Answers to Review Questions

1. 0.0142g KCl x 1 mol KCl x 1.0 ml solution x 1000 ml = 1.91 x 10-5 mol/ liter
   100 ml soln    74.5 g KCl  100 ml soln        1 liter

1.91 x 10-5 mol Cl-  x 35.5 g Cl- x 1 liter soln x 106 = 0.678 ppm
liter soln               mol Cl-         103 g soln

2. 0.0100 g Pb x 331.2 g Pb(NO3)2 x 100 ml = 1.598 g Pb(NO3)2
         ml          207.2 g Pb

 3. 1.00 microg Pb x 1 g Pb x 331.2 g Pb(NO3)2 x 100 ml = 1.598 x 10-4 g Pb(NO3)2
     ml            106 microg Pb   207.2 g Pb

Too little to weigh accurately. Make 100 ml of a solution using 100 times as much Pb(NO3)2. Then dilute 1 ml of this to 100 ml to obtain the desired solution.

4. (5 ppm /1.94) x 1.02 = 2.63 ppm in the sample.

Assuming that the response of the instrument is linear in the range covered by the sample and standard.

5. All the Cu in the 25 ml extract was originally contained in 1 liter of sample.

4.8 ppm x 25/1000 = 0.12 ppm

  1. Answer 11.5 (Remember 1000 liter = 1 m3)
  2. 10 mg found in 11.5 m3 of air. 0.87 mg/m3
  3. 5 g/106 g (or ml) x 25 ml = 1.25 x 10-4 g in 11.5 m3. Answer 1.09 x10-5 g/m3

    4. or 10.9 ug/m3

  4. Loss on drying 2.323 g. % volatiles =69.8. So a 100 g sample contains

    5. 69.8 g volatiles x (2.45 g water/2.574 g volatiles) = 66.5 g water

    Totals 100-69.8 = 30.2 % solids, 66% water, 3.8% other volatiles

  5. Because we mixed equal volumes of the sample and standard, each of their concentrations will be half of the original concentration and the total concentration will be the sum of these two. You can use the equation for standard addition, but you should be able to do this from scratch. The instrument response is proportional to concentration so you can write 23.4 = kx, where x is the sample conc. and k is the proportionality constant. Also, 24.9 = k (.5x + .5(1.5)). k, of course remains the same. These two equations with two unknowns, can be solved simultaneously to get the answer, 1.3 ppm. Using the standard addition equation, of course you get the same answer, but thinking it out for yourself is good too! (Corrected answer)

  6. Calculate mol NaOH in original solution. Subtract mol of HCl used to titrate excess NaOH to obtain mol H+ in air. Calculate moles of air. Calculate mol acid /106 mol air. Ans: 7.7