This question is describing two chemical equations whereby the concentration of ammonia has to be determined. The first reaction is between 25.00 mL of ammonia and 50.00 mL of 0.100-M HCl whose excess was neutralized with 21.50 mL of 0.050-M Na₂CO₃ and thus, the concentration ammonia in the cloudy solution was determined as 0.114 M.
First of all we need to go over the titration of the excess HCl with Na₂CO₃ by writing the chemical equation it takes place when they react:
[tex]2HCl+Na_2CO_3\rightarrow 2NaCl+CO_2+H_2O[/tex]
Whereas the mole ratio of HCl to Na₂CO₃ is 2:1 and the volume of the HCl leftover is determined as follows:
[tex]V_{HCl}^{leftover}=\frac{2*0.050M*21.50mL}{0.100M} =21.5mL[/tex]
Next, we infer that the consumed volume of HCl by the ammonia solution was:
[tex]V_{HCl}^{consumed}=50.00mL-21.50mL=28.5 mL[/tex]
Then, we write the chemical equation that takes place between ammonia and HCl:
[tex]HCl+NH_3\rightarrow NH_4Cl[/tex]
Whereas the mole ratio is now 1:1, which means that the concentration of ammonia was:
[tex]M_{NH_3}=\frac{28.5mL*0.100M}{25.00mL}\\\\ M_{NH_3}=0.114M[/tex]
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