Search found 1542 matches
- Wed Feb 14, 2024 3:08 pm
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 1254
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 a...
- Tue Feb 13, 2024 4:08 am
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 1254
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.
- Tue Feb 13, 2024 3:46 am
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 1254
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...
- Sun Feb 11, 2024 11:24 pm
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 1254
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
- Sun Feb 11, 2024 3:56 am
- Forum: Chemistry forum
- Topic: Chloroauric acid
- Replies: 6
- Views: 591
Re: Chloroauric acid
You need about 355 ml solvent for the 0.7 g.
But read how to make it
What is aqua Regina it is 3 part HCl and one part HNO3 .
https://en.m.wikipedia.org/wiki/Chloroauric_acid
You have to do boiling until all HNO3 is evaporated as NO2, later dissolve the product again in conc. HCl to get 355 ml
But read how to make it
What is aqua Regina it is 3 part HCl and one part HNO3 .
https://en.m.wikipedia.org/wiki/Chloroauric_acid
You have to do boiling until all HNO3 is evaporated as NO2, later dissolve the product again in conc. HCl to get 355 ml
- Sun Feb 11, 2024 3:40 am
- Forum: Chemistry forum
- Topic: Chloroauric acid
- Replies: 6
- Views: 591
Re: Chloroauric acid
I thing to figure out what is aqua Regina and the calculati8n you have to do by yourself.
Calculate the moles of Gold you have then covert this to the volume of the acid to get 0,01 M
Calculate the moles of Gold you have then covert this to the volume of the acid to get 0,01 M
- Sun Feb 11, 2024 2:52 am
- Forum: Chemistry forum
- Topic: Chloroauric acid
- Replies: 6
- Views: 591
Re: Chloroauric acid
Dissolve in aqua Regina.
- Sat Feb 10, 2024 1:21 pm
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 1254
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.
- Sat Feb 10, 2024 8:35 am
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 1254
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
- Fri Feb 09, 2024 1:52 pm
- Forum: Chemistry forum
- Topic: Has anyone tried chloroform?
- Replies: 1
- Views: 265
Re: Has anyone tried chloroform?
Sorry we dont discuss this kind of things.
- Fri Feb 09, 2024 7:07 am
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 1254
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.
- Mon Feb 05, 2024 11:56 am
- Forum: Chemistry forum
- Topic: Exercise on pH
- Replies: 1
- Views: 487
Re: Exercise on pH
Where did you get the ka value from? The value is wrong. Potassium nitrate is a salt made by nitric acid and potassium hydroxide. Both are strong chemicals, acid and base. Also pH should be near neutral. Ka of HNO3 = 20,89, pKa =-1,32 Normaly pH = 14 - 1/2 * (pkB - lg [c anion]) If the value is righ...
- Fri Feb 02, 2024 3:43 pm
- Forum: Chemistry forum
- Topic: Electrochemistry
- Replies: 3
- Views: 1015
Re: Electrochemistry
You are right the concentration is less as 2 M
But its 2 mol/l - 8 *10^(-9) mol/l = 1,999999992 mol/l ~ 2 mol/l
So its near the same.
But its 2 mol/l - 8 *10^(-9) mol/l = 1,999999992 mol/l ~ 2 mol/l
So its near the same.
- Fri Feb 02, 2024 12:39 pm
- Forum: Chemistry forum
- Topic: Electrochemistry
- Replies: 3
- Views: 1015
Re: Electrochemistry
Why do you think the concentration of SO4 2- is wrong. You have 2 M H2SO4 so the value is right.
E =E0 + 0,059V/ z * log c Pb 2+
c Pb 2+ = ksp(PbSO4) / c SO4 2-
cPb 2+ = 1,6 *10^-8 (mol/l)^2 / 2 mol/l = 8 *10^(-9) mol/l
E = -0,126 V+ 0,059/2 * log (8*10^(-9))
E = -0,126V -0,238 V
E = -0,36 V
E =E0 + 0,059V/ z * log c Pb 2+
c Pb 2+ = ksp(PbSO4) / c SO4 2-
cPb 2+ = 1,6 *10^-8 (mol/l)^2 / 2 mol/l = 8 *10^(-9) mol/l
E = -0,126 V+ 0,059/2 * log (8*10^(-9))
E = -0,126V -0,238 V
E = -0,36 V
- Tue Jan 30, 2024 11:25 pm
- Forum: Chemistry forum
- Topic: Stoichiometry problem
- Replies: 5
- Views: 1676
Re: Stoichiometry problem
Why not give yours first?