To find the concentration of the carbonic acid (H₂CO₃), we can use the stoichiometry of the reaction and the data given.
Given:
-
The reaction is:
\( 2 \text{NaOH} + \text{H}_2\text{CO}_3 \rightarrow \text{Na}_2\text{CO}_3 + 2 \text{H}_2\text{O} \) -
Volume and concentration of NaOH:
\( V_{\text{NaOH}} = 50 \text{ mL} = 0.050 \text{ L} \)
\( C_{\text{NaOH}} = 0.30 \text{ M} \) -
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