Question

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 1.13×10−2 M HCl. The NH3 reacts with HCl as follows: NH3(aq)+HCl(aq)→NH4Cl(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 14.5 mL of 5.86×10−2 M NaOH to reach the equivalence point.
1)How many grams of NH3 were drawn into the acid solution?
2)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.)
3)Is this manufacturer in compliance with regulations?

Answer 1

Answer:

1) A total of 5.91×10-3 grams were drawn into the HCl solution

2) 84.3 ppm of NH3 were in the air

3) As 84.3 ppm is higher than 50 ppm established in the regulation, the manufacturer does not comply with it.

Explanation:

The problem shows the following process: an amount of air with NH3 passes through a HCl solution. The NH3 reacts with HCl reducing the concentration of the latted and finally the remaining HCl is titrated with NaOH.

The process to solve this problem should go as follows:

a) Calculate the amount of the remaining HCl that was titrated with the NaOH at the end.

b) Calculate the amount of HCl that reacted with NH3, using the data from a)

c) Calculate the amount of NH3 present in air using the data from b)

d) Calculate the grams of NH3 using the data from c) to solve question 1)

e) Calculate the number of moles of air

f) Calculate the ppm of NH3 in air using the data from c) and e) to solve questions 2) and 3)

So, let's proceed:

a) To do this we need to take a look at the chemical equation of the HCl and NaOH reaction:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

And we see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, being n the number of moles:

n(NaOH) = n(HCl)

(5.86×10-2 M)×(14.5×10-3 L) = n(HCl)

n(HCl) = 8.50×10-4 moles of HCl

b) So, as we now have the amount of the remaining HCl, we need to find out how much HCl reacted with NH3 in the first place, so we need to substract the number of moles found in a) from the number of moles we had initially:

n(HCl)_reacted with NH3 = n(HCl)_initial - n(HCl)

n(HCl)_reacted with NH3 = (1.13×10-2 M)×(106×10-3 L) - 8.50×10-4

n(HCl)_reacted with NH3 = 3.49×10-4 moles of HCl

c) Now, as we know the amount of HCl that reacted with NH3, we can calculate the amount of NH3 that was drawn into the solution using the chemical equation (fortunately, the equation is already balanced):

NH3(aq) + HCl(aq) → NH4Cl(aq)

And we can see 1 mole of NH3 reacts with 1 mole of HCl, so we can conclude that 3.49×10-4 moles of HCl have reacted with 3.49×10-4 moles of NH3.

As we are being asked by the grams, we must convert that using the molar mass of NH3 that is 17 g/mol (N=14, H=1), so:

grams of NH3 = (3.49×10-4 mol)×(17 g/mol) = 5.91×10-3 grams of NH3

d) Now we must calculate the number of moles of air in order to be able to calculate the parts per million of NH3:

In this case we have to notice that we have passes air at a rate of 10.0 liters per minute and we have done it by 10 minutes, that means that the total amount of air (in liters) we have passed through the solution is:

liters of air = 10 min × 10 L/min = 100 L

e) That volume of air can be converted into moles using the information from question 2):

moles of air = (100 L) × (1.2 g/L) × (1 mol/29 g) = 4.14 moles

f) We can calculate now using the information from c) and e) as follows:

ppm of NH3 in air = number of moles of NH3 / number of moles of air × 1000000

ppm of NH3 = (3.49×10-4 mol)/(4.14 mol)×1000000 = 84.3 ppm

In conclusion, the manufacturer does not comply with the regulation of maximum 50ppm of NH3.