It takes 50mL of 0.30M of sodium hydroxide (

Question

N
a
O
H
) to neutralize 15mL of carbonic acid (
H
2
C
O
3
) for the following chemical reaction:

2
N
a
O
H

+

H
2
C
O
3

→

N
a
2
C
O
3

+

2
H
2
O


The concentration of the carbonic acid is ___________.

(1 point)
Responses

0.50 M
0.50 M

0.60 M
0.60 M

0.96 M
0.96 M

0.24 M

Answer 1

To find the concentration of carbonic acid (\( H_2CO_3 \)), we need to start with the stoichiometry of the neutralization reaction provided:

\[ 2 NaOH + H_2CO_3 \rightarrow Na_2CO_3 + 2 H_2O \]

From the balanced equation, we can see that 2 moles of sodium hydroxide react with 1 mole of carbonic acid.

Step 1: Calculate moles of \( NaOH \)

First, we convert the volume of sodium hydroxide from mL to L:

\[ 50 , \text{mL} = 0.050 , \text{L} \]

Now we can calculate the number of moles of \( NaOH \):

\[ \text{Moles of } NaOH = \text{Concentration} \times \text{Volume} = 0.30 , \text{mol/L} \times 0.050 , \text{L} = 0.015 , \text{moles} \]

Step 2: Calculate moles of \( H_2CO_3 \)

According to the balanced equation, 2 moles of \( NaOH \) correspond to 1 mole of \( H_2CO_3 \). Therefore, the moles of \( H_2CO_3 \) can be calculated as follows:

\[ \text{Moles of } H_2CO_3 = \frac{1}{2} \times \text{Moles of } NaOH = \frac{1}{2} \times 0.015 = 0.0075 , \text{moles} \]

Step 3: Calculate the concentration of \( H_2CO_3 \)

Next, we need to calculate the concentration of \( H_2CO_3 \) using the volume of the carbonic acid solution, which is 15 mL or 0.015 L:

\[ \text{Concentration of } H_2CO_3 = \frac{\text{Moles of } H_2CO_3}{\text{Volume in L}} = \frac{0.0075 , \text{moles}}{0.015 , \text{L}} = 0.50 , \text{M} \]

Thus, the concentration of carbonic acid is:

0.50 M