% w/w of CaCO3 in Chalk (calculation check for backward titration)?

Question

his is the method I used:

2.0312 g of chalk / suspended in 100cm3 of water and 50cm3 of HCl(excess). Mixture is boiled for 1 minute then cooled and then its titrated against NaOH.

the Molarity of the HCL is 0.9931
the Molarity of the NaOH is 0.9931

the result of the titration is 9.9 (NaOH in the buirette and the excess HCl in the flask)

My calculation:

HCl(excess) + NaOH ---> xxxxxxx

moles of NaOH = (0.9931x10.00)/1000 = 9.931x10^-3 moles
1:1 ration with HCl so moles of HCl = 9.931x10^-3 moles

Moles of the HCl used at the start of the experiment = (0.9931x50.00)/1000 = 0.49655 moles

Amount of HCl used in the equation below = 0.49655 - (9.931x10^-3) = 0.039724 moles

CaCO3 + 2HCl ------> xxxxxxxx + excess HCl


its a 1:2 ration so 0.039724 / 2 = 0.019862 moles of CaCO3
Mr of CaCO3 = 100.087
Mass of CaCO3 = 100.087 x 0.019862 =1.988g
therefore Percentage w/w is 97.87%

does it look right?

titration is 10.00 not 9.9

Answer 1

The calculation may be simplified as follows :
- First calculate the volume of HCl remaining unused after the rection-

N1V1 = N2V2
Here N1 = normality of NaOH =0.9931N
V1 = 9.9 ml.
N2 = 0.9931 N
V2 = Vol. remaing used to be known;

or 0.9931*9.9=0.9931*V2

or V2 =9.9 ml.

It means volume of HCl used in the reaction = 50-9.9 =40.1ml
Normality of 40.1 ml HCl =40.1*0.9931=0.398233N
Equivalent normality of CaCO3 =0.398233N
Weight of CaCO3 = 0.398233*50
or = 19.91165 gm/litre
or =1.991165 gm/100 cc

or Percentage of CaCO3 =(1.991165/2.0312 )*100
=98.029 %

Answer 2

hi :) of course! XD bypass on an escalator, haha XD And bypass on one that is going down, and you will walk up it, and you will by no potential get to the suited (till you certainly run particularly rapid, lol). And that way, you're walking forward and backward on a similar time, lol =P

Answer 3

I have really checked your work and it is perfect

Answer 4

YES!!!!