pH of buffer = 11.31
Kb of dimethyl amine = 5.4 x 10-4
Ka of dimethyl ammonium = 10-14/(5.4 x 10-4) = 1.8 x 10-11
pKa = -log(1.8 x 10-11) = 10.73
Moles of dimethyl amine = 0.50 M * 0.250 L = 0.125 mol
According to Henderson Hasselbalch equation:
pH = pKa + log[moles of dimethyl ammonium bromide]/[moles of dimethylamine]
11.31 = 10.73 + log[moles of dimethyl ammonium bromide]/[0.125 mol]
100.58 = [moles of dimethyl ammonium bromide]/[0.125 mol]
3.80 = [moles of dimethyl ammonium bromide]/[0.125 mol]
moles of dimethyl ammonium bromide = 0.475 mol
mass of dimethyl ammonium bromide = 0.475 mol*molar mass
= 0.475 mol * 126 g/mol
= 59.85 g
Answer : Mass of dimethyl ammonium bromide (in 2 significant figures) = 60 g
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