# A 120.0 mL buffer solution is

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

the answer is not 0.084g!

Answer:

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

moles of NH₃ = ([NH₃])(volume of buffer solution in L)= (0.255)(120×10⁻³)= 0.0306moles of NH₄⁺ = ([NH₄⁺])(volume of buffer solution in L)= (0.4)(120×10⁻³)= 0.048

From the stoichiometry of the reaction 1 mol of NH3 reacts with1 mol of HCl and produces 1 mol of NH4Cl

moles of NH₃ remaining after the reaction= 0.0306-xmoles 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 x2.8x= 0.00708x = 0.00253 = moles of HCl addedMolecular weight of HCl =  36.5 g/mol

Mass of HCl = moles x molecular weight

= 0.00253 moles x 36.5 g/mol = 0.092345 gif more HCl added to solution than 0.092345 g then its pH wouldfall below 9.00.

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