Answer:
0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH
Explanation:
The pKa's of phosphoric acid are:
H₃PO₄/H₂PO₄⁻ = 2.1
H₂PO₄⁻/HPO₄²⁻ = 7.2
HPO₄²⁻/PO₄³⁻ = 12.0
To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻
To have a 50mM solution of phosphoures we need:
2L * (0.050mol / L) = 0.10 moles of H₃PO₄
0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4
To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:
H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺
H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺
Using H-H equation we can find the amount of NaOH added:
pH = pKa + log [A⁻] / [HA] (1)
Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻
pH = 7.40
pKa = 7.20
[A-] + [HA] = 0.10moles (2)
Replacing (2) in (1):
7.40 = 7.20 + log 0.10mol - [HA] / [HA]
0.2 = log 0.10mol - [HA] / [HA]
1.5849 = 0.10mol - [HA] / [HA]
1.5849 [HA] = 0.10mol - [HA]
2.5849[HA] = 0.10mol
[HA] = 0.0387 moles = H₂PO₄⁻ moles
That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles
The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles
And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles
Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH
Then, you need to dilute both solutions to 2.00L with distilled water.
