Question

# A chemistry graduate student is given 250. mL of a 0.50 M dimethylamine ((CH3)2NH) solution. Dimethlyamine...

A chemistry graduate student is given 250. mL of a 0.50 M dimethylamine ((CH3)2NH) solution. Dimethlyamine is a weak base with Kb=5.4*10^-4. What mass of (CH3)2NH2Br should the student dissolve in the (CH3)2NH solution to turn it into a buffer with pH=11.31?
You may assume that the volume of the solution doesn’t change when the (CH3)2NHBr is dissolved in it. Be sure your answer has a unit symbol and round it to 2 significant digits

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

#### Earn Coins

Coins can be redeemed for fabulous gifts.