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A 0.723 g sample of chalk (largely calcium carbonate) was dissolved in 20.0 cm3 of 1.00 M hydrochloric acid. When the reaction was complete, the remaining acid was titrated with 0.500 M sodium hydroxide solution. The titre was 12.00 cm3.
What would be used to measure out the 20.0 cm3 of hydrochloric acid?
What feature of the first reaction between the acid and the chalk could lead to an error and how could this be avoided?
Assuming only the calcium carbonate reacts with the acid, what percentage of the chalk sample is calcium carbonate?
The 20.0 cm3 must be measured out with a pipette. Whilst it is an excess of acid, we need to know exactly how much acid is present for the calculation. Consequently, we need to be precise with our measurement.
The reaction produces carbon dioxide gas, so there is effervescence. It is important that no acid spray is allowed to escape, otherwise this will affect our titre value. Using a conical flask for this reaction should limit loss of spray as the sides of the flask slope inwards.
Number of moles of NaOH = 0.500 × 12.00/1000
|
HCl(aq) + NaOH(aq) ⇒ NaCl(aq) + H2O(l) |
HCl º NaOH
\ Number of moles of HCl titrated = 0.500 × 12.00/1000 = 0.00600
Number of moles of HCl originally = 1.00 × 20.0/1000 = 0.0200
\ Number of moles of HCl reacted with CaCO3 = 0.0200 - 0.00600 = 0.0140
|
2HCl(aq) + CaCO3(s) ⇒ CaCl2(aq) + H2O(l) + CO2(g) |
CaCO3 º 2HCl
\ Number of moles of CaCO3 = ½ × 0.0140
CaCO3 = 100
\ Mass of CaCO3 = ½ × 0.0140 × 100 = 0.700 g
Percentage of calcium carbonate = 0.700/0.723 × 100 = 96.8%