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. Check with pH paper. Store in a flask and adjust pH as needed.

Answer 1

V = %00 mL

pH = 10 buffer

Total Molarity = 0.2 M

then

[NH3] + [NH4Cl] = 0.2

if V= 500 mL then :

mmol balance:

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

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

apply henderson hasselbalch equation

pOH = pKb + log([NH4+]/[NH3])

do a mmol balance:

pOH = 4.75 + log(mmol of NH4+ / mmol of NH3)

if pH = 10, then pOH = !4-10 = 4

4 =  4.75 + log(mmol of NH4+ / mmol of NH3)

mmol of NH4+ / mmol of NH3 = 10^(4-4.75) = 0.1778

mmol of NH4+ = mmol of NH3 * 0.1778

mmol of NH3 + mmol of NH4Cl = 100

solve for either

let mmol of NH4+ = x

mmol of NH3 = y

x = y * 0.1778

x+ y = 100

substitute

y * 0.1778 +  y = 100

1.1778y = 100

y = 100/1.1778 = 84.9040 mmol of NH3

for NH4+ = 0.1778*84.9040 = 15.09593 mmol of NH4+

so:

mas of NH3 required --> mol*MW = (84.9040 *10^-3) (17) = 1.443368 g of NH3

from the 28% mix...

C = mass / V

V = mass/C = 1.443368/0.28 = 5.1548 mL of NH3 mixutre

for NH4Cl

mass = mol*MW = (15.09593 *10^-3)(53.491 ) = 0.807496 g of NH4Cl

add all in 1 container

add water until Volume marks 500 mL