Question & Answer: Explain how you would prepare 500 ml of a buffer with a pH of 7.50 using 6 M solutions of any…..

The pH of buffer is 7.50 hence we choose NaH2PO4/ Na2HPO4,because the pK_a (7.20 ) which is neatest within unit of the pH (7.50 ).

Here we use NaH2PO4/ Na2HPO4as follows:

Na2HPO4 This has a pKa of 7.2

Now,

Use Henderson-Hasslebach equation as below:

pH = pKa + log base/acid

7.50 = 7.20 + log [HPO4 2− ]/ [H2PO4− ]

7.50 = 7.20 + log [HPO4 2− ]/ [H2PO4− ]

log [HPO4 2− ]/ [H2PO4− ] = 0.3

[HPO4 2− ]/ [H2PO4− ] = 10^0.3

[HPO4 2− ] /[H2PO4− ] = 1.995 —1

Since volume (V= 500 mL) is same for both [HPO4 2− ] and [H2PO4− ], from eqn(1)

[(moles of HPO4 2−) / V] / [(moles of H2PO4− ) / V] = 1.995

OR

[(moles of HPO4 2−)] / [(moles of H2PO4− )] = 1.995

[(moles of HPO4 2−)] = [(moles of H2PO4− )] * 1.995   —- 2

We need to Prepare 500 mL of 6 M phosphate buffer.

V = 500 mL = 500 mLx(1L / 1000mL) = 0.500L

Hence moles of [HPO4 2− ] and [H2PO4− ] in the phosphate buffer = VxM = 0.500Lx 6mol/L

= 3.0 mol

Hence   [HPO4 2− ] + [H2PO4− ] = 3.0 mol ——(3)

=>  [(moles of H2PO4− )] * 1.995   + (moles of H2PO4− )= 3.0 mol [from eqn(2)]

== > [(moles of H2PO4− )] * 2.995   = 3.0 mol

=>[(moles of H2PO4− )]   = 1.00 mol

Hence moles of HPO4 2− = 3.00 – (moles of H2PO4−)

= 3.00 – 1.00 = 2.00 mol

Mass of (NaH2PO₄) =1.00 mol x (119.96g/mol) = 119.96 g

Mass of (Na2HPO₄) = 2.00 mol x (141.96g/mol) = 283.92 g

Hence we need to add 119.96 g of NaH2PO₄ and 283.92 g of Na2HPO4 and dilute them to 5000 mL to prepare a buffer of pH 7.50