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Titration question - help needed please
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.
Please help!
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.
Please help!
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
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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