To find the concentration of the vinegar used in the reaction with sodium bicarbonate (NaHCO₃), we'll follow these steps:
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Calculate the number of moles of NaHCO₃ in 2.2 g of baking soda:
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The molar mass of NaHCO₃ (sodium bicarbonate) is calculated as follows:
- Na: 23 g/mol
- H: 1 g/mol
- C: 12 g/mol
- O (3 oxygens): 3 × 16 g/mol = 48 g/mol
- Total = 23 + 1 + 12 + 48 = 84 g/mol
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Now, calculate the moles of NaHCO₃: \[ \text{moles of NaHCO₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.2 \text{ g}}{84 \text{ g/mol}} \approx 0.0262 \text{ mol} \]
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Calculate the number of moles of NaHCO₃ that reacted with vinegar:
- The concentration of the NaHCO₃ solution is 0.1 mol/dm³, and we need to convert the volume from cm³ to dm³: \[ 60 \text{ cm}^3 = 60 \text{ mL} = 0.060 \text{ dm}^3 \]
- Calculate the moles of NaHCO₃ in that volume: \[ \text{moles of NaHCO₃} = \text{concentration} \times \text{volume} = 0.1 \text{ mol/dm}^3 \times 0.060 \text{ dm}^3 = 0.006 \text{ mol} \]
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Determine the molar ratio in the reaction:
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The balanced reaction between NaHCO₃ and acetic acid (vinegar) is: \[ \text{NaHCO₃} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \]
- This shows a 1:1 molar ratio.
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Therefore, the moles of vinegar (acetic acid, CH₃COOH) that reacted is also 0.006 mol.
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Calculate the concentration of the vinegar:
- The volume of vinegar used is 75 cm³, which is equivalent to: \[ 75 \text{ cm}^3 = 0.075 \text{ dm}^3 \]
- The concentration (C) of vinegar (acetic acid) can be calculated using the formula: \[ C = \frac{\text{moles}}{\text{volume}} = \frac{0.006 \text{ mol}}{0.075 \text{ dm}^3} \approx 0.08 \text{ mol/dm}^3 \]
Conclusion: The concentration of the vinegar used is approximately 0.08 mol/dm³.