# Problem: Federal regulations set an upper limit of 50 parts per million (ppm) of NH3 in the air in a work environment [that is, 50 molecules of NH3(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 106 mL of 1.11×10−2 M HCl. The NH3 reacts with HCl as follows:NH3(aq) + HCl(aq) → NH4+(aq)After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 14.3 mL of 5.86×10−2 M NaOH to reach the equivalence point.How many ppm of NH3 were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol under the conditions of the experiment.)

###### FREE Expert Solution

The reaction titration:

NH3(aq) + HCl(aq) → NH4Cl(aq)

82% (490 ratings) ###### Problem Details

Federal regulations set an upper limit of 50 parts per million (ppm) of NH3 in the air in a work environment [that is, 50 molecules of NH3(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 106 mL of 1.11×10−2 M HCl. The NH3 reacts with HCl as follows:

NH3(aq) + HCl(aq) → NH4+(aq)

After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 14.3 mL of 5.86×10−2 M NaOH to reach the equivalence point.

How many ppm of NH3 were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol under the conditions of the experiment.)