Question

In: Chemistry

In our lab experiment, we filtered CaCl and CaOH2 solutions that we prepared the previous week...

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.

Solutions

Expert Solution

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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