Question

how would you prepare a liter of 0.52M sodium phosphate buffer ar pH 7.8?

Answer 1

Given that

[A^-] + [HA] = 0.52 M.....(1)

Also, from literature, pKa = 7.21

From Henderson Hasselbach equation

pH = pKa + log[A^-]/[HA]

7.8 = 7.21+log[A^-]/[HA]

[A^-]/[HA] = 3.89

[A^-] = 3.89 [HA]

From (1)

3.89[HA] + [HA] = 0.52 M

[HA] = 0.106 M

So, [A^-] = 3.89*0.106 = 0.414 M

Moles of [A^-] = Molarity *volume = 0.414 M *1 L = 0.414 moles = moles of KH₂PO₄

Moles of [HA] = 0.106 M * 1.0 L = 0.106 moles = Moles of K₂HPO₄

Molar mass of KH₂PO₄ = 136 g/mol

Molar mass of K₂HPO₄ = 174 g/mol

Mass of KH₂PO₄ = 0.414 mol * 136 g/mol = 56.3 g

Mass of K₂HPO₄ = 0.106 mol * 174 g/mol = 18.5 g

Thus to prepare a buffer solution, add 56.3 g of KH₂PO₄ and 18.5 g of K₂HPO₄ in enough water to make a litre solution.