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# Titration question - help needed please watch

1. 0.5g of impure ammonium chloride is warmed with an excess of sodium hydroxide solution. The ammonia liberated is absorbed in 25cm3 of 0.2mol dm-3 sulphuric acid. The excess of sulphuric acid requires 5.64cm3 of 0.2 moldm-3 sodium hydroxide solution for titration. Calculate the percentage of ammonium chloride in the original sample.

Here are the calculations I have done:

NH4Cl + NaOH + NH3 + H2O + NaCl

2NH3 + H2SO4 + (NH4)2SO4

Moles of sulphuric acid = (25x0.2)/1000 = 0.005moles

H2SO4 + NaOH + H2O + NaHSO4

Number of moles of sodium hydroxide = (5.64x0.2)/1000 = 0.001128 moles

Amount of sulphuric acid reacted = 0.005-0.001128=0.003872

What should I do next? I do not understand what it means when it says ammonia liberated is absorbed into sulphuric acid.

2. So now you know the number of moles of sulphuric acid reacted, you know that twice this number of moles gives you the number of moles of NH3 produced. Number of moles of NH3 = 2*0.003872 = 0.007744

Since 1 mole NH4Cl -> 1 mole NH3

Therefore the number of moles of NH4Cl in the samle is 0.007744

This weighs 0.007744*(14+4*1+35.5) = 0.41304g

Therefore percentage by mass = (0.41304/5) *100 = 83% to 2sf

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Updated: February 4, 2006
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