Question

How many mL of 0.655 M phosphoric acid solution are required to neutralize 15 mL of 1.284 M `NaOH` solution?

Answer 1

Well, you need to know that phosphoric acid is a DIACID in aqueous solution... The third proton is rarely removed.

Explanation:

And we write the stoichiometric equation as...

`H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) + 2H_2O(l)`

And so .................

`n_(NaOH)=15*mLxx10^-3*L*mL^-1xx1.284*mol*L^-1=0.01926*mol`...and so we need HALF of this molar quantity, i.e. `9.63xx10^-3*mol`, with respect to `H_3PO_4`.

And so we take the quotient....`(9.63xx10^-3*mol)/(0.655*mol*L^-1)=14.7xx10^-3*L=14.7*mL` to get the volume required for equivalence.

Answer 2

`V=9.80 mL`

Explanation:

Moles of `NaOH` `= 1.284/1000 xx 15`

Moles of `NaOH` `= 0.01926`

Reaction will be -
`H_3PO_4 + 3NaOH rarr Na_3PO_4 +3H_2O`

So, `0.01926 /3` moles are required to neutralize `H_3PO_4`

Moles required to neutralize `H_3PO_4 = 0.00642`

Let the volume required to neutralize `H_3PO_4` be `V` mL.

Moles of `H_3PO_4` `= 0.655/1000xxV`

`0.00642 = 0.655/1000xxV`

`V= 0.00642xx 1000/0.655`

`V= 6.42/0.655`

`V=9.80 mL`