Kb = 1.8 x 10-5
pKb = -log(Kb) = -log(1.8*10-5) = 4.74
pOH = 14 - pH
= 14 - 8.680 = 5.32
pOH = pKb + Log([salt]/[base])
5.32 = 4.74 + log([NH4Br] / [ NH3]
log([NH4Br] / [ NH3] = 5.32-4.74
log([NH4Br] / [ NH3] =0.58
[NH4Br] / [ NH3] = ANTILOG(0.58)
[NH4Br] / [ NH3] = 3.802
given [NH3] = 0.253M
[NH4Br] / 0.253 = 3.802
[NH4Br] = 3.802 X 0.253 = 0.962M
Volume = 0.500L
n = molarity x volume
= 0.962 x 0.500 = 0.481mol
molar mass of NH4Br = 97.942gmol-1
mass of NH4Br = n x molar mass
= 0.481 x 97.942
= 47.11g
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