Question

Calculate the appropriate amount of 28 wt% aqueous NH3 and solid NH4Cl to be mixed together to yield 500 mL of a ~0.2 M pH 10 buffer.

Answer 1

V = 500 mL required

at pH = 10; total molarity = 0.2 M

then

apply buffer equation

pH = pKa + log(NH3/NH4+)

10 = 9.25 + log(NH3/NH4+)

NH3/NH4+) = 10^(10-9.25) = 5.6234

NH3/NH4+ = 5.6234

now...

if 28 %w/w

then

assume a basis of 1 liter solution, so 1000 mL --> 1000 g oslution, of which 28% is NH3

280 g of NH3

mol = mass/MW = 280/17 = 16.470 mol

[NH3] = mol/V = (16.470)/1L = 16.470 M

concentration of mix is 16.470M which is pretty high

mmol of NH3 + mmol of NH4Cl = 0.2*500 = 100 mmol

Equation 1:

mmol of NH3/ mmol of NH4+ = 5.6234

Equation 2:

mmol of NH3 + mmol of NH4Cl =  100 mmol

now.. solve for each

mmol of NH3 = 5.6234*mmol of NH4+

substittue in

5.6234*mmol of NH4+ + mmol of NH4Cl =  100 mmol

mmol of NH4+ = (100)/(5.6234+1) = 15.0979 mmol

mmol of NH4Cl = 15.0979

mass of NH4Cl = mmol*MW = 15.0979*53.4915 = 807.6 mg of NH4Cl required

calculate...

mmol of NH3 = 5.6234*mmol of NH4+

mmol of NH3 = 5.6234*15.0979 = 84.901

from

M = mmol/mL

mL = mmol/M = 84.901 / 16.470 = 5.1548 mL of NH3 mis required

then...

instructions:

1) weight 807.6 mg of NH4Cl salt

2) add 5.1548 mL of NH3 in beaker

3) add some water, up to V = 200 mL

4) mix

5) add weighed mass of NH4Cl

6) mix

7) add water up to V = 500 mL

8) stirr

finally

pH = 10 approx and V = 500 mL