If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course), what volumes of each would you mix before diluting to a final volume of 2.00 L to prepare 2.00 L of pH 7.40 buffer with a final total concentration of 50 mM of phosphorous contains species (e. g. so that [H3PO4] [H2PO4 - ] [HPO4 2- ] [PO4 3- ]

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.

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.