Example 3.25: Calculation of Parts per Million and Parts per Billion Concentrations
According to the EPA, when the fraction by mass of lead in tap water reaches
15 ppb, certain remedial actions must be taken. What is this "concentration" in ppm? At this level, what mass of lead (μg) would be contained in a typical glass of water (
300 mL)?
Solution
\(\mathrm{ppm}\) \(= \frac{1 }{ 1000000}\)
\(\mathrm{ppb}\) \(= \frac{1 }{ 1000000000}\)
\(V_{\mathrm{glass}}\) \(= 300\ \mathrm{mL}\)
\(f_{\mathrm{lead}}\) \(= 15. \cdot \mathrm{ppb}\)
\(\ \ \ =15. \cdot \frac{1 }{ 1000000000}\)
\(\ \ \ =1.5\times 10^{-8}\)
\(f_{\mathrm{lead}}\) \(= \dfrac{f_{\mathrm{lead}}}{\mathrm{ppm}}\mathrm{\ \mathrm{ppm}}\)
\(\ \ \ =\dfrac{1.5\times 10^{-8}}{\frac{1 }{ 1000000}}\mathrm{\ \mathrm{ppm}}\)
\(\ \ \ =\dfrac{1.5\times 10^{-8}}{\frac{1 }{ 1000000}}\mathrm{\ \mathrm{ppm}}\)
\(\ \ \ =0.015\mathrm{\ \mathrm{ppm}}\)
The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. However, only the volume of solution (
300 mL) is given, so we must use the density to derive the corresponding mass. We can assume the density of tap water to be roughly the same as that of pure water (~1.00 g/mL), since the concentrations of any dissolved substances should not be very large.
\(ρ_{\mathrm{solution}}\) \(= 1.00\ \frac{\mathrm{g}}{\mathrm{mL}}\)
\(m_{\mathrm{solution}}\) \(= ρ_{\mathrm{solution}} \cdot V_{\mathrm{glass}}\)
\(\ \ \ =1.00\ \frac{\mathrm{g}}{\mathrm{mL}} \cdot 300\ \mathrm{mL}\)
\(\ \ \ =300.\ \mathrm{g}\)
\(m_{\mathrm{lead}}\) \(= f_{\mathrm{lead}} \cdot m_{\mathrm{solution}}\)
\(\ \ \ =1.5\times 10^{-8} \cdot 300.\ \mathrm{g}\)
\(\ \ \ =4.5\times 10^{-6}\ \mathrm{g}\)
Finally, convert this mass to the requested unit of micrograms:
\(m_{\mathrm{lead}}\) \(= 4.5\times 10^{-6}\ \mathrm{g}\)
\(\ \ \ =4.5\ \mathrm{μg}\)