Question

A student was asked to determine the concentration of ammonia, a volatile substance used in the clinical setting as a respiretory stimulant to prevent fainting. First the student pipetted 25.00 mL of the cloudy ammonia solution into a 250.0 mL conical flask. 50.00 mL of 0.100 mol L' HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution. The excess (unreacted) HCI was then titrated with 0.050 mol L- Na2CO3(aq). 21.50 mL of Na2CO3(aq) was required. Calculate the concentration of the ammonia in the cloudy ammonia solution.​

Answer 1

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:

$2HCl+Na_2CO_3\rightarrow 2NaCl+CO_2+H_2O$

Whereas the mole ratio of HCl to Na₂CO₃ is 2:1 and the volume of the HCl leftover is determined as follows:

$V_{HCl}^{leftover}=\frac{2*0.050M*21.50mL}{0.100M} =21.5mL$

Next, we infer that the consumed volume of HCl by the ammonia solution was:

$V_{HCl}^{consumed}=50.00mL-21.50mL=28.5 mL$

Then, we write the chemical equation that takes place between ammonia and HCl:

$HCl+NH_3\rightarrow NH_4Cl$

Whereas the mole ratio is now 1:1, which means that the concentration of ammonia was:

$M_{NH_3}=\frac{28.5mL*0.100M}{25.00mL}\\\\ M_{NH_3}=0.114M$

Learn more:

• (Titration) brainly.com/question/15687419
• (Titration) brainly.com/question/25328286