Question

how many grams of acetic acid (mw 60.05 g/mol) and NaAcetate (MW 82.03 g/mol) are needed to prepare 1.0l of a 250 mM buffer of a pH 5.0 solution (pka 4.75)?

Answer 1

pH of buffer = pKa + log[salt]/[acid]

number of moles of buffer = 250 mM * 1.0L

                          = 250 * 10^-3 mol/L * 1.0 L

                          = 0.25 moles

mass of acetic acid = 0.25 * 60.06 = 15.0125 g

mass of sodium acetate = 0.25 * 82.03 = 20.5075 g