How to prepare 100 millimolar potassium phosphate buffer?
Answers
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.
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