Question

if you were asked to make up a pH=6.86, 0.1M phosphate buffer, how would you do this?

Using the henderson-hasselbach equation, plug in the pH and the Pka for sodium phosphate monobasic (6.86)

for this situtation, what is the log(A-/HA)=

Answer 1

Answer.

The Henderson-Hasselbalch equation is

Given that,

pH = 6.86

pKa = 6.86 (for monobasic sodium phosphate)

molarity = 0.1 M of phosphate

putting the values in given Henderson-Hasselbach equation, we get

which means that we need to form a solution of which consists of equal amount of acid and it's conjugate base.

Let volume of solution be 1L.

Using moles = concentration*volume, we have

moles = 0.1*1 = 0.1 moles

[A^-] + [HA] = 0.1 moles

using equation (1), we have

[HA] = 0.1/2 = 0.05 moles

Therefore,

[A^-] = 0.05 moles

Hence, we need 0.05 moles of dibasic and 0.05 moles of monobasic.

For given situation,