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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

0.24 M
0.24 M

0.50 M
0.50 M

0.60 M
0.60 M

0.96 M
0.96 M
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1 answer

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

Given:

  1. The reaction is:
    \( 2 \text{NaOH} + \text{H}_2\text{CO}_3 \rightarrow \text{Na}_2\text{CO}_3 + 2 \text{H}_2\text{O} \)

  2. Volume and concentration of NaOH:
    \( V_{\text{NaOH}} = 50 \text{ mL} = 0.050 \text{ L} \)
    \( C_{\text{NaOH}} = 0.30 \text{ M} \)

  3. Volume of H₂CO₃:
    \( V_{\text{H}_2\text{CO}_3} = 15 \text{ mL} = 0.015 \text{ L} \)

Step 1: Calculate moles of NaOH.

\[ \text{Moles of NaOH} = C_{\text{NaOH}} \times V_{\text{NaOH}} = 0.30 , \text{mol/L} \times 0.050 , \text{L} = 0.015 , \text{mol} \]

Step 2: Use stoichiometry to find moles of H₂CO₃.

According to the balanced reaction, 2 moles of NaOH react with 1 mole of H₂CO₃. So the moles of H₂CO₃ can be calculated from the moles of NaOH:

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

Step 3: Calculate the concentration of H₂CO₃.

Now, we can find the concentration of H₂CO₃ using the volume of H₂CO₃.

\[ C_{H_2CO_3} = \frac{\text{moles of } H_2CO_3}{V_{H_2CO_3}} = \frac{0.0075 , \text{mol}}{0.015 , \text{L}} = 0.50 , \text{M} \]

The concentration of carbonic acid (H₂CO₃) is 0.50 M.

Thus, the correct response is:

0.50 M