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 + H O
The concentration of sodium hydroxide is _____.
(1 point)
2
0.08 M
0.125 M
0.10 M
0.04 M

Answer 1

To find the concentration of sodium hydroxide (NaOH), we first need to calculate the amount of moles of hydrochloric acid (HCl) used in the reaction.

1. **Calculate the moles of HCl:**
\[
\text{Moles of HCl} = \text{Concentration} \times \text{Volume}
\]
\[
\text{Moles of HCl} = 0.10 \, \text{M} \times 0.040 \, \text{L} = 0.004 \, \text{moles}
\]

2. **Use the stoichiometry of the reaction:**
The balanced chemical equation is:
\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]
From the balanced equation, we see that the mole ratio of HCl to NaOH is 1:1. Therefore, the moles of NaOH will also be 0.004 moles.

3. **Calculate the concentration of NaOH:**
We know the volume of NaOH is 50 mL, which is 0.050 L. Now we can calculate the concentration:
\[
\text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of NaOH}} = \frac{0.004 \, \text{moles}}{0.050 \, \text{L}} = 0.08 \, \text{M}
\]

So, the concentration of sodium hydroxide is **0.08 M**.