Hey Malovane ...I think I'm in your class. I too have the same lab to do tomorrow(you may have been Tuesday). Your write up reads verbatum from my lab manual.
)) School starts with a K....teacher is JA?? Right?
Here's what I've done for the pre-lab and I may be making an error - perhaps someone can point out what I'm doing wrong.
First I identified the moles of HCl initially present which was 0.04
Then I figured out how many moles of NaOH reacted in the titration.
NaOH and HCl are 1/1 so there was 0.00132 moles HCl in excess present.
Which means that 0.04 moles HCl minus 0.00132 moles HCl leaves 0.03868 moles hcl that reacted with the ca(oh)2
I figured that the ca(oh)2 is 2/1 ration which gave me 0.07736 moles .
0.500g divided by 0.07736 moles = 93.28 which is NOT the molar mass of Ca(OH)2..?!!!!