Question

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0 -mL aliquot and adding 50.0 \(\mathrm{mL}\) of 0.213 \(\mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required 13.21 \(\mathrm{mL}\) of 0.103 \(\mathrm{M}\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) . Sulfuric acid has two acidic hydrogens.

Short answer

The molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 0.00929 M.

Step-by-step solution

Calculate the moles of excess OH- ions

First, we find out the moles of excess \(\mathrm{OH}^{-}\) ions after the reaction. We have the volume and molarity of \(\mathrm{HCl}\) for neutralization of \(\mathrm{OH}^{-}\). n(HCl) = V(HCl) × M(HCl) n(HCl) = Volume of HCl × Molarity of HCl n(OH-) = n(HCl) (Since HCl and OH- ions react in 1:1 ratio) n(OH-) = 13.21 mL × 0.103 M Convert mL to L: n(OH-) = 0.01321 L × 0.103 M n(OH-) = 0.00136033 moles

Calculate the moles of NaOH initially added

Next, we need to find out the moles of \(\mathrm{NaOH}\) initially added. n(NaOH) = V(NaOH) × M(NaOH) n(NaOH) = Volume of NaOH × Molarity of NaOH n(NaOH) = 50 mL × 0.213 M Convert mL to L: n(NaOH) = 0.050 L × 0.213 M n(NaOH) = 0.01065 moles

Calculate the moles of OH- ions that reacted with H2SO4

Now, we will subtract the excess \(\mathrm{OH}^{-}\) ions to find out the moles of \(\mathrm{OH}^{-}\) ions that reacted with the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Since sulfuric acid has two acidic hydrogens, it reacts with two moles of OH- ions per mole. n(OH-)_{reacted} = n(NaOH) - n(OH-)_{excess} n(OH-)_{reacted} = 0.01065 moles - 0.00136033 moles n(OH-)_{reacted} = 0.00928967 moles

Calculate the moles of H2SO4

We have the moles of \(\mathrm{OH}^{-}\) ions that reacted with the \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Since each mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with two moles of \(\mathrm{OH}^{-} \) ions: n(H2SO4) = n(OH-)_{reacted} / 2 n(H2SO4) = 0.00928967 moles / 2 n(H2SO4) = 0.004644835 moles Finally, now we have the moles of the original \(\mathrm{H}_{2} \mathrm{SO}_{4}\) sample.

Calculate the molarity of H2SO4

We have the moles and volume of the original \(\mathrm{H}_{2} \mathrm{SO}_{4}\) sample. M(H2SO4) = n(H2SO4) / V(H2SO4) M(H2SO4) = 0.004644835 moles / 0.500 L M(H2SO4) = 0.00928967 M The molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 0.00929 M.

Key concepts

Molarity Calculation

Understanding molarity is crucial when dealing with solution chemistry. Molarity (M) represents the concentration of a solute in a solution and is defined as the number of moles of solute per liter of solution. In practice, it allows us to know how much of a substance is present in a given volume of liquid.
To calculate molarity:

• First, determine the number of moles of the solute. This involves using the formula: \( n = V \times M \), where \( n \) is the number of moles, \( V \) is the volume in liters, and \( M \) is the molarity.
• Next, divide these moles by the volume of the solution in liters to find the molarity, \( M = \frac{n}{V} \).

This method is applied in our example, where sulfuric acid (\( \mathrm{H}_2\mathrm{SO}_4 \)) is the solute whose molarity is determined after reacting with sodium hydroxide (\( \mathrm{NaOH} \)) and hydrochloric acid (\( \mathrm{HCl} \)). Calculations like this assist in understanding the strength and concentration of acids and bases in solutions.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in chemical reactions. It is based on the balanced chemical equation and is foundational in understanding how different chemicals interact.
In the context of the acid-base titration with \( \mathrm{H}_2\mathrm{SO}_4 \), understanding stoichiometry becomes vital. Each molecule of sulfuric acid can donate two hydrogen ions (\( \mathrm{H}^+ \)), meaning every mole of \( \mathrm{H}_2\mathrm{SO}_4 \) can neutralize two moles of hydroxide ions (\( \mathrm{OH}^- \)). This 1:2 relation is critical in the calculations:

• Determine the moles of \( \mathrm{NaOH} \) added initially through titration.
• Calculate the excess \( \mathrm{OH}^- \) that remained after reacting with \( \mathrm{H}_2\mathrm{SO}_4 \).
• Ascertain the moles of \( \mathrm{OH}^- \) that actually reacted with the \( \mathrm{H}_2\mathrm{SO}_4 \).

This stoichiometry-driven approach leads us to find the number of moles of \( \mathrm{H}_2\mathrm{SO}_4 \) that initially reacted, which is crucial for accurate molarity calculations.

Solution Chemistry

Solution chemistry focuses on how substances dissolve, react, and interact within a liquid medium. In acid-base titration, this is particularly relevant as it underlies the neutralization reactions and solubility principles you observe.
During the titration process:

• A known concentration of \( \mathrm{NaOH} \) is slowly added to the \( \mathrm{H}_2\mathrm{SO}_4 \) solution, neutralizing the acid.
• The reaction produces water and an excess of hydroxide ions if extra base is added.
• To determine how much base was in excess, \( \mathrm{HCl} \) is carefully introduced until neutrality is again achieved.

This sequence involves a dynamic interplay of the components in solution. It's important to understand how ionic compounds like these behave in water, dissociating into their respective ions and participating in reactions that are influenced by the concentration and nature of each ion present. This knowledge is essential to predicting outcomes and ensuring accurate measurements in chemical reactions.