Back titration question:
20.00mL of cloudly ammonia was pipetted into a 250mL volumetric flask. 100mL of 0.5866M HCl in excess was added. The volume was made up to 250mL with distilled water. A burette was filled with 0.1194M NaOH (standardised) solution. The HCl and ammonia solution was titrated with the NaOH, with the mean titre being 22.75mL.
The manufacturer of the cloudy ammonia claims that the detergent contains 45.2gL-1 ammonia as ammonium hydroxide (NH4OH).
a) Calculate the concentration in gL-1 of NH4OH in the cloudly ammonia sample
b) Provide possible explainations for differences between the calculated results and the manufacturers claims
In back titration, the reason people find it hard is because the way they approach these questions is to try do everything in one go or an illogical process or they have poor layout for their answer. Being specific in terms of what moles you are calculating such as excess or initial or reacted here is really important
To do back titration you should do a backwards approach. Meaning start analysing the question from the bottom first and then move up towards the top
Following this approach first summarise the sentence: The HCl and ammonia solution was titrated with the NaOH, with the mean titre being 22.75mL.
NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l)
V(NaOH) = 22.75 mL
Then summarise the sentence before it:
C(NaOH) = 0.1194 M
n(NaOH) = CV = (0.1194)(22.75/1000) = 2.71635 x 10^-3 moles
This is the NaOH reacted with the excess HCl, so therefore:
n(HCl) excess in 20.00 mL aliquot = n(NaOH) = 2.71635 x 10^-3 moles
Then continue with that approach:
Because only 20.00 mL aliquots were taken, this means that we need to change it to 250 mL. To do this just think about the ratio:
n(HCl) excess is 1.00 mL = (2.71635 x 10^-3)/20.00 = 1.358175 x 10^-4 moles
so n(HCl) excess in 250.0 mL = (1.358175 x 10^-4)x250.0 = 0.033954375 moles
Other reaction occurring is:
NH4OH(aq) + HCl(aq) --> NH4Cl(aq) + H2O(l)
n(HCl) initial = CV = (0.5866)(0.100) = 0.05866 moles
n(HCl) reacted with NH4OH = n(HCl) initial - n(HCl) excess
n(HCl) reacted with NH4OH = 0.05866 - 0.033954375
n(HCl) reacted with NH4OH = 0.024705625 moles
n(NH4OH) = n(HCl) reacted with NH4OH
n(NH4OH) = 0.024705625 moles
m(NH4OH) = n x MM = (0.024705625)(14.01+5x1.008+16) = 0.8659321563 g
This is NH4OH in 20.00 mL of the cloudy ammonia.
Therefore, g/L conc = (0.8659321563) / (20.00 x 10^-3) = 43.2966... g/L = 43.3 g/L