Question

A solution is prepared by dissolving 20.0 g of K₂HPO₄ and 25.0 g of KH₂PO₄ in enough water to produce 1.0 L of solution. What is the pH of this buffer? For phosphoric acid (H₃PO₄), K_a2 = 6.2

Answer 1

Applying Handerson's equation, we get

pH = pK_a + log{[K₂HPO₄]/[KH₂PO₄]}

Now, pK_a = -logK_a = -log(6.2*10^-8) = 7.21

Molar mass of K₂HPO₄ = 174 g/mole

Thus, moles of K₂HPO₄ = mass/molar mass = 20/174 = 0.115

Molar mass of KH₂PO₄ = 136 g/mole

Thus, moles of KH₂PO₄ = mass/molar mass = 25/136 = 0.184

Let there be 1 litre of the solution

Therefore, [KH₂PO₄] = Moles/volume in litres = 0.184 M

[[K₂HPO₄] = Moles/volume in litres = 0.115 M

thus, pH = 7.21 + log(0.115/0.184) = 7.01