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Chapter 10: Problem 18

The weak electrolyte \(\mathrm{NH}_{3}(g)\) does not obey Henry's law. Why? \(\mathrm{O}_{2}(g)\) obeys Henry's law in water but not in blood (an aqueous solution). Why?

Short Answer

Expert verified
NH3 does not obey Henry's law because it partially ionizes and reacts with water, forming NH4+ and OH- ions. O2 obeys Henry's law in water as it is non-reactive and has low solubility. However, in blood, O2 binds with hemoglobin, making its solubility nonlinear and not solely dependent on partial pressure, thus not obeying Henry's law.

Step by step solution

01

Ammonia is a weak electrolyte that can ionize when interacting with water as follows: \[ \mathrm{NH}_{3}(aq) + H_{2}O \rightleftharpoons NH_{4^{+}}(aq) + OH^{-}(aq) \] Since ammonia ionizes partially in water, it participates in a chemical reaction and thus cannot be considered a non-reactive gas in this context. #Step 2: Explain why NH3(g) does not obey Henry's law#

As ammonia partially dissolves into water and forms ions, it violates the condition that the gas should not react with the solvent. Thus, the behavior of ammonia in water does not correspond with Henry's law. #Step 3: Analyze the behavior of O2(g) in water and blood#
02

To start, we look at the solubility of oxygen in water. Oxygen has low solubility in water and does not react with it. Therefore, it obeys Henry's law in this case. However, when we consider oxygen in blood, the situation changes. Blood is an aqueous solution primarily composed of water, but it also contains hemoglobin, a protein that binds to oxygen. #Step 4: Explain why O2(g) obeys Henry's law in water but not in blood#

Oxygen is a non-reactive gas with low solubility in water, so it follows Henry's law when dissolved in it. However, in blood, oxygen binds to the hemoglobin molecules present in red blood cells. This binding makes the solubility of oxygen in blood nonlinear, as it no longer depends only on the partial pressure of oxygen above the solution. Consequently, oxygen does not obey Henry's law in blood due to the presence of hemoglobin.

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Most popular questions from this chapter

The vapor pressure of a solution containing 53.6 g glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) in \(133.7 \mathrm{g}\) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 113 torr at \(40^{\circ} \mathrm{C}\) Calculate the vapor pressure of pure ethanol at \(40^{\circ} \mathrm{C}\) assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.

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