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18 Chemistry of the Environment Solutions to Exercises = = = 2.5119 10⁻⁶ = 3 10⁻⁶ M Also, + [HCO₃⁻] + = 1.0 X 10⁻⁵ M We now have 3 equations in 3 unknowns, so we can solve explicitly for one. Solve for [HCO₃⁻] (since it appears in both expressions), then substitute to find and [CO₃²⁻]. 1.0 10⁻⁵ = Kₐ₁ + [HCO₃⁻] + 1.0 10⁻⁵ = 5.8416[HCO₃⁻] + [HCO₃⁻] + 2.2294 10⁻⁵ [HCO₃⁻] [HCO₃⁻]= 6.8416 =1.4616 = 1.5 10⁻⁶ Note that [CO₃²⁻] is very small compared to and [HCO₃⁻]. (2.5119 10⁻⁶)(1.4616 10⁻⁷ 10⁻⁶) = 8.5383 10⁻⁶ = 8.5 10⁻⁶ M 4.3 (5.6 2.5119 10⁻⁶ = 3.2586 10⁻¹¹ = 3.3 10⁻¹¹ M Check. 1.5 10⁻⁶ M + 8.5 10⁻⁶ M + 3.3 X 10⁻¹¹ M = 1.0 M (b) In order to test for sulfur-containing species, we must first remove the various forms of carbonate. One method is to exploit the solubility differences between carbonate and sulfate salts. Most sulfates are soluble, while most carbonates are not. However, values for carbonates are relatively large, and [CO₃²⁻] in the raindrop is very small. Precipitating insoluble carbonates will shift the acid dissociation equilibria to the right, but precipitation may not be the best method for effectively removing carbonates. A different method involves removing carbonates as CO₂(g). Heating the rainwater will decrease the solubility of CO₂(g), which will bubble off as a gas. Slightly acidifying the solution will encourage this process, by shifting the acid dissociation equilibria toward and CO₂(g). After removal of carbonates, sulfates are precipitated with Ba²⁺ (aq). The amount of precipitate is small, but it does cause turbidity in the solution. Turbidity is detected by instrumental methods which measure light scattering by colloids. 575