Here is a sample calculation:
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Data:
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Mass of 50.0 mL vinegar: 51.0 g
Number of mL of vinegar to reach endpoint: 23.0 mL
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Calculations:
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Density of vinegar:
51.0 g/ 50.0 mL = 1.02 g/mL
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Number of grams of vinegar added:
(1.02 g/mL) x (23.0 mL) = 23.5 g
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Mass of C2H4O2Â added:
(23.5 g) x (0.0500) = 1.18 g
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Molecular Mass of C2H4O2Â = 60.0 amu
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Number of moles of C2H4O2Â added:
(1.18 g/1) x (1 mole/60.0 grams) = 0.0197 moles
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Number of moles of ammonia that exist in solution:
(0.0197 moles C2H4O2Â /1)x(1 mole ammonia/1 mole C2H4O2) =
0.0197 moles ammonia
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Concentration of ammonia:
0.0197 moles ammonia / 0.0100 L = 1.97 M