Chemistry - Back titrations - worked example
Back titrations - worked example
In a laboratory mix-up, a container of sodium chloride NaCl became
uniformly contaminated with sodium carbonate Na2CO3.
In order to determine the extent of the contamination, the following
procedure was followed. A 4.00 g sample of the salt was added to 500 mL of
0.100 M HCl(aq).
When the reaction
was complete, the amount of hydrochloric acid remaining was determined by
titrating a 25.00 mL aliquot of the solution with 0.0500 M NaOH(aq).
40.50 mL of the NaOH(aq) was required to reach the endpoint of the
Determine the percentage, by mass, of Na2CO3 in the contaminated sample.
The important aspect of such an analysis is that the contaminant be
accurately quantified. To ensure that all the contaminant Na2CO3 reacts
when an excess of acid is used.
It then becomes a matter of realising that _n_(HCl) actually reacting with
the contaminant Na2CO3 will need to be determined via:
Tagging the equation with the relevant data helps simplify the
The skill now is to think back through the steps required to determine the
amount of Na2CO3 reacting in and consequently the_ m_(Na2CO3) in the
original salt sample.
The key is to pick up on the necessary 'adjustments' e.g. realising that
only 25.00 mL of the 500 mL of original solution is analysed for excess
acid and thus the need to adjust back to 500 mL
A possible calculation procedure to follow is:
Consider the calculations in four parts, ie determining:
(i) The _n_(HCl) remaining after the reaction with Na2CO3.
(ii) The _n_(HCl) reacting with Na2CO3.
(iii) The_m_(Na2CO3) in the contaminated sample.
(iv) The %, by mass, Na2CO3 in the contaminated sample.
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