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 1.00 * 102 ml of 0.0105 M HCI. The NH3 reacts with HCI according to: 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 13.1 ml_ of 0.0588 MNaOH to reach the equivalence point, (a) How many grams of NH3 were drawn into the acid solution? (b) 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.) (c) Is this manufacturer in compliance with regulations?
Solution: Step1: a) Volume of NaOH solution used = 13.1 mL = 1000 L = 0.0131 L Molarity of the NaOH solution = 0.0588 M Therefore, number of moles of NaOH = volume of solution in litres x molarity = 0.0131 L x 0.0588 M = 0.00077 mol Since, NaOH is monobasic, so, number of moles of HCl reacted with NaOH = 0.008 mol Volume of HCl solution used = 1.00 x 10 mL = 100 mL = 100 L = 0.100 L 1000 Molarity of the HCl solution = 0.0105 M Therefore, number of moles of HCl = volume of solution in litres x molarity = 0.100 L x 0.0105 M = 0.00105 mol Step2: Now, the reaction between HCl and NH takes plac3in the following manner: HCl(aq) + NH (aq) ------> NH Cl(aq) 3 4 It is clear from the above equation that, 1 mol of HCl requires 1 mol of NH for 3 neutralization. Thus, number of moles of NH drawn fr3 air = number of moles of HCl reacted with NH 3 = total number of moles of HCl - number of moles of HCl reacted with NaOH = 0.00105 - 0.00077 = 0.00028 mol Step3: Now, 1 mol of NH = mola3mass of NH 3 Therefore, 0.00028 mol NH = 0.00028 x molar mass of NH 3 3 = 0.00028 x 17.03 g [molar mass of NH = 17.3 g/mol] = 0.0048 g Thus, 0.0048 g of NH were 3awn into the acid solution. Step4: 23 b) 1 mol of NH = 6.32 x 10 molecules of NH 3 23 Therefore, 0.00028 mol of NH = 0.00023x 6.022 x 10 molecules of NH 3 = 0.0017 x 10 molecules of NH 3 = 1.7 x 10 molecules of NH 3 Step5: Again, Volume of air drawn through acid solution in 1 min = 10 L Therefore, volume of air drawn through acid solution in 10 min = 10 x 10 L = 100 L Density can be defined as mass per unit volume. i.e., Density = mass / volume Therefore, mass = density x volume Given, density of air = 1.20 g/L Therefore, mass of air drawn = 1.20 x 100 g = 120 g Step6: Now, molar mass of air = 29.0 g/mol Therefore, number of moles of air = mass of air / molar mass = 120 / 29.0 = 4.14 mol Therefore, number of molecules of air = number of moles x Avogadro’s number 23 = 4.14 x 6.022 x 10 molecules = 24.9 x 10 molecules = 2.49 x 10 molecules Step7: Thus, ppm of NH in a3 = (number of molecules of NH / number of m3ecules of air) x 10 6 20 24 6 = ( 1.7 x 10 / 2.49 x 10 ) x 10 = 0.683 x 10 2 = 68.3 ppm. Step8: c) Since the amount of NH obtained by the manufacturer(68.3 ppm) is greater than the 3 upper limit of NH set3y the Federal regulations(50 ppm), therefore, the manufacturer is not in compliance with regulations. --------------------