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### Topic: calculate the percentage purity of the potassium hydroxide  (Read 27431 times)

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#### biomed77

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##### calculate the percentage purity of the potassium hydroxide
« on: February 04, 2006, 11:17:23 AM »
16.0g of an impure sample of potassium hydroxide was dissolved in water and made up to 1.0dm3 in a volumetric flask. if 25cm of this solution required 50cm3 of 0.5mol dm-3 sulphuric acid for neutralisation, calculate the percentage purity of the potassium hydroxide (assuming the impurities are natural substances.

what i did is 50 x 0.5 / 25 = 1
and then 16.0 / 56 = 0.286
so purity = 0.286 / 1 x 100 = 28.6%

is that possible??  can someone correct me if wrong?

thanks )
« Last Edit: February 08, 2006, 05:56:17 PM by Mitch »

#### Borek

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##### Re:percentage purity
« Reply #1 on: February 04, 2006, 12:06:01 PM »
IMHO question doesn't make sense. Are you sure numbers you have copied are correct?

Note that sulfuric acid is diprotic.
« Last Edit: February 04, 2006, 12:07:52 PM by Borek »
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#### jdurg

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##### Re:percentage purity
« Reply #2 on: February 04, 2006, 10:47:40 PM »
I got the same purity as Borek so there is definitely an error there in the numbers you provided.
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#### biomed77

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##### Re:percentage purity
« Reply #3 on: February 05, 2006, 09:17:49 AM »
16.0g of an impure sample of potassium hydroxide was dissolved in water and made up to 1.0dm^3 in a volumetric flask. if 25cm^3 of this solution required 50cm^3 of 0.5mol dm^-3 sulphuric acid for neutralisation, calculate the percentage purity of the potassium hydroxide (assuming the impurities are natural substances.)

this is the original question that i have to solve...why the question does nt make sense and which numbers are incorrect???
this is the exact copy of the question maybe my solution is wrong, can you please help me???

#### Borek

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##### Re:percentage purity
« Reply #4 on: February 05, 2006, 10:08:49 AM »
maybe my solution is wrong

Your solution was wrong too, but it doesn't change the fact that if you do calculations properly you will end with purity of the KOH in the range of 700% - which doesn't make sense.

Start with the reaction equation (write it here!) and calculate number of moles of KOH in the 25 mL sample. Then calculate number of moles in the 1L, then convert it to mass of KOH. Show every step of your calculations.

Note that the question wording is incorrect at least for one more (besides wrong result) reason. Impurities are not natural substances, but neutral substances.
« Last Edit: February 05, 2006, 10:11:51 AM by Borek »
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#### biomed77

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##### Re:percentage purity
« Reply #5 on: February 05, 2006, 10:17:01 AM »
2KOH + H2SO4 ---->  K2 SO4 + 2H2O

25 / 56 =2.24mol

so in 1L 17.86

THEN WHAT?

#### Borek

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##### Re:percentage purity
« Reply #6 on: February 05, 2006, 10:46:52 AM »
2KOH + H2SO4 ---->  K2 SO4 + 2H2O

OK

Quote
25 / 56 =2.24mol

No idea what you did - are you dividing 25mL by 56g? Not to mention the fact that on my calculator 25/56=0.446 and not 2.24.
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#### biomed77

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##### Re:percentage purity
« Reply #7 on: February 05, 2006, 11:25:48 AM »
sorry i have been lazy..

i ll try again

formulae mass = 56  16.0g in 56 = 0.286 in 10cm^3 therefore 2.86 in 100cm^3 or 28.6 in 1000cm^3 so molarity = 28.6

25 x 28.6 / 1000= 0.715

#### Borek

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##### Re:percentage purity
« Reply #8 on: February 05, 2006, 01:13:59 PM »
formulae mass = 56

OK

Quote
16.0g in 56 = 0.286

Thats OK, although it has nothing to do with things I have asked you to calculate

Quote
in 10cm^3 therefore 2.86 in 100cm^3 or 28.6 in 1000cm^3 so molarity = 28.6

And here I am completely at loss.

Either start again - doing exactly what you were asked to do two posts earlier, or I will be not able to help.
« Last Edit: February 05, 2006, 01:14:17 PM by Borek »
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#### biomed77

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##### Re:percentage purity
« Reply #9 on: February 08, 2006, 08:10:52 AM »

Start with the reaction equation (write it here!) and calculate number of moles of KOH in the 25 mL sample. Then calculate number of moles in the 1L, then convert it to mass of KOH. Show every step of your calculations.

2KOH + H2SO4 ------> K2 SO4 + 2H2O

first it was my tutor's mistake that he wrote in the paper 0.5 mol , it is actually 0.05mol.

so 0.05 x 50 / 1000 = 2.5 x 10^-3 = 0.0025
0.0025 / 0.0025 = 1M
what do i do next?
« Last Edit: February 08, 2006, 09:16:07 AM by jdurg »

#### Borek

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##### Re:percentage purity
« Reply #10 on: February 08, 2006, 09:21:29 AM »
first it was my tutor's mistake that he wrote in the paper 0.5 mol , it is actually 0.05mol.

Happens.

Quote
2KOH + H2SO4 ------> K2SO4 + 2H2O

OK

Quote
so 0.05 x 50 / 1000 = 2.5 x 10^-3 = 0.0025

0.0025 mole of H2SO4 was used. OK

Quote
0.0025 / 0.0025 = 1M

I can only guess what you are doing here - I assume you are trying to calculate concentration of KOH. First, you forgot that sulphuric acid is diprotic, second, 0.0025L is not the same as 25mL.
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#### jdurg

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##### Re:percentage purity
« Reply #11 on: February 08, 2006, 09:25:27 AM »
Also remember that 1.0 dm^3 = 1000 cm^3.

The first thing you need to do is calculate how many moles of Sulfuric acid were used to neutralize the solution of your unknown concentration.  (Remember that the H2SO4 will be donating two moles of H+ for every moles of H2SO4).  The number of moles you come up with will be how many moles of OH- are in your 25 mL sample.

Once you've gotten the number of moles in your 25 mL sample, you will then need to figure out how many moles are in your 1000 mL sample.  Again, when that is figured out you'll have the total number of moles of OH- in your unknown powder which will be equal to the number of moles of KOH in your powder.  Now figure out the mass of KOH you have, divide it by the total mass of the sample and you have your percentage KOH.

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#### biomed77

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##### Re:percentage purity
« Reply #12 on: February 08, 2006, 01:37:17 PM »
so 2 x 0.025 = 0.05 moles
0.05 x 25 / 1000 =
0.05 x 0.025 = 1.25 x 10^-3 = 0.00125 mol

is it right so far, i am really struggling with it...

#### biomed77

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##### Re:percentage purity
« Reply #13 on: February 08, 2006, 02:13:52 PM »
i finally got it.....i think !

formulae mass: 56moles  so 16.0g in 56 = 0.286mol

moles = volume x concentration
0.286 x 25 / 1000 = 7.15 x 10^-3 = 0.00715  for potassium hydroxide

0.05 x 50 / 1000 = 2.5 x 10 ^-3  = 0.0025  for sulphuric acid

0.0025 x 2 = 5 x 10 ^-3 = 0.005

so, 0.005 /0.00715 x 100 = 69.9 = 70%

hope it is correct, if yes thanks for the guideness..

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