Question

1) How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.82? Kb for ammonia is 1.8 x 10^-5.

2) A beaker with 150 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.80 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

1) we have formula pOH =pkb + log [NH4Cl]/[NH3]

where pkb = -log Kb = -log ( 1.8x10^-5) = 4.745 , pOH = 14-pH = 14-8.82 = 5.18

hence 5.18 = 4.745 + lg [NH4Cl] / ( 0.8)

[NH4Cl] = 2.17816

NH4Cl moles = M x V = 2.17816 x 2.4 = 5.2276

NH4Cl mass = moles x molar mass = 5.2276 x 53.491 = 279.63 g

2) pH = pka + log [acetate ]/[acetic acid]

5 = 4.74 + log [acetate ]/[acetic acid]

[acetate ] = 1.82 [ acetic acid] ...........(1)

given [acetate] + [acetic acid] = 0.1   ............(2)

by (1) ( 2) we get [acetic acid] =0.03546 , [acetate] = 0.064539

Moles of acetic acid = M x V = 0.03546 x 150/1000 = 0.005319

acetate moles = 0.064539 x 0.15 = 0.00968

HCl moles added= H+ moles = M x V = 0.49 x 6.8/1000 = 0.003332

H+ reacts with acetate and frms acetic acid

Hence acetic acid moles = 0.005319+0.003332 = 0.008651

acetate moles = 0.00968 - 0.003332 =0.006348

pH = 4.74 + log ( 0.0046348/0.008651)

= 4.47

pH change = 5-4.47 = 0.53