Question

Suppose that \(27.34 \mathrm{mL}\) of standard \(0.1021 \mathrm{M} \mathrm{NaOH}\) is required to neutralize \(25.00 \mathrm{mL}\) of unknown \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Calculate the molarity and the normality of the unknown solution.

Short answer

The molarity of the unknown H2SO4 solution is 0.05578 M, and the normality is 0.11156 N.

Step-by-step solution

Calculate moles of NaOH

First, we'll find the moles of the NaOH solution using its given volume and molarity. We know the molarity of NaOH is 0.1021 M and the volume is 27.34 mL. We can convert molarity (moles/L) into moles using the formula: moles of NaOH = Molarity × Volume Since the volume is given in mL, we need to convert it to L by dividing by 1000: \( volume_{NaOH} (L) = \frac{27.34 \, mL}{1000} = 0.02734 \, L\) Now we can find the moles of NaOH: \( moles_{NaOH} = 0.1021 M * 0.02734 L = 0.002789 \, moles \)

Find moles of H2SO4

Next, we'll use the balanced chemical equation to determine the moles of H2SO4 since we have found the moles of NaOH. H2SO4 + 2NaOH → Na2SO4 + 2H2O From the balanced chemical equation, 1 mole of H2SO4 reacts with 2 moles of NaOH. Hence, \(moles_{H2SO4} = \frac{moles_{NaOH}}{2}\) Therefore, \(moles_{H2SO4} = \frac{0.002789 \, moles}{2} = 0.0013945 \, moles\)

Calculate the molarity of H2SO4

Now that we have found the moles of H2SO4, we can calculate its molarity using the formula: Molarity = moles of solute/volume of solution (in L) First, we'll convert the volume of H2SO4 from mL to L: \(volume_{H2SO4} (L) = \frac{25.00 \, mL}{1000} = 0.025 \, L\) Now we can find the molarity: \(M_{H2SO4} = \frac{0.0013945 \, moles}{0.025 \, L} = 0.05578 \, M\)

Calculate the normality of H2SO4

Finally, we'll find the normality of the H2SO4 solution, which represents the number of equivalent moles per liter. Since each mole of H2SO4 has 2 acidic protons (H+ ions), normality is twice the molarity: Normality = Molarity * n-factor where n-factor = number of replaceable H+ ions in the acid For H2SO4, n-factor = 2 Thus, \(Normality_{H2SO4} = 0.05578 \, M * 2 = 0.11156 \, N\) In conclusion, the molarity of the unknown H2SO4 solution is 0.05578 M, and the normality is 0.11156 N.

Key concepts

Molarity

Molarity is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute (\(mol\)) per liter of solution (\(L\)). The formula to calculate molarity (\(M\)) is:

• \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

In the original exercise, we calculated the molarity of \(\text{H}_{2}\text{SO}_{4}\). By using the number of moles of \(\text{H}_{2}\text{SO}_{4}\)and the volume of the solution, we found that the molarity was \(0.05578\, \text{M}\). This value tells you how concentrated the sulfuric acid is in the solution.

Normality

Normality, similar to molarity, is a measure of concentration but focuses on the reactive capacity of a solution. It accounts for the number of equivalents (reactive capacity) in a chemical reaction. The formula for normality (\(N\)) is:

• \( N = Molarity \times n\text{-factor} \)

The \(n\text{-factor}\) represents the number of \(\text{H}^+\) ions a molecule can donate or accept. For \(\text{H}_{2}\text{SO}_{4}\), the \(n\text{-factor}\) is 2 because each molecule can donate two hydrogen ions. Knowing this, we can calculate the normality by doubling the molarity, as shown in the solution. The normality of the \(\text{H}_{2}\text{SO}_{4}\)sample was found to be \(0.11156\, \text{N}\), which reflects the solution's reactivity.

Acid-Base Reaction

An acid-base reaction is a chemical process where an acid reacts with a base to produce water and a salt. In these reactions, acids donate protons (\(\text{H}^+\) ions), while bases accept them. The reaction, therefore, neutralizes the solution. The standard equation for a neutralization reaction is:

• \( \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} \)

In our exercise, \(\text{H}_{2}\text{SO}_{4}\) (sulfuric acid) reacted with \(\text{NaOH}\) (sodium hydroxide) to produce water and sodium sulfate (\(\text{Na}_{2}\text{SO}_{4}\)). These reactions are fundamental to understanding titration because they allow us to measure how much reactant is needed to completely neutralize an acid or base.

Chemical Equation

A chemical equation represents a chemical reaction with the reactants on the left and the products on the right. Accurate chemical equations are vital for understanding and solving stoichiometry problems. For the titration in the exercise, the balanced chemical equation is:

• \(\text{H}_{2}\text{SO}_{4} + 2\text{NaOH} \rightarrow \text{Na}_{2}\text{SO}_{4} + 2\text{H}_{2}\text{O} \)

This equation shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide. The balanced equation ensures that atoms are conserved in the reaction, which forms the basis for calculating the moles of each substance involved.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction using the balanced chemical equation. It allows you to determine the amount of one substance based on another, using mole ratios derived from the equation. In our example:

• The balanced equation shows that 2 moles of \(\text{NaOH}\) are required to react with 1 mole of \(\text{H}_{2}\text{SO}_{4}\).
• This relationship was crucial in determining the moles of sulfuric acid present, based on the moles of sodium hydroxide used.

Knowing these relationships helps us calculate concentration values like molarity and normality. Stoichiometry provides the math behind the chemistry, enabling precise measurement in reactions.