volumetric analysis calculations

[biomed77]

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

[Borek]

3*0.15*10=4.5mol

mmol

4.5=0.1

Huh?

45mL

Surprisingly - looks correct.

[biomed77]

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!

[Borek]

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

[darien2009]

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

[darien2009]

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