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November 25th, 2018, 09:07 AM   #1
Joined: Nov 2012
From: The Netherlands

Posts: 37
Thanks: 2

Help with a Titration Issue

Hi All

I am the father of my 16 year Old Son. We are in the Netherlands , so please do not mind the language translation in the problem description.
also, I can not render the text here, co Calcium carbonate is caco3
and 10 to the power minus four is : 10 -4
But I hope that would be forgiven .

He has come to me with a titration issue which I can not resolve myself.
He has done his Bit, but we both think that the approach has not been good .

This is the Problem description and the solution that my son has tried to achieve:


In a beaker 100 ml HCL, a 1 gram paper with Caco3 was dipped inside .
from the 100 ml solution , (a reaction will take place) and now we took 10 ml out into a new a beaker ''B''


The amount of gram of paper that we used: 1.00 grams
The molarity of hydrochloric acid (HCl) used: 0.09
The molarity of sodium hydroxide solution (NaOH) used: 0.12

1) we used 8.30 ml (8.3 x 10 -3 L) of Naoh in a buret to dip inside beaker B

Question is: what is mass % of Caco3 from the 1 gram paper.

what we have done :
Calculation for the number of moles of H₃O⁺ in the 100ml solution before paper which came in(got dipped in):
Mol H₃O⁺ 0.09 mol/lts

X = 0.09 mole x 0.1 L1.00 L = 9 x 10 -3 mole HCL

Thus 9 x 10 -3 mol H₃O⁺ ions are contained in a solution of 100 ml hydrochloric acid

Calculation for the number of moles OH- in the 100 ml solution before paper came in:
Mol OH-0.12 X
used= Liter OH- 1.00 8.30 x 10 -3

X = 0.12 mole x 8.3 x 10 -3 L1.00 L = 9.96 x 10 -4 moles

Thus 9.96 x 10 -4 mol OH - ions are contained in a solution of 8.30 ml hydrochloric acid. We used 8.3 NaOh ml from the buret .
So there are also 9.96 x 10 -4 mol of H₃O⁺ ions in a solution of 8.30 ml of hydrochloric acid, because the ratio between the reaction of the H₃O⁺ ions and the OH ions is 1: 1. See below:

OH- (ag) + H₃O⁺ (aq) → 2H2O (l)

In order to know how many moles of H₃O⁺ ions have reacted with the OH - ions, you have to subtract the number of moles of H₃O⁺ ions contained in the 100 ml hydrochloric acid solution with the number of moles of H₃O⁺ ions to which you have titrated (Beaker B) .
Thus 9 x 10 -3 moles - 9.96 x 10 -4 mole = 8.004 x 10 -3 moles. So 8,004 x 10 -3mol H₃O⁺ ions have reacted.

To determine the amount of mole of CaCO₃ you must divide the number of reacted moles of H₃O⁺ ions by 2, because the reaction H₃O⁺ reacts with CaCO₃ in the ratio 2: 1. See the comment below:

CaCO₃ (s) + 2H₃O⁺ (aq) → Ca²⁺ (aq) + CO₂ (g) + 3H₂O (l)

Thus 8.004 x 10 -3 mol ÷ 2 = 4.002 x 10 -3mol. So there are 4.002 x 10 -3 moles of CaCO₃ parts in a 10 ml solution. If we round this in the correct number of significant digits you will get 4.002 x 10 -3mol CaCO3 parts in a 10 ml solution.

Calculation for the number of grams of CaCO₃ in 4.002 x 10 -3 mol CaCO₃:
Mol CaCO3 1.00 4.002 x 10 -3
Gram CaCO3 100.09 X

X = 4.002 x 10 -3 mol x 100.09 gram1.00 mol = 0.40 gram ( in Beaker B)

So there is 0.40 grams of CaCO₃ in 4.00 x 10 -3 moles of CaCO3.

now what we have done is:
0.4 gms in 10 ml (Beaker B), gives 4 gm in the initial 100 ml HCL solution ,but this cant be true beacause the paper was 1 gram , and we can not have 4 grams of caco3 on the paper (as the solution to this problem will always be < 1 gram).

where have we gone wrong here ?

+++++++++++++++End Snip ++++++++++++++++++++++++++++++++++

Thanks for any assistance !

hello_math is offline  
November 26th, 2018, 12:07 PM   #2
Joined: Nov 2012
From: The Netherlands

Posts: 37
Thanks: 2

Hi All
Thanks to all.
ITMT, we have resolved this issue.
Just for the records , we consulted :

and came to know why were we approaching this in a wrong manner.

Thanks all.
hello_math is offline  

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