Question

A \(10.00-\mathrm{mL}\) sample of sulfuric acid from an automobile battery requires \(35.08 \mathrm{mL}\) of \(2.12 \mathrm{M}\) sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? Sulfuric acid contains two acidic hydrogens.

Short answer

The molarity of the sulfuric acid is approximately \(3.719 \mathrm{M}\).

Step-by-step solution

Write the balanced chemical equation for the reaction

The neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) can be written as: \(H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\) This equation shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide to form one mole of sodium sulfate and two moles of water.

Calculate the moles of sodium hydroxide used

We are given the volume and the molarity of the sodium hydroxide solution. The volume is \(35.08 \mathrm{mL}\), and the molarity is \(2.12 \mathrm{M}\). To calculate the moles of sodium hydroxide used, we will use the formula: Moles of NaOH = Volume of NaOH solution (in L) × Molarity of NaOH solution First, we need to convert the volume of NaOH solution from milliliters to liters: \(35.08 \mathrm{mL} = 35.08 × 10^{-3} \mathrm{L} = 0.03508 \mathrm{L}\) Now we can calculate the moles of NaOH used: Moles of NaOH = \(0.03508 \mathrm{L} × 2.12 \mathrm{M} = 0.07437 \mathrm{mol}\)

Calculate the moles of sulfuric acid

From the balanced chemical equation, we know that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. Therefore, the moles of sulfuric acid can be calculated by dividing the moles of sodium hydroxide by 2: Moles of H2SO4 = \(\frac{0.07437 \mathrm{mol}}{2} = 0.037185 \mathrm{mol}\)

Calculate the molarity of sulfuric acid

Now that we have the moles of sulfuric acid, we can calculate its molarity using the given volume of the sample. Recall that the molarity is defined as moles of solute per liter of solution. The volume of the sulfuric acid solution is \(10.00 \mathrm{mL}\), which is equal to \(0.01000 \mathrm{L}\). Therefore, the molarity of sulfuric acid can be calculated as follows: Molarity of H2SO4 = \(\frac{0.037185 \mathrm{mol}}{0.01000 \mathrm{L}} = 3.719 \mathrm{M}\) The molarity of the sulfuric acid is approximately \(3.719 \mathrm{M}\).

Key concepts

Neutralization Reaction

A neutralization reaction is a specific type of chemical reaction in which an acid and a base react together to form water and a salt. This process usually involves the transfer of a proton (H+) from the acid to the base. A well-known example of a neutralization reaction is the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), which forms water and sodium chloride (NaCl).
In the given exercise, a neutralization reaction occurs between sulfuric acid (H2SO4), which is the acid, and sodium hydroxide (NaOH), which is the base. The products of this reaction are sodium sulfate (Na2SO4) and water (H2O). Understanding neutralization reactions is vital because it allows us to predict the outcome of the interaction between acids and bases and serves as the foundation for many applications including treatment of acid spills, manufacturing of various chemicals, and acid-base titrations.

Chemical Equation Balancing

Balancing a chemical equation is crucial because it ensures that the law of conservation of mass is satisfied. This law states that matter cannot be created or destroyed in an ordinary chemical reaction. In essence, what goes into a reaction must come out, and the number of atoms of each element must be the same on both sides of the equation.
In the context of the problem, the balanced equation is essential for determining the molarity of the sulfuric acid. It indicates that two moles of sodium hydroxide react with one mole of sulfuric acid. Without a balanced equation, stoichiometric calculations, which are needed to find the molarity, cannot be accurately performed. To balance an equation, one must adjust the coefficients (the numbers before the chemical formulas) to ensure that the number of atoms of each element is the same on both sides of the equation.

Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It is based on the balanced chemical equations and uses the relationship between moles, mass, and molecular weight to predict the outcome of the reactions.
When tackling exercises like the given one, stoichiometry is used to determine the moles of reactants or products. For instance, to find the molarity of sulfuric acid, we use stoichiometry to relate the volume and molarity of sodium hydroxide solution to the amount of sulfuric acid. This calculation hinges on understanding the mole ratio between the acid and the base, which comes from the balanced chemical equation.

Key Steps in Stoichiometry

• Conversion of the reactant's volume to liters (if given in milliliters).
• Use of molarity to find the number of moles of the reactant.
• Application of mole ratios to relate the moles of one substance (sodium hydroxide) to another (sulfuric acid).
• Calculation of the molarity of the unknown using the moles of solute and the volume of solution.

Acid-Base Titration

Acid-base titration is an analytical procedure used to determine the concentration of an unknown acid or base by neutralizing it with a standard solution of known concentration. The point at which neutralization occurs is known as the equivalence point and can often be detected using an indicator or pH meter.
In our exercise, the amount of sulfuric acid is unknown, and we determine its molarity by titrating it with a known volume and concentration of sodium hydroxide. The step-by-step solution calculates the moles of sodium hydroxide used, then uses the stoichiometry of the balanced equation to find the moles of sulfuric acid. Using the volume of the acid sample, we then calculate the molarity. This process is a key example of how acid-base titration and the principles of stoichiometry are applied in a laboratory to find the concentration of an acid or base solution.