Phosphoric acid is a triprotic acid with the following pKa values:pKa1=2.148, pKa2=7.198, pKa3=12.375You wish to prepare 1.000 L of a 0.0500 M phosphate buffer at pH 7.540. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?Mass NaH2PO4Mass Na2HPO4What other combination of phosphoric acid and/or its salts could be mixed to prepare this buffer? (Check all that apply).H3PO4 and NaH2PO4H3PO4 and Na2HPO4H3PO4 and Na3HPO4NaH2PO4 and NA3PO4Na2HPO4 and NA3PO4PLEASE SOLVE AND EXPLAIN

So, H₂PO₄⁻ is the acid form, and its concentration is the same as NaH₂PO₄, and HPO₄⁻² is the conjugated base, and its concentration is the same as Na₂HPO₄ (stoichiometry is 1:1 for both).

So, the number of moles of these salts are:

NaH₂PO₄ = 0.01563 M * 1.000 L = 0.01563 mol

Na₂HPO₄ = 0.03436 M* 1.000 L = 0.03436 mol

The molar masses are, Na: 23 g/mol, H: 1 g/mol, P: 31 g/mol, and O = 16 g/mol, so:

NaH₂PO₄ = 23 + 2*1 + 31 + 4*16 = 120 g/mol

Na₂HPO₄ = 2*23 + 1 + 31 + 4*16 = 142 g/mol

The mass is the number of moles multiplied by the molar mass, so:

NaH₂PO₄ = 0.01563 mol * 120 g/mol = 1.876 g

Na₂HPO₄ = 0.03436 mol * 142 g/mol = 4.879 g

To prepare this buffer, it's necessary to have in solution the species H₂PO₄⁻ and HPO₄⁻², so it can be prepared for mixing the combination of:

H₃PO₄ and Na₂HPO₄ (the acid is triprotic so, it will form the H₂PO₄⁻ , and the salt Na₂HPO₄ will dissociate in Na⁺ and HPO₄²⁻);

H₃PO₄ and Na₃HPO₄ (same reason).

The other combinations will not form the species required.

jayilych4real jayilych4real · Answer 2 · 2022-03-31T14:50:50+02:00

The mass of each salt which will be added to the mixture is:

NaH₂PO₄ = 1.876 g
Na₂HPO₄ = 4.879 g

The other combinations of phosphoric acid and/or its salts which could be mixed to prepare this buffer are:

H₃PO₄ and Na₂HPO₄;
H₃PO₄ and Na₃HPO₄

Calculations and Parameters:

If we want to have a buffer at 7.540, then the acid has to be in the second ionization which means that for the buffer capacity, we would use pKa2 = 7.198

7.540 = 7.198 + log[A⁻]/[HA]
log[A⁻]/[HA] = 0.342
[A⁻]/[HA] =
[A⁻]/[HA] = 2.198
[A⁻] = 2.198*[HA]

The concentration of the acid is conjugated to base and must be equal to the concentration of the buffer 0.0500 M, so:

[A⁻] + [HA] = 0.0500
2.198*[HA] + [HA] = 0.0500
3.198*[HA] = 0.0500
[HA] = 0.01563 M

[A⁻] = 0.0500 - 0.01563

[A⁻] = 0.03436 M

The mixed reaction is: NaH₂PO₄ + Na₂HPO₄ → HPO₄⁻² + 3Na + H₂PO₄⁻

The second ionization is:

H₂PO₄⁻ ⇄ HPO₄⁻² + H⁺

Because H₂PO₄⁻ is the acid form, its concentration is the same as NaH₂PO₄, and HPO₄⁻² is the conjugated base, and its concentration is the same as Na₂HPO₄ (stoichiometry is 1:1 for both).

Therefore, the number of moles of these salts are:

NaH₂PO₄ = 0.01563 M * 1.000 L = 0.01563 mol
Na₂HPO₄ = 0.03436 M* 1.000 L = 0.03436 mol

The molar masses are, Na: 23 g/mol, H: 1 g/mol, P: 31 g/mol, and O = 16 g/mol, so:

NaH₂PO₄ = 23 + 2*1 + 31 + 4*16 = 120 g/mol
Na₂HPO₄ = 2*23 + 1 + 31 + 4*16 = 142 g/mol

NaH₂PO₄ = 0.01563 mol * 120 g/mol = 1.876 g
Na₂HPO₄ = 0.03436 mol * 142 g/mol = 4.879 g

It is important to have the solution of the species H₂PO₄⁻ and HPO₄⁻² in order to prepare it for buffering, so it can be prepared for mixing the combination of:

H₃PO₄ and Na₂HPO₄ (the acid is tricrotic so, it will form the H₂PO₄⁻ ,

The salt Na₂HPO₄ will dissociate in Na⁺ and HPO₄²⁻);

The other combinations will not form the species required.

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