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A chemical analysis has revealed the following composition for the rain falling on the city of Thule: 1.38 mg/L of HNO3, 3.21 mg/L of H2SO4, and 0.354 mg/L of HCl. HNO3, H2SO4 and HCl are strong acids, so you can assume that the H is fully released when these acids are dissolved in water. a. Calculate the pH of the rain water. Assume that the impact of carbon dioxide on pH is negligible. b. Based on the pH you obtained in question a, please comment if it would be reasonable to neglect the effect of carbon dioxide

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Answer:

pH = 4.19

Explanation:

Let's assume we have 1 L of rain, then we'll have

  • 1.38 mg of HNO₃
  • 3.21 mg of H₂SO₄
  • 0.354 mg of HCl

Now let's convert the mass into moles:

  • 1.38 mg HNO₃ ÷ 63mg/mmol = 0.0219mmol = 2.190*10⁻⁵ mol HNO₃
  • 3.21 mg H₂SO₄ ÷ 98mg/mmol = 0.0328mmol = 3.275*10⁻⁵ mol H₂SO₄
  • 0.354 mg HCl ÷ 36.46mg/mmol = 9.709*10⁻³mmol = 9.709*10⁻⁶ mol HCl

One mol of each of these strong acids is equal to one mol of H⁺, so the total number of H⁺ moles in 1 L of rain is:

  • 2.190*10⁻⁵ + 3.275*10⁻⁵ + 9.709*10⁻⁶ = 6.436*10⁻⁵ mol H⁺

The molar concentration of H⁺, or [H⁺], is:

  • [H⁺] = 6.436*10⁻⁵ mol H⁺ / 1 L =  6.436*10⁻⁵ M
  • pH = -log[H⁺] = 4.19

According to EPA, typical acid rain has a pH between 4.2 and 4.4, so in this case it is reasonable to neglect the effect of carbon dioxide, because the calculated pH value is not significantly different than the pH of typical acid rain.

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