Question

How many moles of NH4Cl must be added to 1.0 L of 1.0 M solution of NH3 to prepare a buffer solution with a pH of 9.00 ? of 9.50 ? Kb = 1.8 x 10 -5

Answer 1

  pOH = pKb + log ([NH₄⁺] / [NH₃])

pH = 9.0

pH + pOH = 14

Rewritting the above one we get the pOH

pOH = 14-9.0 = 5

pKb = - log (Kb)

pKb = - log(1.8*10^-5) = 4.745

[NH₃] = 1.0 M

5 = 4.745 + log (([NH4+] /1)

0.255 = log (([NH₄⁺] /1)

10^0.255 = [NH₄⁺]

[NH₄⁺] = 1.8

1.8 moles of [NH₄⁺] ion per litre of solution will have the pH of 9.0

pOH = pKb + log ([NH₄⁺] / [NH₃])

pH = 9.50

pOH = 14-9.5 = 4.5

pKb = - log (Kb)

pKb = - log(1.8*10^-5) = 4.745

[NH₃] = 1.0 M

4.5 = 4.745 + log (([NH₄⁺] /1)

0.245 = log (([NH₄⁺] /1)

10^0.245 = [NH₄⁺]

[NH₄⁺]  = 1.75

1.75 moles of [NH₄⁺] ion per litre of solution will have the pH of 9.50