SO3 in the mixture

Divide the mass of K2SO3 by the total mass of the mixture, and then multiply by 100 to find the percentage by mass of K2SO3:(0.1037 g K2SO3 / 0.486 g mixture) * 100 = 21.33%

Key concepts

Molarity

When dealing with solutions in chemistry, one of the foundational concepts is molarity, defined as the number of moles of solute per liter of solution. It can be expressed using the formula:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Molarity allows chemists to quantify the concentration of a solution and is essential for dilutions and stoichiometric calculations. In the example given, we see molarity in action when learning that the concentration of KOH is \(0.100\ M\). This becomes the starting point for stoichiometric calculations to understand the composition of the mixture being analyzed.

Titration

Titration is an analytical technique used to determine the unknown concentration of a solution. It involves the gradual addition of a solution of known concentration (the titrant) into a solution of unknown concentration (the analyte) until the reaction reaches its endpoint, indicated by a color change or a change in electrical conductivity.
In our example, titration is performed using \(0.100\ M \text{KOH}\) to find out how much of the original \(\text{HCl}\) did not react with the \(\text{K}_2\text{SO}_3\). The volume of \(\text{KOH}\) needed to reach the end point tells us about the excess \(\text{HCl}\) in the mixture, which we then relate back to the amount of \(\text{K}_2\text{SO}_3\) in the original sample.

Stoichiometric Calculations

Stoichiometric calculations are a series of steps used to predict the amounts of reactants or products involved in a chemical reaction. It operates on the principle of the conservation of mass, where the total mass of the reactants equals the total mass of the products.
In our step by step solution, stoichiometric calculations help us establish the relationship between the moles of \(\text{KOH}\) used in the titration and the moles of \(\text{K}_2\text{SO}_3\) that were in the original mixture. Understanding this relationship is key to solving the problem and requires a solid grasp of chemical equations and molar ratios.

Molecular Mass

Another integral concept in this exercise is the molecular mass (or molar mass), expressed in grams per mole (g/mol). It represents the mass of one mole of a compound and is computed by summing the atomic masses of all the atoms in the molecule based on the periodic table.
For instance, the molecular mass of \(\text{K}_2\text{SO}_3\) is used in the final steps of our problem-solving strategy to convert moles of \(\text{K}_2\text{SO}_3\) into grams. This conversion is critical when we attempt to find out what percentage of the original mixture was \(\text{K}_2\text{SO}_3\).

Percentage Composition

Lastly, percentage composition is a quantitative expression that depicts the proportion of each element within a compound or a mixture. In this context, it's the percentage of the mixture that is \(\text{K}_2\text{SO}_3\), which can be calculated from the mass of \(\text{K}_2\text{SO}_3\) in the mixture relative to the total mass of the mixture, multiplied by 100.
This concept answers the core question of the exercise: 'What was the percentage by mass of \(\text{K}_2\text{SO}_3\) in the original mixture?' By using the mass obtained from the stoichiometric calculations and the total mass of the mixture, we are able to solve for the percentage composition -- in this case, 21.33%.