Question

What is the pH of a one liter solution that is 0.100 M in NH₃ and 0.100 M in NH₄ Cl after 1.2 g of NaOH has been added? K_b for NH₃ is 1.8 × 10^-5

Answer 1

we know that

moles = mass / molar mass

so

moles of NaOH = 1.2 / 40 = 0.03

now

moles = molarity x volume (L)

so

moles of NH3 = 0.1 x 1 = 0.1

moles of NH4Cl = 0.1 x 1 = 0.1

now

the reaction is

OH- + NH4+ ---> NH3 + H20

we can see that

moles of NH4+ reacted = moles of NaOH added = 0.03

moles of NH4+ remaining = 0.1 - 0.03 = 0.07

moles of NH3 formed = moles of NH4+ reacted = 0.03

new moles of NH3 = 0.1 + 0.03 = 0.13

now

according to hasselbach henderson equation

pOH = pKb + log [ salt / base]

also

pKb = -logKb

so

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

so

pOH = -log 1.8 x 10-5 + log [ 0.07 / 0.13]

pOH = 4.476

now

pH = 14 - poH

so

pH = 14 - 4.476

pH = 9.524

so

the pH of the solution is 9.524