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### Topic: Titration of mixture of NaOH and KOH  (Read 275 times)

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#### kobir

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##### Titration of mixture of NaOH and KOH
« on: August 01, 2019, 01:37:09 AM »
Hello,
I need to verify the concentration of NaoH and KOH mixture (100g).
The composition of the formulation is 40%  of NaOH and 10% of KOH (both wt/wt).
Accroding to the theoretical calculation the total OH moles are:
n (NaOH) = ~1 mole
n (KOH) = 10/56= 0.1786 moles
n (Total) = 1.1786 moles = moles of OH = required moles of HCl.

I Performed a titration with 1N of HCL with pH meter .The volume of the HCl was 17.95ml.
That means that the 0.01795 moles of HCl were needs instead of 1.1786 moles.

i'm aware that carbonates affects the results but the deviation is huge.

I will be happy to have a help on this subject.
Kobi

#### AWK

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##### Re: Titration of mixture of NaOH and KOH
« Reply #1 on: August 01, 2019, 01:55:02 AM »
Did you take the whole mixture?
AWK

#### kobir

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##### Re: Titration of mixture of NaOH and KOH
« Reply #2 on: August 01, 2019, 03:00:30 AM »
No, i titrated 1ml of it (diluted by water).
That explains the difference.
so how should i calculate it?
Thanks!

#### AWK

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##### Re: Titration of mixture of NaOH and KOH
« Reply #3 on: August 01, 2019, 04:50:26 AM »
1 ml of 100 ml?

AWK

#### kobir

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##### Re: Titration of mixture of NaOH and KOH
« Reply #4 on: August 01, 2019, 07:48:25 AM »
Yes,
i've already  manage to calculate it correctly. thank you.