Question

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

0.04 M
0.04 M

0.10 M
0.10 M

0.125 M
0.125 M

0.08 M

Answer 1

To determine the concentration of sodium hydroxide (NaOH), we can use the concept of moles and stoichiometry from the balanced chemical equation:

\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the balanced equation, we see that 1 mole of HCl reacts with 1 mole of NaOH.

1. **Calculate the moles of HCl**:

The concentration (\(C\)) and volume (\(V\)) of HCl are given as:
- Concentration of HCl: 0.10 M
- Volume of HCl: 40 mL = 0.040 L

Using the formula for moles (\(n = C \cdot V\)):
\[
n_{\text{HCl}} = 0.10 \, \text{M} \times 0.040 \, \text{L} = 0.004 \, \text{moles}
\]

2. **Calculate the moles of NaOH**:

Since the reaction shows a 1:1 ratio:
\[
n_{\text{NaOH}} = n_{\text{HCl}} = 0.004 \, \text{moles}
\]

3. **Determine the concentration of NaOH**:

We know the volume of NaOH used is 50 mL, which is equal to 0.050 L. We can now find the concentration (\(C_{\text{NaOH}}\)):

Using the moles we calculated:
\[
C_{\text{NaOH}} = \frac{n_{\text{NaOH}}}{V_{\text{NaOH}}}
\]
\[
C_{\text{NaOH}} = \frac{0.004 \, \text{moles}}{0.050 \, \text{L}} = 0.08 \, \text{M}
\]

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