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.