Titration of mixture of KOH and K2CO3
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Titration of mixture of KOH and K2CO3
Weigh 0.7341 g of technical KOH, which also contains K2CO3, quantitatively transfer it to a 250 mL volumetric flask, dissolve and make up to the mark. Prepare two Erlenmeyer flasks, pipette 50 mL of the prepared solution into each, add the indicator phenolphthalein to the first, and methyl orange to the second and titrate each with HCl solution. With the indicator phenolphthalein, the titrant consumption is 26.1 mL, and with methyl orange 33.7 mL. The concentration of the Determine the HCl by weighing out 0.1250 g of the KIO3 primary standard, adding KI, Na2S2O3 and the appropriate indicator, and titrating with the sol. HCl to overflow, consumption is 35.7 mL. Calculate the mass percentage of potassium in the sample.
Please help me with this assignment, I always get too high a score (over 100%).
Thank you!
Please help me with this assignment, I always get too high a score (over 100%).
Thank you!
- ChenBeier
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Re: Titration of mixture of KOH and K2CO3
First get concentration of HCl
Develop the redoxreaction iodate iodine and the consumed H+.
IO3- + 5 I- + 6 HCl => 3 I2 + 3 H2O + 6 Cl-
With the two first titration you get K2CO3 pH 8.2 and the mixture with KOH pH 4.5 , the difference is KOH
From that calculate mass of K.
Develop the redoxreaction iodate iodine and the consumed H+.
IO3- + 5 I- + 6 HCl => 3 I2 + 3 H2O + 6 Cl-
With the two first titration you get K2CO3 pH 8.2 and the mixture with KOH pH 4.5 , the difference is KOH
From that calculate mass of K.
Re: Titration of mixture of KOH and K2CO3
That's how I calculated it:
c(HCl) = (6*0.1250)/(0.0357*214)= 0.09817 M
n(K2CO3) = 0.0261*0.09817 = 0.002562 mol
V(HCl for KOH) = 33.7 - 26.1 = 7.6 mL
n(KOH) = 0.0076 * 0.09817 = 0.000746 mol
m(K2CO3) = 0.002562 mol *138.21 g/mol * 5 = 1.7705 g
1.7705 g is much more than 0.7341 g. So where did I go wrong?
c(HCl) = (6*0.1250)/(0.0357*214)= 0.09817 M
n(K2CO3) = 0.0261*0.09817 = 0.002562 mol
V(HCl for KOH) = 33.7 - 26.1 = 7.6 mL
n(KOH) = 0.0076 * 0.09817 = 0.000746 mol
m(K2CO3) = 0.002562 mol *138.21 g/mol * 5 = 1.7705 g
1.7705 g is much more than 0.7341 g. So where did I go wrong?
- ChenBeier
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Re: Titration of mixture of KOH and K2CO3
It was asked for potassium not for potassium carbonate or hydroxide.
The moles for K2CO3 are wrong
K2CO3 + 2 HCl => 2 KCl + H2O + CO2
For 1 K2CO3 2 HCl needed or 1 HCl correspond to 1/2 K2CO3
n (K2CO3) =0,0261 *0,09817/2 = 0,001281 mol
The moles for K2CO3 are wrong
K2CO3 + 2 HCl => 2 KCl + H2O + CO2
For 1 K2CO3 2 HCl needed or 1 HCl correspond to 1/2 K2CO3
n (K2CO3) =0,0261 *0,09817/2 = 0,001281 mol
Re: Titration of mixture of KOH and K2CO3
But still, if we consider the aliquot, the mass is too high.
- ChenBeier
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Re: Titration of mixture of KOH and K2CO3
Yes something doesn't fit.
The titration values are wrong or the method to get molarity of HCl.
The titration values are wrong or the method to get molarity of HCl.
- ChenBeier
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Re: Titration of mixture of KOH and K2CO3
Values for KOH and K2CO3 have to be exchangend.
See here an example with sodiumhydroxide and Carbonate
https://www.linkedin.com/pulse/analytic ... oda-kartal
See here an example with sodiumhydroxide and Carbonate
https://www.linkedin.com/pulse/analytic ... oda-kartal
- ChenBeier
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Re: Titration of mixture of KOH and K2CO3
The solution
p Value is sum of KOH and 1/2 K2CO3 or KHCO3 ( ml HCl consumed Phenolphthalein)
m value is sum of KOH and K2CO3 ( ml HCl consumed Methylorange)
n (KOH) = 2 p - m = 2*(KOH + 1/2 K2CO3) - (KOH + K2CO3) = KOH
n ( K2CO3) = 2*(m-p) = 2*(KOH + K2CO3 - (KOH + 1/2 K2CO3)) = K2CO3
n KOH = 2 * 26,1 - 33,7 = 18,5 ml this times molarity HCl
0,0185 l * 0,09817 mol/l = 0,001816 mol KOH
times molar mass 56 g/mol = 0,101 g
Times 5 = 0,508 g in 250 ml
n(K2CO3) = 2 * (33,7 - 26,1) = 15,2 ml times molarity HCl
0,0152 l * 0,09817 mol/l = 0,00149 mol K2CO3
times molar mass 138 g/mol = 0,205 g
times 5 is 1,025 g in 250 ml
What means in sum its 1,533 g. Its almost double what was dissolved
(0.7341 g).
So one number is not correct.
p Value is sum of KOH and 1/2 K2CO3 or KHCO3 ( ml HCl consumed Phenolphthalein)
m value is sum of KOH and K2CO3 ( ml HCl consumed Methylorange)
n (KOH) = 2 p - m = 2*(KOH + 1/2 K2CO3) - (KOH + K2CO3) = KOH
n ( K2CO3) = 2*(m-p) = 2*(KOH + K2CO3 - (KOH + 1/2 K2CO3)) = K2CO3
n KOH = 2 * 26,1 - 33,7 = 18,5 ml this times molarity HCl
0,0185 l * 0,09817 mol/l = 0,001816 mol KOH
times molar mass 56 g/mol = 0,101 g
Times 5 = 0,508 g in 250 ml
n(K2CO3) = 2 * (33,7 - 26,1) = 15,2 ml times molarity HCl
0,0152 l * 0,09817 mol/l = 0,00149 mol K2CO3
times molar mass 138 g/mol = 0,205 g
times 5 is 1,025 g in 250 ml
What means in sum its 1,533 g. Its almost double what was dissolved
(0.7341 g).
So one number is not correct.
Re: Titration of mixture of K….
Can we use alternative titration methods to determine the concentration of potassium in a mixture, such as using an unconventional indicator or a non-aqueous solvent, to enhance accuracy and precision?
- ChenBeier
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Re: Titration of mixture of KOH and K2CO3
In this case convert all to KCl. pH 7.
Probably the determination of molarity of HCl is also wrong. If it was only the half instead 0.0098 M only 0.0049 M then it would more or less fit.
Probably the determination of molarity of HCl is also wrong. If it was only the half instead 0.0098 M only 0.0049 M then it would more or less fit.
Re: Titration of mixture of K...
Can we use titration to determine the concentration of a mixture of potassium (K) isotopes, and if so, what challenges would we face during the process? How would the presence of multiple isotopes affect the accuracy of our results in this unique scenario?
- ChenBeier
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Re: Titration of mixture of KOH and K2CO3
You cannot determine different Isotops of one element by titration.
If you check for potassium then you will get all Isotopes together. The Isotopes are different by the amount of neutrons in the atom, this is not possible to figure out by chemical analysis. The chemical behaviour is driven by the amount of the electrons in the shell. Potassium has only one in the outer shell, what can be spend to other atoms.
If you check for potassium then you will get all Isotopes together. The Isotopes are different by the amount of neutrons in the atom, this is not possible to figure out by chemical analysis. The chemical behaviour is driven by the amount of the electrons in the shell. Potassium has only one in the outer shell, what can be spend to other atoms.