Question

A chemist uses 0.25 L of 2.00 M H2SO4 to completely neutralize a 2.00 L of solution of NaOH. The balanced chemical equation of the reaction is given below.

2NaOH + H2SO4 Right arrow. Na2SO4 + 2H2O

What is the concentration of NaOH that is used?
0.063 M
0.25 M
0.50 M

Answer 1

To determine the concentration of the NaOH solution used, we can start by calculating the amount of H₂SO₄ that was used in the reaction, and then use stoichiometry to find the concentration of NaOH.

1. **Calculate moles of H₂SO₄:**
- Molarity (M) = moles/volume (L)
- Given 0.25 L of 2.00 M H₂SO₄:
\[
\text{Moles of H₂SO₄} = 2.00 \, \text{M} \times 0.25 \, \text{L} = 0.50 \, \text{moles}
\]

2. **Use the balanced reaction to find moles of NaOH:**
- The balanced equation states that 1 mole of H₂SO₄ reacts with 2 moles of NaOH:
\[
H₂SO₄ + 2 \, \text{NaOH} \rightarrow \text{Na₂SO₄} + 2 \, \text{H₂O}
\]
- Therefore, 0.50 moles of H₂SO₄ will react with:
\[
0.50 \, \text{moles H₂SO₄} \times 2 \, \text{moles NaOH/mole H₂SO₄} = 1.00 \, \text{mole NaOH}
\]

3. **Calculate the concentration of NaOH:**
- We know there are 1.00 mole of NaOH in a 2.00 L solution:
\[
\text{Concentration (M)} = \frac{\text{moles}}{\text{volume (L)}}
\]
\[
\text{Concentration of NaOH} = \frac{1.00 \, \text{moles}}{2.00 \, \text{L}} = 0.50 \, \text{M}
\]

Thus, the concentration of NaOH used is **0.50 M**.