In our lab experiment, we filtered CaCl and CaOH2 solutions that we prepared the
ID: 504460 • Letter: I
Question
In our lab experiment, we filtered CaCl and CaOH2 solutions that we prepared the previous week and transferred both of the filtrates using a 25 ml pipette into two erlenmeyer flasks. We then prepared the standard HCl by pipeting 100 ml of HCl to prepare 250 ml of 0.04M HCl from the 0.1 M HCl solution that we initially had. We pipeted 100 ml of HCl into a 250 ml volumetric flask and diluted to the mark. The buret was filled with the prepared HCl and both the CaCl and CaOH2 were titrated until a yellow endpoint was reached, this titration was repeated three times for each solution.
The data that I obtained was: CaCl= Initial(ml)--Final (ml) --- Volume of HCl(ml)
1.30 --24.2--22.9
0.50 --23.5 --23.0
1.60 --23.2 -- 21.6
Ca(OH)2= initial--final-- volume of HCl
1.00-- 28.3-- 27.3
2.00-- 28.8-- 26.8
2.00-- 28.1 --26.1
Q. Using mean titration volumes, calculate the hydroxide concentration in each of the two saturated Ca(OH)2 solutions.
Explanation / Answer
In our lab experiment, we filtered CaCl and CaOH2 solutions that we prepared the previous week and transferred both of the filtrates using
a 25 ml pipette into two erlenmeyer flasks. We then prepared the standard HCl by pipeting 100 ml of HCl to prepare 250 ml of 0.04M HCl from
the 0.1 M HCl solution that we initially had. We pipeted 100 ml of HCl into a 250 ml volumetric flask and diluted to the mark.
The buret was filled with the prepared HCl and both the CaCl and CaOH2 were titrated until a yellow endpoint was reached,
this titration was repeated three times for each solution.
The data that I obtained was: CaCl= Initial(ml)--Final (ml) --- Volume of HCl(ml)
1.30 --24.2--22.9
0.50 --23.5 --23.0
1.60 --23.2 -- 21.6
Note : HCl do not react with CaCl2, so I assume that it is Ca(OH)2:
The mean volome of HCl = (22.9 + 23+ 21.6)/3 = 22.5 ml
No. of moles of HCl = molarity x volume in L = 0.04 M x 0.0225 L = 0.0009 moles
Balanced equation:
2HCl + Ca(OH)2 --> CaCl2 + 2H2O
The HCl and Ca(OH)2 react in a 2:1 molar ratio.
Therefore, no moles of Ca(OH)2 = no. moles of HCl/2 = 0.0009/2 = 0.00045 moles
We have used 25 mL for the titration, so molarity of the solution is
M = (no. moles)/(volume in L) = 0.00045/0.025L = 0.018 M
Ca(OH)2= initial--final-- volume of HCl
1.00-- 28.3-- 27.3
2.00-- 28.8-- 26.8
2.00-- 28.1 --26.1
Similarly;
The mean volome of HCl = (27.3 + 26.8+ 26.1)/3 = 26.73 ml
No. of moles of HCl = molarity x volume in L = 0.04 M x 0.026733 L = 0.00107 moles
Balanced equation:
2HCl + Ca(OH)2 --> CaCl2 + 2H2O
The HCl and Ca(OH)2 react in a 2:1 molar ratio.
Therefore, no moles of Ca(OH)2 = no. moles of HCl/2 = 0.0009/2 = 0.00053 moles
We have used 25 mL for the titration, so molarity of the solution is
M = (no. moles)/(volume in L) = 0.00053/0.025L = 0.0214 M
Q. Using mean titration volumes, calculate the hydroxide concentration in each of the two saturated Ca(OH)2 solutions.
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