Question

How many grams of solid ammonium bromide should be added to 1.50 L of a 0.118 M ammonia solution to prepare a buffer with a pH of 9.930 ? grams ammonium bromide = ? g

How many grams of solid ammonium chloride should be added to 1.00 L of a 0.291 M ammonia solution to prepare a buffer with a pH of 8.340 ?
grams ammonium chloride =? g

Answer 1

1)

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

POH = 14 - pH

= 14 - 9.93

= 4.07

use formula for buffer

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

4.07 = 4.7447 + log ([NH4Br]/[NH3])

log ([NH4Br]/[NH3]) = -0.6747

[NH4Br]/[NH3] = 0.2115

[NH4Br]/0.118 = 0.2115

[NH4Br] = 0.025

volume , V = 1.5 L

use:

number of mol,

n = Molarity * Volume

= 2.495*10^-2*1.5

= 3.743*10^-2 mol

Molar mass of NH4Br,

MM = 1*MM(N) + 4*MM(H) + 1*MM(Br)

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

= 97.942 g/mol

use:

mass of NH4Br,

m = number of mol * molar mass

= 3.743*10^-2 mol * 97.94 g/mol

= 3.666 g

Answer: 3.67 g

2)

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

POH = 14 - pH

= 14 - 8.34

= 5.66

use formula for buffer

pOH = pKb + log ([NH4Cl]/[NH3])

5.66 = 4.7447 + log ([NH4Cl]/[NH3])

log ([NH4Cl]/[NH3]) = 0.9153

[NH4Cl]/[NH3] = 8.2276

[NH4Cl]/0.291 = 8.2276

[NH4Cl] = 2.3942

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 2.394*1

= 2.394 mol

Molar mass of NH4Cl,

MM = 1*MM(N) + 4*MM(H) + 1*MM(Cl)

= 1*14.01 + 4*1.008 + 1*35.45

= 53.492 g/mol

use:

mass of NH4Cl,

m = number of mol * molar mass

= 2.394 mol * 53.49 g/mol

= 128 g

Answer: 128 g