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Chapter 11: Problem 20

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
\(\mathrm{NH}_{3}(g)\) does not obey Henry's law because it undergoes a chemical reaction with water molecules, forming ionic species upon dissolving, which results in the partial pressure not being directly proportional to the mole fraction in the solution. On the other hand, \(\mathrm{O}_{2}(g)\) obeys Henry's law in water but not in blood because it dissolves without undergoing any reactions in water, but it reacts with hemoglobin in blood, making its partial pressure not directly related to its concentration.

Step by step solution

01

Understand Henry's law

Henry's law states that the partial pressure of a gas in equilibrium with its solute phase is directly proportional to the mole fraction in the solution. Mathematically, it can be represented as \(P = K_Hx\), where P is the partial pressure of the gas, K_H is the Henry's law constant, and x is the mole fraction of the gas in the solution. Henry's law is applicable under certain conditions such as when the gas does not react with the solvent, and the system is in a dilute solution range.
02

Examine NH3 as a weak electrolyte

Ammonia gas, \(\mathrm{NH}_{3}\), is a weak electrolyte, meaning that when it dissolves in water, it partially ionizes and reacts with the water molecules, forming the ammonium (NH4^+) and hydroxide (OH^-) ions. The reaction can be represented as: \( \mathrm{NH}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \leftrightharpoons \mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq) \) This chemical reaction makes it so that the mole fraction of dissolved \(\mathrm{NH}_{3}\) in the solution cannot be directly related to the partial pressure above the solution.
03

Explain why NH3 does not obey Henry's law

The reason why \(\mathrm{NH}_{3}(g)\) does not obey Henry's law is that it reacts with the water molecules and forms ionic species upon dissolving. As a result, the partial pressure (P) is not directly proportional to the mole fraction (x) in the solution, which violates the conditions required for Henry's law to be applicable.
04

Examine O2 behavior in water

Oxygen gas (\(\mathrm{O}_{2}(g)\)) is a neutral, nonreactive gas. When it dissolves in water, it does not undergo any chemical reactions. Therefore, the partial pressure of \(\mathrm{O}_{2}(g)\) is directly proportional to its mole fraction in the solution, which means it obeys Henry's law in water.
05

Examine O2 behavior in blood

Blood is an aqueous solution with a complex composition containing proteins such as hemoglobin. When \(\mathrm{O}_{2}(g)\) dissolves in blood, it binds with hemoglobin to form oxyhemoglobin (HbO2). This binding process can be represented as: \( \mathrm{Hb}(aq) + \mathrm{O}_{2}(aq) \leftrightharpoons \mathrm{HbO}_{2}(aq) \) As there is a chemical reaction involved in the binding process, the partial pressure of \(\mathrm{O}_{2}\) is not directly related to its concentration in the blood.
06

Explain why O2 does not obey Henry's law in blood

The \(\mathrm{O}_{2}(g)\) gas does not obey Henry's law in blood because it undergoes a chemical reaction with hemoglobin, which makes the partial pressure (P) not directly proportional to the mole fraction (x) in the solution. This behavior violates the conditions required for Henry's law to be applicable.

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Most popular questions from this chapter

The vapor pressure of pure benzene is 750.0 torr and the vapor pressure of toluene is 300.0 torr at a certain temperature. You make a solution by pouring “some” benzene with “some” toluene. You then place this solution in a closed container and wait for the vapor to come into equilibrium with the solution. Next, you condense the vapor. You put this liquid (the condensed vapor) in a closed container and wait for the vapor to come into equilibrium with the solution. You then condense this vapor and find the mole fraction of benzene in this vapor to be 0.714. Determine the mole fraction of benzene in the original solution assuming the solution behaves ideally.

Consider an aqueous solution containing sodium chloride that has a density of 1.01 \(\mathrm{g} / \mathrm{mL}\) . Assume the solution behaves ideally. The freezing point of this solution at 1.0 \(\mathrm{atm}\) is $-1.28^{\circ} \mathrm{C}$ . Calculate the percent composition of this solution (by mass).

Which solvent, water or carbon tetrachloride, would you choose to dissolve each of the following? a. \(\mathrm{KrF}_{2}\) b. \(\mathrm{SF}_{2}\) c. \(\mathrm{SO}_{2}\) d. \(\mathrm{CO}_{2}\) e. \(M g F_{2}\) f. \(C H_{2} O\) g. \(C H_{2}=C H_{2}\)

In order for sodium chloride to dissolve in water, a small amount of energy must be added during solution formation. This is not energetically favorable. Why is NaCl so soluble in water?

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C} )\) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C} .\) You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C} )\) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at $25^{\circ} \mathrm{C}\( ) that has a boiling point of \)102.0^{\circ} \mathrm{C} .$ What will happen to the plant cells in the leaf?
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