How many grams of nh4cl (mm = 53.49 g/mol) must be added to 0.250 l of 0.375 m nh3to produce a buffer solution with ph = 9.45? (kb of nh3 is 1.8 * 10−5)?

When we have PH = 9.45 so, we can get POH from:

PH = 14 - POH 

POH = 14 - 9.45

        = 4.55 
So by using H-H equation, we can get [NH4+]:

POH = Pkb + ㏒ [B-/HB-]

 when Kb = 1.8 x 10^-5

∴Pkb = -㏒(1.8 x 10^-5)

          = 4.7

by substitution:

4.55 = 4.7 + ㏒[0.375]/[NH4]

∴[NH4] = 0.53 M

now we need to get the moles of NH4 at 0.25 L

moles NH4 = molarity * volume 

                    = 0.53 M* 0.25L = 0.1325 moles 

now, we can get the mass of NH4 when:

Mass NH4 = moles NH4 * molar mass

                   = 0.1325 moles * 53.49 g/mol

                   = 7 g
   

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