Question

A buffer solution is composed of 1.426g KH2PO4 and 5.191g of Na2HPO (Ka for K2HPO4=6.2x10^-8)

I calculated the pH to be 7.75

Now, I am being asked to calcualte the mass of K2HPO4 needed to decrease the pH by .10

I keep messing it up, help please :)

Answer 1

The dissociation of hydrogen phosphate is given as

H₂PO₄^- (aq) <====> H⁺ (aq) + HPO₄^2- (aq); K_a = 6.2*10^-8

Mass of KH₂PO₄ taken = 1.426 g; mass of Na₂HPO₄ taken = 5.191 g.

Molar mass of KH₂PO₄ = (1*39 + 2*1 + 1*31 + 4*16) g/mol = 136 g mol^-1

Molar mass of Na₂HPO₄ = (2*23 + 1*1 + 1*31 + 4*16) g/mol = 142 g mol^-1

Moles of KH₂PO₄ taken = (1.426 g)/(136 g mol^-1) = 0.0105 mole.

Moles of Na₂HPO₄ taken = (5.191 g)/(142 g mol^-1) = 0.0365 mole.

Let us assume the volume to be V L; [H₂PO₄^-] = (0.0105/V) mol/L, [HPO₄^2-] = (0.0365/V) mol/L.

Given K_a = 6.2*10^-8, pK_a = -log (K_a) = -log (6.2*10^-8) = 7.208

pH = pK_a + log [HPO₄^2-]/[H₂PO₄^-] = 7.208 + log [(0.0365/V mol/L)/(0.0105/V mol/L)] = 7.208 + log (0.0365/0.0105) = 7.208 + log (3.476) = 7.208 + (0.541) = 7.749 ≈ 7.75 (ans)

Now we need to reduce the pH by 0.10; therefore, the desired pH is 7.65. We will have to calculate the mass of KH₂PO₄ keeping the mass of Na₂HPO₄ constant. Therefore moles of Na₂HPO₄ = 0.0365 will remain constant. Let x moles of KH₂PO₄ be required.

pH = pK_a + log (0.0365/x)

===> 7.65 = 7.208 + log (0.0365/x)

===> 0.442 = log (0.0365/x)

===> 0.0365/x = antilog (0.442) = 2.767

===> x = 0.0365/2.767 = 0.0132

Therefore, moles of KH₂PO₄ = 0.0132; mass of K₂HPO₄ taken = (0.0132 mole)*(136 g mol^-1) = 1.7952 g.

Mass of KH₂PO₄ to be added extra = (1.7952 – 1.426) g = 0.3692 g ≈ 0.369 g (ans)