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Ch.18 - Chemistry of the Environment
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All textbooksBrown 14th EditionCh.18 - Chemistry of the EnvironmentProblem 89

Chapter 18, Problem 89

The Henry's law constant for CO2 in water at 25 °C is 3.1x10^-2 M atm-1. (a) What is the soubility of CO2 in water at this temperature if the soltuion is in contact with air at normal atmospheric pressure?

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Hello everyone. Today we have the following problem. The Henry's law constant for carbon dioxide and water at 25 °C is 3.1 times 10 to the negative two molars times atmosphere to the negative one problem. A what is the solubility of carbon dioxide in water at this temperature? If the solution is in contact with air at 1.28 atmospheric pressure. So we can calculate that solubility with using Henry's law that states that the solubility of the gas is equal to Henry's law constant multiplied by the partial pressure of that gas. So we need to calculate the partial pressure of carbon dioxide in the air first. So our partial pressure of carbon dioxide is equal to the mole fraction of carbon dioxide multiply multiplied by its pressure and atmospheres. So we have that molarity B 0.0 or we don't use the molarity yet. But this is a value that we can find in the reference text and this will be multiplied by our atmospheric pressure of 1.28 for a partial pressure of 4.80 times 10 is the negative four atmospheres. We can then plug this into our equation. So our Henry's law constant is 3.1 times 10. That's the negative two molar atmospheres to the negative one, we multiply it by the pressure pressure of gas which is 4.8 zero times 10 to the negative four atmospheres for a value of 1.5 times 10 to the negative fifth molar or enter to c as our answer overall, I hope this helped. And until next time.
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