# Titration problem

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#1
If I have an unknown which has certain concentrations of Carbonate and biCarbonate ions. In addition, I have a ph.ph indicator and methyl orange indicator

-First of all, you used the ph.ph to titrate the unknown with HCL then the volume used from HCL (X) is equivalent to titrate half the concentration of carbonate ions.
-Secondly, you used methyl orange to titrate the same volume of the unknown with HCL and the volume used of HCL (Y) is equivalent to titrate all the concentrations of Carbonate and bicarbonate ions.
-you have molarity of HCL = M(HCl)
Calculate concentrations of carbonate M(CO3) and bicarbonate M(HCO3) ions in the unkown

My attempt :
After first titration :-
2X . M(HCl) = Volume of CO3 . M(CO3) = C

Now, I can get molarity of CO3 in sample and no. of moles of carbonate that titrated to bicarbonate (C)

After second titration :-
(Y-2x) . M(HCl) = C + number of moles of bicarbonate in unknown sample

Now, I can calculate number of moles of bicarbonate in unknown sample and divide it by volume so I can get its Molarity

IS THIS SOLUTION TRUE ??

__________________
0
8 years ago
#2
(Original post by m.maher)
First of all, you used the ph.ph to titrate the unknown with HCL then the volume used from HCL (X) is equivalent to titrate half the concentration of carbonate ions.
This looks to be slightly misleading. It is not half the concentration, it is half of the FULL titration, that is first end point is after

CO32- + H+ -> HCO3-

and not after

CO32- + 2H+ -> H2CO3
0
#3
I think that this question is very misleading too since half the concentration may means that it is used to titrate half the amount of carbonate to bicarbonate or it is used to do half the titration (All the carbonate to bicarbonate) but do you think my way of solving this problem is true if X is volume used to convert half the carbonate to bicarbonate ???
0
8 years ago
#4
Your choice of symbols is rather unfortunate, as each time I look at the formulas I start to wonder concentration of what is denoted by C.

I think your second equation is wrong - Y-2X already takes care of the carbonate presence, so there is no need for C on the right.

But I can be wrong.
0
8 years ago
#5
no
0
#6
Let me express my solution a bit more :
C = no of moles of carbonate titrated to bicarbonate from the first titration
Y-2X = volume of HCL used to titrate (Bicarbonate in the sample + bicarbonate produced from titrated carbonate) That's why I added C moles with M(HCO3) . V(Sample)
0
8 years ago
#7
(Original post by m.maher)
Y-2X = volume of HCL used to titrate (Bicarbonate in the sample + bicarbonate produced from titrated carbonate)
No. If X is a volume required to convert carbonate to bicarbonate, volume required to titrate sum of bicarbonates is Y-X, not Y-2X.
0
X

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