To determine how much carbonic acid (H₂CO₃) is being neutralized in this reaction, we can follow these steps:
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Calculate the moles of NaOH used: First, we need to find the number of moles of NaOH in the solution used for neutralization.
\[ \text{Moles of NaOH} = \text{Concentration (M)} \times \text{Volume (L)} \]
The concentration of NaOH is 0.30 M, and the volume used is 50 mL, which is 0.050 L:
\[ \text{Moles of NaOH} = 0.30 , \text{mol/L} \times 0.050 , \text{L} = 0.015 , \text{mol} \]
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Use the stoichiometry of the reaction: According to the balanced chemical equation:
\[ 2 , \text{NaOH} + \text{H}_2\text{CO}_3 → \text{Na}_2\text{CO}_3 + 2 , \text{H}_2\text{O} \]
From the equation, we see that 2 moles of NaOH react with 1 mole of H₂CO₃. Therefore, we can find the moles of H₂CO₃ that are neutralized:
\[ \text{Moles of H₂CO₃} = \frac{\text{Moles of NaOH}}{2} = \frac{0.015 , \text{mol}}{2} = 0.0075 , \text{mol} \]
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Calculate the concentration of H₂CO₃: We know that the volume of H₂CO₃ used is 15 mL or 0.015 L. We can calculate the molarity (concentration) of H₂CO₃:
\[ \text{Concentration of H₂CO₃} = \frac{\text{Moles of H₂CO₃}}{\text{Volume of H₂CO₃ (L)}} \]
\[ \text{Concentration of H₂CO₃} = \frac{0.0075 , \text{mol}}{0.015 , \text{L}} = 0.50 , \text{M} \]
In summary, the concentration of carbonic acid (H₂CO₃) in the solution is 0.50 M.