[Scheikunde]

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.Since my son cant write this length in english, I am assisting him.
also, I can not render the text here, so for e.g:  Calcium carbonate is caco3 (in the text below)
and 10 to the power minus four is : 10 -4   etc
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:

++++++++++++Snip+++++++++++++++++++++

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''

then

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.

OUR SOLUTION(My Son's):
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 ++++++++++++++++++++++++++++++++++

let me know if some things weren't clear here .

Thanks for any assistance !

Regards

[Borek]

It is quite simple to correctly format your posts with buttons over the edit field (you will find there also a link to the About post formatting... page). CaCO₃, H₃O⁺, 10^-4, no need to reinvent the wheel.

I am afraid your post is way too chaotic to to guarantee any help I feel like you are mistaking substances and solutions several times, making following your line of thinking impossible.

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

That's OK, this is is initial amount of HCl used.

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

No idea what you mean here. 100 mL solution didn't contain OH^- (or rather: it did contain some minute amount, but its calculation can't be done with stoichiometry which you are expected to use). What you have calculated is the number of moles of NaOH in the 8.3 mL used to titrate excess acid.

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 .

Why 8.3 mL of hydrochloric acid, if 8.3 mL was a volume of the NaOH solution? You titrated 10 mL aliquot of acid, not 8.3 mL of it.

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.

Amount makes a bit of sense, but you assign it to a wrong solution.

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.

As I can only guess what you are doing, I bet that's where you went wrong - you forgot you have not titrated whole 100 mL, but 1/10 of it.

[AWK]

Something is wrong in your data.
For 100 ml of 0.09 M HCl you need 100x0.09/0.12 = 75 ml of 0.12 M NaOH, and 10 x 8.3 = 83 ml were used (it means that sample contained acid, not calcium carbonate).
Interchanging concentrations of HCl and NaOH (my guess) you can obtain 22.7 % of CaCO₃ in the sample.

[Scheikunde]

Dear Borek & AWK

Firstly thanks for the replies !
Yes indeed, at hindsight even when I read the issue description , it comes over quiet chaotic .
My apologies for that.
ITMT we have resolved the issue and just for the records our approach to the known and unknown solutions were a bit botched up.
on a short note, we were making mistakes in getting the concentration of the unknown solution.
Thanks again and hopefully till the next mind Jammer.

*and I will render the text from the next time via the helpful tool that was pointed out to me.