Question

What is the pH of a buffer system prepared by dissolving 10.70 grams of NH4Cl and 25.00 mL of 12M NH3 in enough water to make 1.000 L of solution? Kb= 1.80 x 10^-5 for NH3. Calculate the change in pH after addition of 50.00 mL of 0.15 NaOH.

Answer 1

we know that

moles = mass / molar mass

so

moles of NH4Cl = 10.70 / 53.941

moles of Nh4Cl = 0.2

now

moles = molarity x volume (L)

so

moles of NH3 taken = 12 x 25 x 10-3 = 0.3

now

for buffers

pOH = pkb + log [salt / acid ]

also

pKb = -log Kb

so

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

pOH = -log 1.8 x 10-5 + log [ 0.2 / 0.3]

pOH = 4.57

pH = 14 - pOH

pH = 14 - 4.57

pH = 9.43

2)

now

moles of NaOH added = 0.15 x 50 x 10-3 = 0.0075

now

OH- + NH4+ --> NH3 + H20

so

moles of NH4+ reacted = moles of NaOH added = 0.0075

moles of NH3 formed = moles of NaOH added = 0.0075

so

finally

moles of NH4+ = 0.2 - 0.0075 = 0.1925

moles of NH3 = 0.3 + 0.0075 = 0.3075

now

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

so

pOH = -log 1.8 x 10-5 + log [ 0.1925 / 0.3075]

pOH = 4.54

so

pH = 14 - 4.54

pH = 9.46

so

the new pH is 9.46

pH change = 9.46 - 9.43

pH change = 0.03

so

the change in pH is 0.03