To find the concentration of carbonic acid (H₂CO₃), we can use the stoichiometry of the reaction and the molarity of the sodium hydroxide (NaOH).
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Determine the number of moles of NaOH used: \[ \text{Moles of } \text{NaOH} = \text{Molarity} \times \text{Volume (in L)} \] Given that the molarity of NaOH is 0.30 M and the volume is 50 mL (which is 0.050 L): \[ \text{Moles of NaOH} = 0.30 , \text{M} \times 0.050 , \text{L} = 0.015 , \text{moles} \]
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Use the stoichiometry from the balanced equation: \[ 2 \text{NaOH} + \text{H}_2\text{CO}_3 \rightarrow \text{Na}_2\text{CO}_3 + 2 \text{H}_2\text{O} \] From the reaction, 2 moles of NaOH react with 1 mole of H₂CO₃. Therefore, the number of moles of H₂CO₃ is: \[ \text{Moles of } \text{H}_2\text{CO}_3 = \frac{\text{Moles of NaOH}}{2} = \frac{0.015 , \text{moles}}{2} = 0.0075 , \text{moles} \]
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Calculate the concentration of H₂CO₃: \[ \text{Concentration of } \text{H}_2\text{CO}_3 = \frac{\text{Moles of } \text{H}_2\text{CO}_3}{\text{Volume of } \text{H}_2\text{CO}_3 \text{ (in L)}} \] Here the volume of H₂CO₃ is 15 mL (which is 0.015 L): \[ \text{Concentration of } \text{H}_2\text{CO}_3 = \frac{0.0075 , \text{moles}}{0.015 , \text{L}} = 0.50 , \text{M} \]
So the concentration of carbonic acid (H₂CO₃) is 0.50 M.