Answer:
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
