Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: biomed77 on January 26, 2006, 05:07:38 AM
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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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3*0.15*10=4.5mol
mmol
4.5=0.1
Huh?
45mL
Surprisingly - looks correct.
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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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Theres no way for 4.5 to equal 0.1, just like 2 never equals 3.
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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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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