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#### biomed77

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##### volumetric analysis calculations
« on: January 25, 2006, 11:07:38 PM »

if 10cm^3 of 0.15mol dm^-3 phosphoric acid was titrated against 0.1mol dm^-3 sodium hydroxide, what volume of sodium hydroxide would be required for complete neutralisation?

the ratio is 3:1  so 3*0.15*10=4.5mol
4.5=0.1
45mL

is it right?/ thanks
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#### Borek

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##### Re:volumetric analysis calculations
« Reply #1 on: January 26, 2006, 12:05:10 AM »

3*0.15*10=4.5mol

mmol

Quote
4.5=0.1

Huh?

Quote
45mL

Surprisingly - looks correct.
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#### biomed77

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##### Re:volumetric analysis calculations
« Reply #2 on: January 26, 2006, 12:25:04 AM »

why r u surprised borek?

what i meant was....  4.5=0.1   so 4.5 divided with 0.1= 45mL         yes???

what about the other problem was it right???/ i appreciate your help, thank you ever so much!
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#### Borek

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##### Re:volumetric analysis calculations
« Reply #3 on: January 26, 2006, 01:36:07 AM »

Theres no way for 4.5 to equal 0.1, just like 2 never equals 3.
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#### darien2009

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##### Re: volumetric analysis calculations
« Reply #4 on: June 22, 2009, 06:39:14 AM »

Hi,
solution:
H3PO4+ 3NaOH ----> Na3PO4 +3H2O

step 1: calculate number of moles of H3PO4
M=number of mole/volume
0.15= number of mole/10
number of mole(H3PO4) =1.5 moles

from equation above:
1mole of H3PO4 react with 3mole of NaOH
1.5moles of H3PO4 react with x moles of NaOH

x= 1.5*3/1
x (number of moles NaOH) =4.5mole

step2: calculate volume of NaOH

M=number of mole/volume
10=4.5/volume
volume of NaOH =1.5/10
volume of NaOH = 0.45cm3
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#### darien2009

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##### Re: volumetric analysis calculations
« Reply #5 on: June 22, 2009, 06:51:06 AM »

I am sorry, it exist wrong in calculate  step 2
step2:

M=number of mole/volume
0.1= 4.5/volume
volume of NaOH = 4.5/0.1
= 45cm3
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