Question

A ${\mathrm{KMnO}}_4($ aq) solution is to be standardized by titration against ${\mathrm{As}}_2{\mathrm{O}}_3(\mathrm{s}).$ A $0.1078\mathrm{g}$ sample of ${\mathrm{As}}_2{\mathrm{O}}_3$ requires $22.15\mathrm{mL}$ of the ${\mathrm{KMnO}}_4($ aq) for its titration. What is the molarity of the ${\mathrm{KMnO}}_4($ aq)? $5{\mathrm{As}}_2{\mathrm{O}}_3+4{\mathrm{MnO}}_4^{-}+9{\mathrm{H}}_2\mathrm{O}+12{\mathrm{H}}^{+}⟶$ $10{\mathrm{H}}_3{\mathrm{AsO}}_4+4{\mathrm{Mn}}^{2+}$

Short answer

The molarity of the ${\mathrm{KMnO}}_4$ solution is $0.01967\mathrm{M}$.

Step-by-step solution

Calculate moles of As2O3

First, calculate the number of moles of ${\mathrm{As}}_2{\mathrm{O}}_3$ using its molar mass. The molar mass of ${\mathrm{As}}_2{\mathrm{O}}_3$ is $197.84\mathrm{g}/\mathrm{mol}$. The moles can be calculated as $\frac{0.1078\mathrm{g}}{197.84\mathrm{g}/\mathrm{mol}}=0.0005447\mathrm{mol}$

Use stoichiometry to calculate moles of KMnO4

From the balanced chemical equation, 5 mol of ${\mathrm{As}}_2{\mathrm{O}}_3$ reacts with 4 mol of ${\mathrm{KMnO}}_4$. Utilizing this mole-to-mole ratio, you can calculate the moles of ${\mathrm{KMnO}}_4$ as $(4/5)\ast 0.0005447\mathrm{mol}=0.00043576\mathrm{mol}$

Calculate the molarity of KMnO4

The molarity $(M)$ of a solution is defined as the number of moles of solute divided by the volume of solution in liters. So, convert the given volume from mL to L, $22.15\mathrm{mL}=0.02215\mathrm{L}$. Then, calculate the molarity as $\frac{0.00043576\mathrm{mol}}{0.02215\mathrm{L}}=0.01967\mathrm{M}$

Key concepts

Titration Process

Titration is an essential laboratory method used to determine the concentration of an unknown solution. In this process, one solution (the titrant) is slowly added to a known quantity of another solution (the analyte) until a specific reaction is completed. The point at which the reaction is complete is called the equivalence point, which is usually indicated by a color change, thanks to an indicator present in the solution. The titration process involves several steps:

• Adding a measured volume of a standard solution to the analyte.
• Carefully mixing the solutions until the reaction reaches the equivalence point.
• Calculating the concentration of the unknown solution based on the volume and concentration of the titrant used.

In our exercise, the ${\mathrm{KMnO}}_4$ solution is titrated to standardize its concentration. The endpoint is reached when ${\mathrm{As}}_2{\mathrm{O}}_3$ has completely reacted, allowing us to determine the molarity of the ${\mathrm{KMnO}}_4$ solution.

Stoichiometry in Chemistry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the conservation of mass and the principle that atoms are neither created nor destroyed during a chemical reaction.To understand stoichiometry, it's essential to use the coefficients in a balanced chemical equation, which tell us the ratio in which substances react and form products. In our example, the reaction equation is: $$5{\mathrm{As}}_2{\mathrm{O}}_3+4{\mathrm{MnO}}_4^{-}+9{\mathrm{H}}_2\mathrm{O}+12{\mathrm{H}}^{+}\to 10{\mathrm{H}}_3{\mathrm{AsO}}_4+4{\mathrm{Mn}}^{2+}$$ The coefficients show that 5 moles of ${\mathrm{As}}_2{\mathrm{O}}_3$ react with 4 moles of${\mathrm{MnO}}_4^{-}$. By using these ratios, you can calculate how much of each reactant is needed and how much product will be formed, crucial for determining unknown concentrations like the molarity of ${\mathrm{KMnO}}_4$.

Chemical Reaction Equations

Chemical equations are symbolic representations of chemical reactions, showing the reactants transforming into products. Each element in the equation is represented by its chemical symbol, and coefficients indicate the number of atoms or molecules involved.A well-balanced chemical equation obeys the law of conservation of mass, meaning the number of atoms for each element is the same on both sides of the equation. This balance is key to using the equation for quantitative predictions, such as determining reactants or products needed or formed. In the provided chemical reaction for our exercise,

• Reactants: ${\mathrm{As}}_2{\mathrm{O}}_3$, ${\mathrm{MnO}}_4^{-}$, ${\mathrm{H}}_2\mathrm{O}$, and ${\mathrm{H}}^{+}$.
• Products: ${\mathrm{H}}_3{\mathrm{AsO}}_4$ and ${\mathrm{Mn}}^{2+}$.

Each molecule remains balanced, ensuring there is neither loss nor gain of atoms during the reaction, which is essential for the titration process that relies on accurate stoichiometric calculations.

Mole Concept in Chemistry

The mole is a fundamental unit in chemistry, used to express amounts of a chemical substance. One mole of any substance contains $6.022\times {10}^{23}$ entities (Avogadro's number), whether they are atoms, molecules, ions, or other particles.Using the mole concept allows chemists to calculate and compare quantities of substances based on their molecular or atomic weight. For instance, if you know the molar mass of ${\mathrm{As}}_2{\mathrm{O}}_3$ is 197.84 $\mathrm{g}/\mathrm{mol}$, you can calculate the moles present in a specific mass using the formula:$$\text{moles}=\frac{\text{mass (g)}}{\text{molar mass}(\mathrm{g}/\mathrm{mol})}$$In the exercise, the mole concept helps determine how much ${\mathrm{KMnO}}_4$ is needed to react with a known amount of ${\mathrm{As}}_2{\mathrm{O}}_3$, providing the foundation for calculating the solution's molarity.