What volume of 0.130 M NaOH is needed to react with 94.0 mL of 0.240 M phosphoric acid?
2007-04-24 10:43:41 · 3 answers · asked by wonderer 1 in Science & Mathematics ➔ Chemistry
Lancenigo di Villorba (TV), Italy
MY ANSWER IS : 0.52 liter.
The reaction interested results an ACID-BASE's One
H3PO4(aq) + 3 NaOH(aq) ---> Na3PO4(aq) + 3 H2O(aq)
ABOUT CHEMICAL STUFFs
As you know, the ACID is H3PO4 or ORTHOPHOSPHORIC ACID.
H3PO4 is a TRI-VALENT ACID, e.g. it may lose all its own three Hydrogen Atoms as Hydrogen Ions ; nonetheless it is also a WEAK ACID, e.g. it overcomes to Equilibrium Conditions during its Ionic Dissociation.
Moreover, the BASE is NaOH or SODIUM HYDROXIDE.
NaOH is a MONO-VALENT BASE, e.g. it may lose its own Hydroxide Ion ; nonetheless it is a STRONG BASE, e.g. it may react completely.
CALCULATIONs
The Stoichiometry of the reaction is
nH3PO4 : nNaOH = (-1) : (-3)
so I have to compare the Molar Availability of the Reactants
(H3PO4's Moles) = 94.0 * 0.240 = 22.56E-3 mol
(NaOH's Moles) = (H3PO4's Moles) * 3 = 67.68E-3 mol
(NaOH's Volume) = 67.68E-3 / 0.130 = 520.6E-3 liter
I hope this could be clear.
2007-04-24 11:05:53 · answer #1 · answered by Zor Prime 7 · 0⤊ 0⤋
94/1000 x 0.24 x 3 x 1000/0.13
2007-04-24 11:04:50 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋
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2016-12-04 19:34:25 · answer #3 · answered by ? 4 · 0⤊ 0⤋