A solution is prepared by dissolving 20.0 g of K 2 HPO 4 and 25.0 g of KH 2 PO 4

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), Ka2 = 6.2

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), Ka2 = 6.2

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), Ka2 = 6.2 X 10^-8.

Explanation / Answer

Applying Handerson's equation, we get

pH = pKa + log{[K2HPO4]/[KH2PO4]}

Now, pKa = -logKa = -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

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