Question

What mass of ammonium chloride should be added to 2.45 L of a 0.165 M NH3 in order to obtain a buffer with a pH of 9.65?

Answer 1

The value of ammonia's base dissociation constant, Kb,

Kb=1.8 x 10⁻⁵

Using the Henderson - Hasselbalch equation

Here the pOH of a buffer solution that contains ammonia, a weak base, and the ammonium ion, is conjugate acid

pOH = pKb + log([conjugated acid / [base])

pOH = pKb + log([NH₄⁺] / [NH₃])

Calculating the pOH of the target solution

pOH = 14 - pH

pOH = 14 - 9.65 = 4.35 ...............................(1)

Use the base dissocaition constant to find pkb

pKb=−log Kb

pkb = - log (1.8 x 10^-5)

pkb = 4.74 ...................................(2)

Putting values into the Henderson - Hasselbalch equation to get

pOH = pKb + log([NH₄⁺] / [NH₃])

4.35 = 4.74+ log ([NH₄⁺] / [NH₃])

log ([NH₄⁺] / [NH₃]) = 4.35−4.74

log ([NH₄⁺] / [NH₃])= - 0.39

([NH₄⁺] / [NH₃]) = 10^-0.39

([NH₄⁺] / [NH₃]) = 0.40738 ..................................(3)

This means that the ratio between the concentration of the conjugate acid , which is the ammonium chloride, and the concentration of the base must be equal to 0.40738

The concentration of the conjugate acid will thus be

[NH4⁺] = [NH3] x (0.40738)

[NH4⁺] = (0.165 M) (0.40738) = 0.06722 M………………..(4)

This means that you will need to provide the solution with

Conc. = no. of moles x volume in litres

n = C x V

no. of moles of ammonium ion [NH4⁺] = 0.067022 M x 2.45 L= 0.16468 moles [NH4⁺] ………(5)

As ammonium chloride will dissociates in aqueous solution to give

NH4Cl(aq]→ NH4⁺ (aq] + Cl−(aq] ………………….(6)

So, number of moles of ammonium chloride will be equal to the number of moles of ammonium ions.

nNH4Cl= n[NH4⁺]= 0.16468 moles ……………..(7)

molar mass of NH₄Cl = 53.49g/mol

mass of NH4Cl = 0.16468moles x 53.49 g/mole

mass of NH4Cl = 8.808 g NH4Cl……………………….(8)

Therefore 8.808 g of NH4Cl should be added to 2.45L of 0.165M NH3 to attain pH 9.65