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Concentration: Parts per Million and Parts per Billion

Chemistry Introductory Concepts. Concentration: Parts per Million and Parts per Billion. J. Flint Baumwirt Granada Hills Charter High School/CSUN Math, Science and Technology Magnet. Concentration Units.

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Concentration: Parts per Million and Parts per Billion

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  1. Chemistry Introductory Concepts Concentration: Parts per Million and Parts per Billion J. Flint BaumwirtGranada Hills Charter High School/CSUN Math, Science and Technology Magnet

  2. Concentration Units • The concentration units parts per million (ppm) and parts per billion (ppb) fine use when dealing with extremely dilute solutions. • Environmental chemists frequently use such units in specifying the concentration of trace pollutants or toxic chemicals in air and water samples. The units are closely related to percent concentration units (x 100).

  3. Concentration units • Because amounts of solute and solution present may be stated in terms of either mass or volume, there are three different forms for each unit: • mass-mass (m/m), volume-volume (v/v), and mass-volume (m/v).

  4. Parts per million • A part per million (ppm) is one part of solute per million parts of solution. In terms of defining equations, we can write: ppm (m/m) = mass solute x 106 mass solution ppm (v/v) = volume solute x 106 volume solution or ppm (m/v) = mass solute(g) x 106 volume solution (mL) A part per billion (ppb) is one part of solute per billion parts of solution. Here the factor is 109 instead of the factor of 106 for parts per million.

  5. Example: • A sample of water upon analysis was found to contain 6.3 x 10-3 grams of lead per 375 mL of solution. What is the lead concentration in a) ppm (m/v) and b) ppb (m/v)?

  6. Solve for ppm • Set up: • Solution: ppm (m/v) = mass solute(g) x 106 volume solution (mL) ppm (m/v) = 6.3 x 10-3 g x 106 = 17 ppm 375 mL

  7. Solve for ppb • Set up: • Solution: ppb (m/v) = mass solute(g) x 109 volume solution (mL) ppb (m/v) = 6.3 x 10-3 g x 109 = 17,000 ppb 375 mL

  8. Extension Problem • The ozone, O3content of smog is approximately 0.5 ppm (v/v). At this concentration, how much O3 (in mL), would be present in a sample of smog the size of a 2 L soda pop container? • Note that this is a volume/volume problem and the units be in units of mL.

  9. Set up and Solution • Given info : 0.5 ppm and 2 L = 2000 mL solutionSet up and solve for the volume of the solute • 0.5 ppm = volume solute x 1062000 mL solution • Volume of solute = 0.5 ppm x 2000 mL 106 Answer: 0.001 mL O3

  10. ApplicationBrady Problem #5.72 • Mercury is an extremely toxic substance that deactivates enzyme molecules that promotion biochemical reactions. • A 25.0 g sample of tuna fish taken from a large shipment was analyzed for this substance and found to contain 2.1 x 10-5 mole of Hg. By law foods having a mercury content above 0.50 ppm cannot be sold. Determine whether this shipment must be confiscated.

  11. Set up and Solution • This problem should be recognized as a mass/mass ppm problem and should be set up to solve for grams of Hg grams of sample 2.1 x 10-5 mol Hg x 200.59 g x 106 ppm = 170 ppm 25.0 g sample 1 mole Hg This is above maximum allowed Hg content of 0.50 ppm so YES the shipment must be confiscated.

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