Problem 30 - Introductory Chemistry Atoms First

Question

A \(30.00-\mathrm{mL}\) volume of aqueous sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), is titrated using \(0.200 \mathrm{MaOH}\) and an indicator that turns color only after all the sulfuric acid protons have reacted with \(\mathrm{OH}^{-}\) ions. The solution turns color when \(48.32 \mathrm{~mL}\) of base has been added. What is the molar concentration of the sulfuric acid?

Step-by-Step Solution

Answer

The molar concentration of the aqueous sulfuric acid is 0.1611 M.

Step 1: Balanced equation

The balanced chemical equation for the titration between sulfuric acid and sodium hydroxide is: \[2 \mathrm{NaOH} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{H}_{2} \mathrm{O}\] Step 2: Calculate the number of moles of NaOH added during titration

Step 2: Moles of NaOH

To calculate the moles of NaOH added, we will use the given volume and molarity of NaOH. Moles of NaOH = Volume of NaOH × Molarity of NaOH = \(48.32 \, \mathrm{mL} \times 0.200 \, \mathrm{M}\) To convert mL to L, we will divide by 1000: \[48.32 \, \mathrm{mL} \times \frac{1 \, \mathrm{L}}{1000 \, \mathrm{mL}} = 0.04832 \, \mathrm{L}\] So, moles of NaOH = \(0.04832 \, \mathrm{L} \times 0.200 \, \mathrm{M} = 0.009664 \, \mathrm{mol}\) Step 3: Use stoichiometry to find the moles of H2SO4

Step 3: Moles of H2SO4

From the balanced equation, we know that 2 moles of NaOH react with 1 mole of H2SO4. So we can find the moles of H2SO4 using the following relation: \[\frac{\text{moles of H2SO4}}{\text{moles of NaOH}} = \frac{1}{2}\] Thus, moles of H2SO4 = \(\frac{1}{2} \times 0.009664 \, \mathrm{mol} = 0.004832 \, \mathrm{mol}\) Step 4: Calculate the molar concentration of H2SO4

Step 4: Molar concentration of H2SO4

To calculate the molar concentration of H2SO4, we will use the given volume and the moles of H2SO4 that we found in the previous step. Molarity of H2SO4 = \(\frac{\text{moles of H2SO4}}{\text{volume of H2SO4}}\) To convert the volume of H2SO4 to liters, we will divide by 1000: \[30.00 \, \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.03000 \, \mathrm{L}\] So, Molarity of H2SO4 = \(\frac{0.004832 \, \mathrm{mol}}{0.03000 \, \mathrm{L}} = 0.1611 \, \mathrm{M}\) Therefore, the molar concentration of the aqueous sulfuric acid is 0.1611 M.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves calculations that use balanced chemical equations to determine the amount of reactants needed or products produced in a given chemical reaction. In our exercise, we relied on the stoichiometric relationship between sulfuric acid (H_2SO_4) and sodium hydroxide (NaOH). Using the balanced equation:\[2 \mathrm{NaOH} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2 \mathrm{H}_2\mathrm{O}\]we know that two moles of NaOH react with one mole of H_2SO_4. This 2:1 ratio enables us to find the moles of acid based on the amount of base used. Stoichiometry ensures the reactants are used efficiently in the correct proportions, avoiding wastage and ensuring complete reactions.

Molarity calculation

Molarity is a way of expressing concentration employing the amount of solute dissolved in a given volume of solution. It is expressed as moles per liter (mol/L). In the context of our titration exercise, we used molarity calculations to determine the concentration of sulfuric acid. Here’s how we did it step-by-step:

• First, we determined the moles of sodium hydroxide from its volume and molarity: \[0.04832 \, \text{L} \times 0.200 \, \text{M} = 0.009664 \, \text{mol NaOH}\]
• Using the stoichiometry from the balanced equation, we calculated the moles of sulfuric acid: \[\frac{1}{2} \times 0.009664 \, \text{mol} = 0.004832 \, \text{mol H}_2\mathrm{SO}_4 \]
• Lastly, we calculated the molarity of sulfuric acid using its moles and volume: \[\frac{0.004832 \, \text{mol}}{0.03000 \, \text{L}} = 0.1611 \, \text{M}\]

The calculation of molarity is crucial in identifying the strength of a solution, allowing chemists to prepare solutions of a precise concentration for various applications.

Chemical equations

Chemical equations represent reactions where substances are transformed into new substances. They feature symbols and formulas to depict the reactants that start the reaction and the products that are formed. A balanced chemical equation ensures that the same quantity of atoms for each element exists on both sides of the reaction, safeguarding the law of conservation of mass.In our titration problem, the chemical equation was:\[2 \mathrm{NaOH} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2 \mathrm{H}_2\mathrm{O}\]This equation tells a story. Two sodium hydroxide molecules react with one sulfuric acid molecule to produce sodium sulfate and water as products. Balancing equations like this one is essential as it makes stoichiometry possible by providing the necessary mole ratios, ensuring that calculations of substances involved in the reaction are accurate and complete.

Acid-base reactions

Acid-base reactions are a prime category of chemical reactions that involve the transfer of protons (H^+). In simple terms, acids donate protons while bases accept them. These reactions often result in the formation of water and salt.In the exercise, sulfuric acid (H_2SO_4) acted as the acid reacting with the base, sodium hydroxide (NaOH), during the titration process. The reaction can be visualized as:\[\mathrm{H}_2\mathrm{SO}_4 + 2 \mathrm{OH}^- \Rightarrow 2 \mathrm{H}_2\mathrm{O} + \mathrm{SO}_4^{2-}\]The neutralization occurs as the H^+ ions from the acid react with the OH^- ions from the base, forming water molecules and resulting in a color change from the indicator. Understanding these reactions is crucial in analytical chemistry techniques like titrations, which help determine unknown concentrations of acidic or basic solutions.