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Concentration of a limewater solution by titration against a standard solution of hydrochloric acid. - page 1
Keywords: Chemistry practical concentration limewater hydrochloric acid titration
By Jenny on 02/07/2009
Level: A Level (Year 13)
Page Number: 1 of 3 pages: 1 2 3Determination of the concentration of a limewater solution by titration against a standard solution of hydrochloric acid.
Plan
Our task is to determine the concentration of a limewater solution. To do this we have available a solution of hydrochloric acid which has a concentration of exactly 2.00 mol dmˉ³. To find the concentration of the alkali limewater I will titrate it against the acidic hydrochloric acid.
Ca(OH)² + 2HCl CaCl² + 2H²O
I will use phenolphthalein as an indicator.
There is 250 cm³ of limewater containing approximately 1g dmˉ³ of calcium hydroxide
Molarity of the limewater:
Moles = Mass / Formula Mass Moles = 1 / 74 0.0135 mol dmˉ³
(This is approximate as it we only know the limewater contains approximately 1g of calcium hydroxide.)
The problem we have is that the 2.00 mol dmˉ³ hydrochloric acid is much more concentrated than the 0.0135 mol dmˉ³ limewater. This means that if you were titrating 25.0 cm³ of the limewater against the hydrochloric acid you would only need a small amount of hydrochloric acid.
Calculating amount of hydrochloric acid required to dilute 25 cm³ of 0.0135 mol dmˉ³ limewater:
The limewater has 0.0135 moles in 1000 cm³.
So, it has 0.0003375 moles in 25 cm³.
We know from the equation for the reaction that hydrochloric acid and limewater react in a ratio of 2:1.
So, 0.000675 moles in 25 cm³ of hydrochloric acid are needed to titrate the limewater.
Or, 0.027 moles in 1000 cm³.
The hydrochloric acid that we have has 2.00 moles in 1000 cm³.
Mass of hydrochloric acid in the solution we have:
Mass = Moles x Formula Mass Mass = 2 x 36.5 73g in 1000cm³
Mass of hydrochloric acid in the solution we want:
Mass = Moles x Formula Mass Mass = 0.027 x 36.5 0.9855g in 13.5cm³
You divide the 2.00 mol dmˉ³ hydrochloric acid by 74.074 to get the 0.027 hydrochloric acid, so you divide 1000 cm³ by 74.074 to get 13.5 cm³.
The amount of limewater being titrated is 25 cm³ not 1000 cm³ (divide by 40), so the amount of hydrochloric acid used is 0.3375 cm³ not 13.5 cm³
0.3375 cm³ of hydrochloric acid required to dilute the 25 cm³ of limewater.
The burette is only accurate to 0.05 cm³ so the two readings have a total error of 0.1 cm³.
0.1 / 0.3375 x 100 = 29.6% error margin
As the acceptable error margin is below 0.5% this is far too high. To solve this problem I will dilute the hydrochloric acid.
To minimise error it is
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