Question

The ammonia/ammonium buffer system is used to optimize polymerase chain reactions (PCR) used in DNA studies. The equilibrium for this buffer can be written as NH4+(aq)??H+(aq)+NH3(aq) Calculate the pH of a buffer that contains 0.045 M ammonium chloride and 0.095 M ammonia. The Ka of ammonium is 5.6×10?10. Express your answer using two decimal places.

Answer 1

Sol:-

given [ NH₄Cl ] = [ NH₄⁺] = 0.045 M

[ NH₃] = 0.095 M

K_a( dissociation constant of acid ) of ammonium ion (NH₄⁺) = 5.6 x 10^-10

because , pK_a = - log K_a

therefore , pK_a of ammonium ion (NH₄⁺) = - log 5.6 x 10^-10

pK_a = - ( -9.2518 )

pK_a = 9.2518

Now , Given Chemical equation of acidic buffer solution is :

              NH₄⁺(aq)     <----------------------->   NH₃(aq)            +           H⁺(aq)

              Acid                                     Conjugate base

by using Henderson - Hasselbalch equation for acidic buffer solution , we have

pH = pK_a + log [ Conjugate base ] / [ Acid ]

pH = pK_a + log [ NH₃ ] / [ NH₄⁺ ]

pH = 9.2518 + log 0.095 M / 0.045 M

pH = 9.2518 + log 2.111

pH = 9.2518 + 0.3245

pH = 9.576