Question

It takes 40 mL of 0.10 M of hydrochloric acid (HCl) to neutralize 50 mL of sodium hydroxide (NaOH) for the following chemical reaction:

HCl + NaOH → NaCl + H2O

The concentration of sodium hydroxide is _____.

(1 point)
Responses

0.08 M
0.08 M

0.04 M
0.04 M

0.125 M
0.125 M

0.10 M
0.10 M
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page 12 of 12

Answer 1

To find the concentration of sodium hydroxide (NaOH), we can use the information given about the neutralization reaction.

1. **Calculate the moles of HCl used**:
- Volume of HCl = 40 mL = 0.040 L
- Molarity of HCl = 0.10 M

\[
\text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.10 \, \text{mol/L} \times 0.040 \, \text{L} = 0.004 \, \text{moles}
\]

2. **Use the stoichiometry of the reaction**:
The reaction between HCl and NaOH is a 1:1 ratio. Therefore, the moles of NaOH will be equal to the moles of HCl.

\[
\text{Moles of NaOH} = 0.004 \, \text{moles}
\]

3. **Calculate the concentration of NaOH**:
- Volume of NaOH = 50 mL = 0.050 L

\[
\text{Molarity of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH}} = \frac{0.004 \, \text{moles}}{0.050 \, \text{L}} = 0.08 \, \text{M}
\]

Thus, the concentration of sodium hydroxide (NaOH) is **0.08 M**.