Question

How would you make a 750.0 mL of a buffer at pH 9.00 if you had 0.48 M ammonia and a bottle of solid ammonium chloride? Kb   = 1.8 x 10 -5 (show all reactions and work, with units

Answer 1

we know that

POH = 14 - pH

so

pOH = 14 - 9

pOH = 5

now

we know that

for buffers

pOH = pKb + log [ salt/ base ]

so

pOH = -log Kb + log [ NH4Cl / NH3]

5 = -log 1.8 x 10-5 + log [ NH4Cl / NH3]

[ NH4Cl / NH3] = 1.8

so

moles of NH4Cl / moles of NH3 = 1.8

take 0.18 mole = 9.612 g of NH4Cl

then

moles of NH3 = 0.1

now

this should be taken from 0.48 M

now

molarity = moles x 1000 / volume(ml)

so

0.48 = 0.1 x 1000 / volume (ml)

volume (ml) = 208.333

so the steps are

1) take 99.612 g of NH4Cl

2) now add 208.33 ml of 0.48 M ammonia

3)

now add water and make the volume to 750 ml