Question

What volume of \(0.151 \mathrm{~N} \mathrm{NaOH}\) is required to neutralize \(24.2 \mathrm{~mL}\) of \(0.125 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) ? What volume of \(0.151 \mathrm{~N} \mathrm{NaOH}\) is required to neutralize \(24.2 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Short answer

The volume of \(0.151 \mathrm{~N} \mathrm{NaOH}\) required to neutralize \(24.2 \mathrm{~mL}\) of \(0.125 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) is approximately \(40.1 \mathrm{~mL}\).

Step-by-step solution

Find the valence factors of H2SO4 and NaOH

In this neutralization reaction, H2SO4 will release 2 hydrogen ions and NaOH will release 1 hydroxide ion. Thus, the valence factor of H2SO4 (n1) is 2, and the valence factor of NaOH (n2) is 1.

Write down the given values and the equation

We have the following information given: - \(V_1 = 24.2 mL\) - \(N_1 = 0.125 N\) - \(N_2 = 0.151 N\) - \(n_1 = 2\) (from step 1) - \(n_2 = 1\) (from step 1) Now, use the equation for neutralization: \( {V_1} * {N_1} * {n_1} = {V_2} * {N_2} * {n_2}\)

Solve the equation for the required volume of NaOH

Replace the given values into the equation and solve for \(V_2\): \( (24.2 \mathrm{~mL}) * (0.125 \mathrm{~N}) * (2) = V_2 * (0.151 \mathrm{~N}) * (1) \) Now, solve for \(V_2\): \(V_2 = \frac{(24.2 \mathrm{~mL}) * (0.125 \mathrm{~N}) * (2)}{(0.151 \mathrm{~N}) * (1)}\) \(V_2 = \frac{6.05 \mathrm{~mL N}}{0.151 \mathrm{~N}}\) \(V_2 = 40.1 \mathrm{~mL}\) The volume of \(0.151 \mathrm{~N} \mathrm{NaOH}\) required to neutralize \(24.2 \mathrm{~mL}\) of \(0.125 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) is approximately \(40.1 \mathrm{~mL}\).

Key concepts

Chemical Stoichiometry

Chemical stoichiometry is a quantitative relationship between reactants and products in a chemical reaction. It's the backbone of chemical equations, providing the proportions of how much of each substance is involved to complete a reaction. In the context of neutralization reactions, stoichiometry allows us to calculate the exact amount of an acid required to react with a given amount of a base, and vice versa.

Here's a simple way to visualize stoichiometry: if you think of a chemical reaction as a recipe, stoichiometry tells you how many cups of flour (reactants) you will need to make a perfect cake (product) without any leftovers. When applied to the textbook exercise, stoichiometry helps in finding out the precise volume of sodium hydroxide (NaOH) needed to neutralize a known volume of sulfuric acid (H2SO4).

Neutralization Reaction

A neutralization reaction is a chemical reaction where an acid and a base react to form water and a salt. It's essentially a double replacement reaction where the hydrogen ions (H+) from the acid combine with the hydroxide ions (OH-) from the base to make water (H2O). The acid's remaining anion then pairs with the base's cation to form a salt.

Understanding the neutralization reaction is crucial for the given textbook exercise, where sulfuric acid (H2SO4), an acid, reacts with sodium hydroxide (NaOH), a base. Since sulfuric acid is diprotic, meaning it can donate two hydrogen ions per molecule, the stoichiometry becomes very important to ensure that both of these hydrogen ions are neutralized by hydroxide ions.

Molarity and Normality

Molarity (M) and normality (N) are both measures of concentration in chemistry. Molarity is the number of moles of solute per liter of solution, which gives us an idea of how concentrated a solution is. Normality extends this concept to account for the number of acid or base equivalents in a solution. For acid-base reactions, one equivalent corresponds to one mole of H+ or OH- ions.

In our exercise, the molarity of H2SO4 is provided, but we also encounter the normality, which is a more relevant measure since it directly relates to the stoichiometry of the acid or base. For H2SO4, the normality can be different from molarity since it's a diprotic acid—one mole of H2SO4 can neutralize two moles of OH- ions. This detail is crucial for calculating the correct volumes during titration, ensuring that the stoichiometric equivalents match between the acid and base.

Acid-Base Titration

Acid-base titration is an analytical technique used to determine the concentration of a given acid or base by reacting it with a solution of known concentration, called a standard solution. It usually involves the gradual addition of the standard into the unknown solution until the reaction reaches the endpoint, which is when the exact stoichiometric equivalents react and the solution is neutralized.

For solving the exercise problem, we essentially perform a titration calculation without the actual experiment. Knowing the normality of the acid and base solutions allows us to use their stoichiometry to find out how much of the sodium hydroxide solution is required to achieve neutralization. The exercise highlights the importance of understanding titration principles in order to solve real laboratory problems through theoretical calculations.