Question

How many grams of solid ammonium bromide should be added to 0.500 L of a 0.164 M ammonia solution to prepare a buffer with a pH of 8.640 ?  

grams ammonium bromide =  g.

Answer 1

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.7447

POH = 14 - pH

= 14 - 8.64

= 5.36

use formula for buffer

pOH = pKb + log ([NH4Br]/[NH3])

5.359999999999999 = 4.7447 + log ([NH4Br]/[NH3])

log ([NH4Br]/[NH3]) = 0.6153

[NH4Br]/[NH3] = 4.1236

[NH4Br]/0.164 = 4.1236

[NH4Br] = 0.6763

volume , V = 0.5 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.6763*0.5

= 0.3381 mol

Molar mass of NH4Br = 1*MM(N) + 4*MM(H) + 1*MM(Br)

= 1*14.01 + 4*1.008 + 1*79.9

= 97.942 g/mol

we have below equation to be used:

mass of NH4Br,

m = number of mol * molar mass

= 0.3381 mol * 97.942 g/mol

= 33.1 g

Answer: 33.1 g