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

1 answer

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.