Question

A 120.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.135 molL−1 in NH4Br. If the same volume of the buffer were 0.255 molL−1 in NH3 and 0.400 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

the answer is not 0.084g!

Answer 1

When the pH fell below 9.00
pOH = 14-9.00
= 5.00
henderson hasselbalch.

pOH= pKb+log([NH₄⁺]/[NH₃])

5.00-4.745 = log([NH₄⁺]/[NH₃])

[NH₄⁺]/[NH₃]= 10^(5.00-4.745) = 1.8

moles of NH₄⁺/moles of NH₃= 1.8

The chemical reaction is

NH₃(aq)+HCl(aq)→NH₄Cl(aq)

Let x be the number of moles of HCl added.

moles of NH₃ = ([NH₃])(volume of buffer solution in L)
= (0.255)(120x10⁻³)
= 0.0306

moles of NH₄⁺ = ([NH₄⁺])(volume of buffer solution in L)
= (0.4)(120x10⁻³)
= 0.048

From the stoichiometry of the reaction 1 mol of NH3 reacts with 1 mol of HCl and produces 1 mol of NH4Cl

moles of NH₃ remaining after the reaction
= 0.0306-x
moles of NH₄⁺ present after the reaction
= (initial moles of NH₄⁺)+( moles of HCl added)
= 0.048+x

(0.048+x)/(0.0306-x)= 1.8

0.048+x = 0.05508 - 1.8 x
2.8x= 0.00708
x = 0.00253 = moles of HCl added
Molecular weight of HCl =  36.5 g/mol

Mass of HCl = moles x molecular weight

= 0.00253 moles x 36.5 g/mol = 0.092345 g

if more HCl added to solution than 0.092345 g then its pH would fall below 9.00.