Question

How many grams of dry NH4Cl need to be added to 1.80 L of a 0.600 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.96? Kb for ammonia is 1.8×10−5

Answer 1

Answer –

We are given, [NH₃] = 0.600 M, volume = 1.80 L

Kb = 1.8*10^-5 , pH = 8.96

First we need to calculate pKb from the Kb

pKb = -log Kb

        = -log 1.8*10^-5

        = 4.74

Now we need to use the Henderson Hasselbalch equation –

pOH = pKb + log [conjugate acid] / [Base]

now we need to calculate pOH from the pH

we know,

pH + pOH = 14

pOH = 14 – pH

         = 14-4.74

          = 9.25

Now plugging the value in the equation-

8.96 = 9.25 + log [NH₄⁺] / (0.600)

8.96-9.25 = log log [NH₄⁺] / (0.600)

-0.295 = log [NH₄⁺] / (0.600)

Taking antilog from both side

0.5066 = [NH₄⁺] /0.600

[NH₄⁺] = 0.5066 *0.600

             = 0.304 M

[NH₄⁺] = [NH₄Cl] = 0.304 M

Now we need to calculate moles of NH₄Cl

Moles of NH₄Cl= molarity of NH₄Cl * volume (L)

                          = 0.304 M * 1.8 l

                          = 0.547 mole of NH₄Cl

Mass of NH₄Cl = 0.547 mole * 53.492 g/mol

                          = 29.3 g of NH₄Cl