Titration Related Calculations

Titration Related Calculations


A lab technician makes up some limewater by stirring 4.27g of calcium hydroxide [Ca(OH)2] with water to make 1dm3 of solution. As it is cloudy she filters the solution. Exactly 25cm3 of 0.1M HCl requires 23.6cm3 of calcium hydroxide solution to neutralise it.
i) Moles of HCl in 25.0cm3: My answer= 0.0025
ii) Moles of Calcium Hydroxide in titre: My answer= 0.0925
iii) Molar and Mass concentration of Calcium Hydroxide solution

I have worked out the Molar concentration by:
—-—- x Number of Moles = 3.92mols dm-3

No. moles= 0.0925
Volume = 23.6 dm3

However im not sure whether this is correct and cannot work out the Mass concentration (which i think the units would be "g per dm-3"???)

Thanks if you get back to me


your units of g/dm3 are correct for gram concentration
I cant see how you arrived at the calcium hydroxide concentration of 3.92.
Your calculation of the hydrochloric acid concentration is fine.
Remember that at the titration end point the number of moles of acid and alkali are equal ifthe stoichiometry of the neutralisation reaction is 1:1
Calcium is in group 2 so the equation for the neutralisation is………….
HINT: the values given for the mass of calcium hydroxide used to make up the solution are not correct because the "solution" was filtered i.e. some of the calcium hydroxide which was weighed out did not dissolve!
Hope this helps and thanks for your contribution to the site Dr Bowker

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