A Basic Buffer NH3 + H2O <------->  NH4+1 + OH-1 Take the -log of both sides A generalized version of this basic Henderson-Hasselbach equation is A chemistry student needs 250 mL of a solution buffered at a pH of 9.00. How many grams of ammonium chloride have to be added to 250 mL of 0.2 mol/L NH3 to make such a buffer? (Volume is assumed not to change.) pH = 9.0      pOH = 5.0 [base] = 0.2         pKb of ammonia = 4.74 (look up from datatable) pOH = pKb +log [cation]                              [base] 5.00 = 4.74 + log [cation]                               (0.2) 0.26 = log [cation]                    (0.2) [cation] = 100.26 X 0.2             = 1.8 * 0.2             = 0.36 mol/L of the NH4+1. But we only need enough for 250 mL  so   0.36 mol  = __x___                                                                  1000 mL     250 mL                                                                  x = 0.09 moles of NH4+1 ions needed Since the NH4+1 comes from NH4Cl then we also need 0.09 moles of NH4Cl. g = n * mm    = 0.09 moles * 53.5 grams/mole    = 4.8 grams of the salt are required.