To find the concentration of the sodium trioxocarbonate solution, we will first determine how many moles of trioxonitrate (V) acid (which is nitric acid, \(HNO_3\)) are present in the solution.
Step 1: Calculate moles of trioxonitrate (V) acid.
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Find the molar mass of \(HNO_3\):
- Hydrogen (H): 1 g/mol
- Nitrogen (N): 14 g/mol
- Oxygen (O): 16 g/mol x 3 = 48 g/mol
\[ \text{Molar mass of } HNO_3 = 1 + 14 + 48 = 63 \text{ g/mol} \]
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Calculate the number of moles of \(HNO_3\):
- Mass of \(HNO_3\) = 0.67 g
- Using the formula:
\[ \text{Moles of } HNO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.67 \text{ g}}{63 \text{ g/mol}} \approx 0.0106 \text{ mol} \]
Step 2: Write the neutralization reaction.
The neutralization reaction between nitric acid and sodium trioxocarbonate (the sodium salt of carbonate) can be represented as:
\[ 2 HNO_3 + Na_2CO_3 \rightarrow 2 NaNO_3 + H_2O + CO_2 \]
From the balanced equation, we see that 2 moles of \(HNO_3\) react with 1 mole of \(Na_2CO_3\). Therefore, the molar ratio of \(HNO_3\) to \(Na_2CO_3\) is 2:1.
Step 3: Calculate moles of sodium trioxocarbonate.
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Find the moles of sodium trioxocarbonate:
- Given that 2 moles of \(HNO_3\) react with 1 mole of \(Na_2CO_3\):
\[ \text{Moles of } Na_2CO_3 = \frac{\text{Moles of } HNO_3}{2} = \frac{0.0106}{2} = 0.0053 \text{ mol} \]
Step 4: Calculate concentration of sodium trioxocarbonate solution.
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Volume of sodium trioxocarbonate solution:
- Volume of the sodium trioxocarbonate solution used = 25 cm³ = 0.025 dm³
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Calculate the concentration of the \(Na_2CO_3\) solution:
- Concentration (\(C\)) is defined as:
\[ C = \frac{\text{moles}}{\text{volume (in L)}} \]
\[ \text{Concentration of } Na_2CO_3 = \frac{0.0053 \text{ mol}}{0.025 \text{ dm}^3} = 0.212 \text{ mol/dm}^3 \]
Final Answer:
The concentration of the sodium trioxocarbonate (Na₂CO₃) solution is approximately 0.212 mol/dm³.