To solve the problem, we need to determine the concentration of the sodium trioxocarbonate (IV) solution used to neutralize the hydrogen chloride (HCl) solution. Let's break down the steps involved in the calculation. ### Step 1: Calculate the concentration of the HCl solution 1. **Determine the number of moles of HCl**: - We know that 0.25 moles of HCl were dissolved in water. 2. **Calculate the concentration of the HCl solution**: - The concentration (C) of a solution is given by the formula: \[ C = \frac{\text{Number of moles}}{\text{Volume in dm}^3} \] - The volume of the HCl solution is 0.50 dm³. - Therefore, the concentration of HCl is: \[ C_{\text{HCl}} = \frac{0.25 \text{ moles}}{0.50 \text{ dm}^3} = 0.50 \text{ mol dm}^{-3} \] ### Step 2: Determine the amount of HCl in the 15.0 cm³ sample 1. **Convert the volume from cm³ to dm³**: - 15.0 cm³ = 0.015 dm³. 2. **Calculate the number of moles of HCl in 15.0 cm³**: - Using the concentration we just calculated: \[ \text{Moles of HCl} = C_{\text{HCl}} \times \text{Volume} = 0.50 \text{ mol dm}^{-3} \times 0.015 \text{ dm}^3 = 0.0075 \text{ moles} \] ### Step 3: Write the neutralization reaction The neutralization reaction between sodium trioxocarbonate (IV) (Na2CO3) and hydrochloric acid (HCl) can be represented as: \[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \] From the balanced equation, we see that 1 mole of Na2CO3 reacts with 2 moles of HCl. ### Step 4: Calculate the moles of Na2CO3 required 1. **Determine the moles of Na2CO3 needed to neutralize 0.0075 moles of HCl**: - According to the stoichiometry of the reaction, 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, the moles of Na2CO3 required is: \[ \text{Moles of Na}_2\text{CO}_3 = \frac{\text{Moles of HCl}}{2} = \frac{0.0075}{2} = 0.00375 \text{ moles} \] ### Step 5: Calculate the concentration of the Na2CO3 solution 1. **Determine the volume of Na2CO3 solution used**: - The volume of Na2CO3 solution used for neutralization is 12.50 cm³, which is equal to 0.0125 dm³. 2. **Calculate the concentration of the Na2CO3 solution**: - Using the formula for concentration: \[ C_{\text{Na}_2\text{CO}_3} = \frac{\text{Number of moles}}{\text{Volume in dm}^3} = \frac{0.00375 \text{ moles}}{0.0125 \text{ dm}^3} = 0.30 \text{ mol dm}^{-3} \] ### Conclusion The concentration of the sodium trioxocarbonate (IV) solution is **0.30 mol dm³**. Therefore, the correct option is **A**. ### Explanation of Other Options - **B. 0.40 mol dm³**: This option is incorrect because it does not match the calculated concentration based on the stoichiometry of the reaction and the moles of HCl neutralized. - **C. 0.50 mol dm³**: This option is also incorrect as it suggests a higher concentration than what was calculated based on the moles of Na2CO3 required for neutralization. - **D. 0.60 mol dm³**: This option is incorrect for the same reasons as above; it overestimates the concentration based on the reaction stoichiometry. ### Revision Summary - The concentration of HCl was calculated to be 0.50 mol dm³. - The moles of HCl in 15.0 cm³ were found to be 0.0075 moles. - The stoichiometry of the reaction indicates that 1 mole of Na2CO3 neutralizes 2 moles of HCl. - The concentration of the Na2CO3 solution was calculated to be 0.30 mol dm³, making option A the correct answer.