Phosphoric acid is a triprotic acid with the following pKa values:

pKa1=2.148, pKa2=7.198, pKa3=12.375

You 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?

a. Mass NaH2PO4
b. Mass Na2HPO4

Mass NaH₂PO₄ = 1.920 g

Mass Na₂HPO₄ = 4.827 g

Explanation:

For a buffer solution we know its pH can be calculated from the Henderson-Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] and [HA] are the concentrations of the weak acid and its conjugate base in the buffer.

We want to prepare a buffer at pH 7.540 so we have chosen salts NaH₂PO₄ and Na₂HPO₄ as the weak acid and conjugate base respectively.

To calculate the mass of these salts we need to compute their ratio in the Henderson- Hasselbach equation .

Now since we are asked to determine the masses of NaH₂PO₄  and  Na₂HPO₄ and we know we want to prepare 1.000 L of a 0.05 M phosphate buffer, we can setup a system of 2 equations with two unknowns from the ratio mentioned above:

pH = pKa + log [A⁻]/[HA]

7.540 = 7.198 + log[HPO₄²⁻] / [H₂PO₄ ⁻]

0.342 = log[HPO₄²⁻] / [H₂PO₄ ⁻]

taking inverse log function to both sides of this equation:

2.198 = [HPO₄²⁻] / [H₂PO₄ ⁻]

but this is also equivalent to

2.198 = mol HPO₄²⁻ / mol H₂PO₄⁻   (M = mol/V)

We also know that in 1 liter of 0.05 M phosphate, we have 0.05 total mol HPO₄²⁻  and H₂PO₄⁻  , thus

mol HPO₄²⁻ + mol H₂PO₄⁻  = 0.05 mol

2.198 = mol HPO₄²⁻ / mol H₂PO₄⁻  

solving this system of equations calling  x = mol HPO₄²⁻ and y = mol H₂PO₄⁻ , we have:

2.198 = x /y    ⇒ x = 2.198y

x + y = 0.05

2.198y + y = 0.05

3.198 y = 0.05 ⇒ y = 0.05 / 3.198 = 0.016

x = 0.05 - 0.016 = 0.034

and the masses can be calculated from the molar masses ( 141.96 g/mol Na₂HPO₄ and 119.98 g/mol NaH₂PO₄

mol HPO₄²⁻ = 0.034 mol x 141.96 g/mol = 4.827 g

mol H₂PO₄⁻ =  0.016 mol x 119.98 g/mol = 1.920 g

the answer to your question is c