Question

How to make a 75ml of 0.05M phosphate buffer, pH 7.0 that is also 0.15M in NaCl?

Answer 1

pH of phosphate buffer = pka2 + log(Na2HPO4/NaH2PO4)

pka2 of H3PO4 = 7.21

total no of mol of buffer = 75*0.05 = 3.75 mmol

Noof mol of Na2HPO4 = x =

No of mol of NaH2PO4 = 3.75-x

7 = 7.21 + log(x/(3.75-x))

x = 1.43 mmol

Noof mol of Na2HPO4 = x = 1.43 mmol

amount of Na2HPO4 = n*Mwt = 1.43*10^-3*141.96 = 0.203 g

No of mol of NaH2PO4 = 3.75-x

                     = 3.75-1.43 = 2.32 mmol

Amount of NaH2PO4 = 2.32*10^-3*119.977 = 0.278 g

so that , take 0.203 g of Na2HPO4 and 0.278 g of NaH2PO4 and add water upto 75 ml.