Question

Before strong acids can be disposed of, they are often neutralized to reduce their corrosiveness. What volume of \(0.251 \mathrm{M} \mathrm{NaOH}\) solution would be required to neutralize \(135 \mathrm{mL}\) of \(0.211 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution?

Short answer

\(227.37\ \textrm{mL}\) of \(0.251\ \textrm{M}\ \mathrm{NaOH}\) solution is required to neutralize \(135\ \textrm{mL}\) of \(0.211\ \textrm{M}\ \mathrm{H}_{2}\mathrm{SO}_{4}\) solution.

Step-by-step solution

Calculate the moles of H₂SO₄

We know the volume and the concentration of the \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution. We can use the formula: Moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) = volume (L) × molarity First, we need to convert the volume of the \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution from milliliters to liters: Volume in L = \(\frac{135\ \textrm{mL}}{1000\ \textrm{mL/L}} = 0.135\ \textrm{L}\) Now, we can calculate the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\): Moles of \(\mathrm{H}_{2}\mathrm{SO}_{4} = 0.135\ \textrm{L}\times 0.211\ \textrm{M} = 0.028485\ \textrm{mol}\)

Determine moles of NaOH required for neutralization

We need to determine the stoichiometry of the reaction between \(\mathrm{H}_{2}\mathrm{SO}_{4}\) and \(\mathrm{NaOH}\). The neutralization reaction is: \(\mathrm{H}_{2}\mathrm{SO}_{4}(aq) + 2\ \mathrm{NaOH(aq)} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}(aq) + 2\ \mathrm{H}_{2}\mathrm{O(l)}\) From the balanced chemical equation, 1 mole of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) reacts with 2 moles of \(\mathrm{NaOH}\). So, the moles of \(\mathrm{NaOH}\) can be calculated as: Moles of \(\mathrm{NaOH} = 2 \times \textrm{moles of}\ \mathrm{H}_{2}\mathrm{SO}_{4} = 2 \times 0.028485\ \textrm{mol} = 0.05697\ \textrm{mol}\)

Calculate the volume of NaOH solution needed

Now that we have the moles of \(\mathrm{NaOH}\) required for neutralization, we can calculate the volume of the \(\mathrm{0.251\ M\ NaOH}\) solution needed using the formula: Volume of \(\mathrm{NaOH}\) solution (L) = \(\frac{\textrm{moles of}\ \mathrm{NaOH}}{\textrm{molarity of}\ \mathrm{NaOH}}\) Volume of \(\mathrm{NaOH}\) solution (L) = \(\frac{0.05697\ \textrm{mol}}{0.251\ \textrm{M}} = 0.22737\ \textrm{L}\) Finally, we need to convert the volume from liters to milliliters: Volume of \(\mathrm{NaOH}\) solution (mL) = \(0.22737\ \textrm{L} \times \frac{1000\ \textrm{mL}}{1\ \textrm{L}} = 227.37\ \textrm{mL}\) So, \(227.37\ \textrm{mL}\) of \(0.251\ \textrm{M}\ \mathrm{NaOH}\) solution is required to neutralize \(135\ \textrm{mL}\) of \(0.211\ \textrm{M}\ \mathrm{H}_{2}\mathrm{SO}_{4}\) solution.

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It helps us determine the right amounts of substances needed and produced in a chemical reaction. This is especially useful when trying to neutralize an acid, like in our problem with \( \mathrm{H}_{2}\mathrm{SO}_{4} \).

To figure this out, we use balanced chemical equations. These equations tell us the ratios in which chemicals react. For example, in our task, we use the equation: \[ \mathrm{H}_{2}\mathrm{SO}_{4}(aq) + 2\ \mathrm{NaOH(aq)} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}(aq) + 2\ \mathrm{H}_{2}\mathrm{O(l)} \]

• 1 mole of \( \mathrm{H}_{2}\mathrm{SO}_{4} \) needs 2 moles of \( \mathrm{NaOH} \) for neutralization.
• This ratio helps us compute how much \( \mathrm{NaOH} \) is required to react with a given amount of \( \mathrm{H}_{2}\mathrm{SO}_{4} \).

Understanding stoichiometry ensures chemicals are neither wasted nor insufficiently mixed when neutralizing an acid, making reactions more efficient and safer.

Molarity

Molarity is a measure to express concentration, specifically the number of moles of solute per liter of solution. Knowing this is crucial in chemistry, as it allows us to understand the strength and capability of a solution to perform reactions.

In the exercise, determining molarity was vital as it explained how potent the \( \mathrm{H}_{2}\mathrm{SO}_{4} \) solution is and how much \( \mathrm{NaOH} \) needs to be used. We are given that the \( \mathrm{H}_{2}\mathrm{SO}_{4} \) solution is 0.211 M and \( \mathrm{NaOH} \) is 0.251 M.

• For the \( \mathrm{H}_{2}\mathrm{SO}_{4} \), we convert the 135 mL to 0.135 L to use in calculations.
• Then, we find it contains about 0.028485 moles by multiplying its volume and molarity.

Understanding and calculating with molarity allows chemists to ensure precise control over the chemical reactions they conduct.

Chemical Reaction Equations

Chemical reaction equations provide a symbolic representation of a chemical reaction. They show the reactants transforming into products, using chemical formulas and states of matter.

In the exercise, the chemical reaction equation outlines how \( \mathrm{H}_{2}\mathrm{SO}_{4} \) and \( \mathrm{NaOH} \) interact during neutralization:\[ \mathrm{H}_{2}\mathrm{SO}_{4}(aq) + 2\ \mathrm{NaOH(aq)} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}(aq) + 2\ \mathrm{H}_{2}\mathrm{O(l)} \]

• Reactants are on the left: \( \mathrm{H}_{2}\mathrm{SO}_{4} \) and \( \mathrm{NaOH} \).
• The arrow points to products: \( \mathrm{Na}_{2}\mathrm{SO}_{4} \) and water.
• Coefficients indicate the number of moles involved: 1 for \( \mathrm{H}_{2}\mathrm{SO}_{4} \) and 2 for \( \mathrm{NaOH} \).

This equation serves as a guide to predict the amount of products formed or reactants needed, ensuring proper chemical balance, crucial for tasks such as acid-base neutralization.