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 solve and explain

NaH₂PO₄ =  1.876 g

Na₂HPO₄ =  4.879 g

Combinations: H₃PO₄ and Na₂HPO₄; H₃PO₄ and Na₃HPO₄

Explanation:

To have a buffer at 7.540, the acid must be in it second ionization, because the buffer capacity is pKa ± 1. So, we must use pKa2 = 7.198

The relation bewteen the acid and its conjugated base (ion), is given by the Henderson–Hasselbalch equation:

pH = pKa + log[A⁻]/[HA], where [A⁻] is the concentration of the conjugated base, and [HA] the concentration of the acid. Then:

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 and it's conjugated base 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 mix reaction is

NaH₂PO₄ + Na₂HPO₄ → HPO₄⁻² + 3Na + H₂PO₄⁻

The second ionization is:

H₂PO₄⁻ ⇄ HPO₄⁻² + H⁺

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.

pure substance: iron,copper

impure substance: water,charcoal