Ph pk + logo [base]/(acid] 7.50 -6.76 + logio [base]/(acid) 7.50-6.76 = log10 [base]/(acid] 10.50-6.76 = [base)/(acid] 5.50: 1 = [pipes-na"): [pipes] = mole pipes-na'/mole pipes this means there are 5.50 mole pipes-na' for each mole of pipes in the buffer at ph 7.50 ..pipes no = 5.5/6.5 of total moles; pipes = 1/6.5 of total moles). 2. calculate the amount (in g) of each compound (e. g. pipes and pipes-na"): mole required in 2 l of 10 mm = 0.02 mol (i. e. 10 mmol l x 2 l) mole pipes required = 0.02 mol (1/6.50) = 3.08 x 10 mol amount (mass) pipes = (3.08 x 10) mol x 302 g mol = 0.93 g mole pipes-nat = 0.02 mol (5.50/6.50) amount (mass) pipes-na = (1.69 x 10? )mol x 324 g mol = 5.48 g 3. weigh 0.93 g pipes and 5.48 g pipes-nat 4. add approximately 1.5 l of dh 0 to the combined compounds and mix until they are completely dissolved 5. make the buffer up to the final volume of 2 l with dh20 6. confirm the ph of 7.50 with a ph meter.