Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\), causing the pH of clean, unpolluted rain to range from about 5.2 to 5.6 . What are the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops?

pH is the negative logarithm of the hydrogen ion concentration, i.e., \(\mathrm{pH} = -\log_{10}([\mathrm{H}^+])\). We can rewrite this equation to solve for the concentration of \(\mathrm{H}^{+}\) ions: \([\mathrm{H}^{+}] = 10^{-\mathrm{pH}}\).

Given the pH range of clean, unpolluted raindrops is between 5.2 and 5.6, we can use the equation from step 1 to find the range of \(\mathrm{H}^{+}\) ion concentration: Lower bound: \([\mathrm{H}^{+}]_{lower} = 10^{-5.6} \approx 2.51 \times 10^{-6} \ \mathrm{M}\) Upper bound: \([\mathrm{H}^{+}]_{upper} = 10^{-5.2} \approx 6.31 \times 10^{-6} \ \mathrm{M}\) So, the range of \(\mathrm{H}^{+}\) ion concentration in the raindrops is approximately \(2.51 \times 10^{-6} \ \mathrm{M}\) to \(6.31 \times 10^{-6} \ \mathrm{M}\).

The ion product constant of water (\(K_w\)) is given by the equation: \(K_w = [\mathrm{H}^+][\mathrm{OH}^{-}]\), where \(K_w = 1.0 \times 10^{-14}\) at 25°C. We can use this equation to find the range of \(\mathrm{OH}^{-}\) ion concentration: Lower bound: \([\mathrm{OH}^{-}]_{lower} = \frac{K_w}{[\mathrm{H}^{+}]_{upper}} = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-6} \ \mathrm{M}} \approx 1.59 \times 10^{-9} \ \mathrm{M}\) Upper bound: \([\mathrm{OH}^{-}]_{upper} = \frac{K_w}{[\mathrm{H}^{+}]_{lower}} = \frac{1.0 \times 10^{-14}}{2.51 \times 10^{-6} \ \mathrm{M}} \approx 3.98 \times 10^{-9} \ \mathrm{M}\) So, the range of \(\mathrm{OH}^{-}\) ion concentration in the raindrops is approximately \(1.59 \times 10^{-9} \ \mathrm{M}\) to \(3.98 \times 10^{-9} \ \mathrm{M}\). In conclusion, the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops are approximately \(2.51 \times 10^{-6} \ \mathrm{M}\) to \(6.31 \times 10^{-6} \ \mathrm{M}\) and \(1.59 \times 10^{-9} \ \mathrm{M}\) to \(3.98 \times 10^{-9} \ \mathrm{M}\), respectively.