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Junior Member
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Oct 31, 2009, 04:15 PM
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check this answer for me , I did it!
standardization of NaOH solution:
a)volume of 6.0M NaOH solution used: 20mL
b)approximate[NaOH] after dilution to 400mL: 0.5M
c)weight of empty flask: 86.525g
d)weight of flask plus KHP: 87.253g
e)weight of KHP: 0.728g
f)moles of KHP: 0.00356mol
g)moles of NaOH: same answer as moles of KHP
h)initial buret reading: 0mL
I)final buret reading: 28.49mL
j)volume of NaOH used up: 28.49mL
k)[NaOH]: 0.00356mole/0.02849L NaOH= 0.12496M
Is it correct? Value of K is correct??
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Uber Member
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Nov 1, 2009, 04:46 AM
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For b), if you used the 6.0 molar NaOH to get it, that would be 0.3 M
No. of moles of NaOH in 20 mL of 6.0 M NaOH = 0.12 mol
No. of moles of NaOH in the 400 mL diluted NaOH = 0.12 mol
No. of moles of NaOH in 1000 mL of diluted NaOH = (0.12/400) * 1000 = 0.3 mol
f) by KHP, you mean potassium hydrogen phosphide? I have never used this compound, but something is strange in that formula. K forms a +1 ion, H forms a +1 ion and P forms -3 ions, giving either KH2P or K2HP. If that's so, all your following work will carry the error.
g) If you took the Mr of KHP as (39.1+1+31) 71.1, then, you have (0.728/71.1) 0.0102 mol.
Btw, is that a practical that you performed or you were just given these values?
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Junior Member
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Nov 1, 2009, 12:00 PM
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weight of KHP is 0.728g
so I think Moles of KHP is = mass/204.23g/mol
so, its 0.728g/204.23g/mol=0.00356mol
so, do you think that my k is correct?
[NaOHJ]=moles/L of NaOH=0.00356mol/0.02849L=0.12496M
in K do I have to use f) or b) ? I think I should use moles for f
please help me
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Uber Member
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Nov 1, 2009, 10:27 PM
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I had to Google what KHP was...
Ok, yes, the answer is good.
However, you still haven't answered my above question,
Originally Posted by me
is that a practical that you performed or you were just given these values?
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Junior Member
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Nov 1, 2009, 10:28 PM
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Ohh.. sorry
Its my experimental values.
Do u think any value have problem?
pleaselet me know and Thank you so much
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Uber Member
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Nov 1, 2009, 10:39 PM
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Yes, for b) I don't understand how you got 0.5M.
You have 0.12 mol of NaOH in 20 mL of 6.0M NaOH
You diluted that 0.12 mol from 20 mL to 400 mL. That gives a concentration of 0.3 M, not 0.5 M
You still have 0.12 mol in 400 mL.
400 mL -> 0.12 mol
1 mL -> 0.12/400 mol
1000 mL -> (0.12/400) * 1000 = 0.3 mol
So, [NaOH] = 0.3 M
Then, I don't know if they do it like that where you are, but I was told that we always had to make averages when titrating. I mean, perform minimum 2 titrations and use the average of the two volumes used.
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Junior Member
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Nov 1, 2009, 10:41 PM
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okay thanks! For B
and yes I did test twice
but I just didn't wrote value of test 2
but l) I got 0.11421M
I added two tests[NaOH] and divided by 2.
Thanks
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Uber Member
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Nov 1, 2009, 10:47 PM
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Oh, OK, things get clearer, phew! Ok, on to the next part, sample 2 now.
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