Question

How to prepare 100mL of phosphatic buffer with PH 6.80 by using p.
1 mol L-1 Na2HPO4

Answer 1

Use Henderson Hasselbalch equation

pH = pKa + log([salt]/[acid])

pKa nearest to 6.8 is 7.2, therefore,

6.8 = 7.2 + log ([salt]/[acid])

-0.4 = log([salt]/[acid])

Antilog (-0.4) = [salt]/[acid]

0.398 = [salt]/[acid]

[Salt] = 0.398[acid]..........….........equation 1

Also, from your question,

[Acid] + [salt] = 0.02......................equation 2

[Acid] + 0.398[acid] = 0.02

1.398[acid] = 0.02

[Acid] = 0.02/1.398

[Acid] = 0.0143M

From equation 1,

[Salt] = 0.398 * 0.0143

[Salt] = 0.00569M

Then you calculate the mass concentrations of the acid and salt

Mass concentration of acid (KH2PO4) = molar concentration * molar mass

= 0.0143 * 136

= 1.945 g/dm3

Mass concentration of salt (K2HPO4) = 0.00569 * 174

= 0.990 g/dm3

The respective mass concentrations are to prepare 1000 ml of the buffer,

But to prepare 100 ml, divide the mass concentrations by 10, to give 0.1945 g/dm3 and 0.099 g/dm3 for acid and base respectively.

Finally, to prepare 100mls of 20mM potassium phosphate buffer pH 6.8,

Weight the respective grammes of acid and salt in a beaker, and measure 100 ml of distilled water to dissolve it. Measure the pH, and adjust if need be.

I believe this is explanatory enough .