Chemistry Forum

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!

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.

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?

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

But still, if we consider the aliquot, the mass is too high.

Yes something doesn't fit.
The titration values are wrong or the method to get molarity of HCl.

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

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.

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?

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.

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?

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.