Question

It takes 50mL of 0.30M of sodium hydroxide (NaOH
) to neutralize 15mL of carbonic acid (H2CO3
) for the following chemical reaction:

2NaOH + H2CO3 → Na2CO3 + 2H2O
The concentration of the carbonic acid is ___________.

(1 point)
Responses

0.60 M
0.60 M

0.24 M
0.24 M

0.96 M
0.96 M

0.50 M

Answer 1

To determine the concentration of carbonic acid (H2CO3), we can use the stoichiometry of the reaction and the information given.

1. **Identify the reaction and stoichiometry:**
The balanced equation is:
\[
2 \, \text{NaOH} + \text{H}_2\text{CO}_3 \rightarrow \text{Na}_2\text{CO}_3 + 2 \, \text{H}_2\text{O}
\]
This tells us that 2 moles of NaOH react with 1 mole of H2CO3.

2. **Calculate the moles of NaOH used:**
We know the concentration and volume of NaOH:
\[
\text{Concentration of NaOH} = 0.30 \, \text{M}
\]
\[
\text{Volume of NaOH} = 50 \, \text{mL} = 0.050 \, \text{L}
\]
The number of moles of NaOH is given by:
\[
\text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.30 \, \text{mol/L} \times 0.050 \, \text{L} = 0.015 \, \text{mol}
\]

3. **Use the stoichiometry to find moles of H2CO3:**
From the reaction, 2 moles of NaOH react with 1 mole of H2CO3. Therefore, the moles of H2CO3 that would react with 0.015 moles of NaOH is:
\[
\text{Moles of H2CO3} = \frac{0.015 \, \text{mol NaOH}}{2} = 0.0075 \, \text{mol H2CO3}
\]

4. **Calculate the concentration of H2CO3:**
We know the volume of the H2CO3 solution used:
\[
\text{Volume of H2CO3} = 15 \, \text{mL} = 0.015 \, \text{L}
\]
The concentration of H2CO3 can be calculated as:
\[
\text{Concentration of H2CO3} = \frac{\text{Moles of H2CO3}}{\text{Volume of H2CO3}} = \frac{0.0075 \, \text{mol}}{0.015 \, \text{L}} = 0.50 \, \text{M}
\]

Therefore, the concentration of carbonic acid (H2CO3) is **0.50 M**.