Question

How many miles of NaOH should be mixed with 1 mile of NaH2PO4 to prepare one liter of a 1M sodium phosphate buffer with a pH of 7.00? (Please explain, a chemical reaction would be nice, the answer should be 0.15 moles)

Answer 1

This is a buffer:

NaH2PO4 = Na+ + H2PO4- (weak acid)

H2PO4- + NaOH = Na + HPO4-(conjugate base) + H2O forms

then...

pH = pKa + log(HPO4-2 / H2PO4-)

pKa = 7.1 for this ionization

initially

H2PO4- = 1

HPO4-2 = 0

after additio nof x mmol of NaOH

H2PO4- = 1 - x

HPO4-2 = 0 + x

substitute in pH

pH = pKa + log(HPO4-2 / H2PO4-)

7.00 = 7.21 + log(x / (1-x))

10^(7-7.21) = x/(1-x)

0.61659 - 0.61659x = x

1.61659x = 0.61659

x = 0.61659 /1.61659 = 0.38 mol of NaOH required

Note that the addition can't be simply 0.15

Proof:

H2PO4- = 1 - x = 1-0.15 = 0.85

HPO4-2 = 0 + x = 0.15

pH = 7.21 + log(0.15 / 0.85 ) = 6.46