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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

An aqueous solution containing 0.250 mole of \(\mathrm{Q},\) a strong electrolyte, in \(5.00 \times 10^{2} \mathrm{g}\) water freezes at \(-2.79^{\circ} \mathrm{C} .\) What is the van't Hoff factor for \(\mathrm{Q} ?\) The molal freezing-point depression constant for water is \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) . What is the formula of \(\mathrm{Q}\) if it is 38.68\(\%\) chlorine by mass and there are twice as many anions as cations in one formula unit of \(\mathrm{Q} ?\)

Calculate the freezing point and the boiling point of each of the following solutions. (Assume complete dissociation.) a. 5.0 \(\mathrm{g} \mathrm{NaCl}\) in 25 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) b. 2.0 \(\mathrm{g} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) in 15 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\)

An aqueous solution is 1.00\(\% \mathrm{NaCl}\) by mass and has a density of 1.071 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\) . The observed osmotic pressure of this solution is 7.83 atm at \(25^{\circ} \mathrm{C}\) . a. What fraction of the moles of NaCl in this solution exist as ion pairs? b. Calculate the freezing point that would be observed for this solution.

Adding a solute to a solvent extends the liquid phase over a larger temperature range. Explain this statement.

What volume of a \(0.580-M\) solution of \(\mathrm{CaCl}_{2}\) contains 1.28 \(\mathrm{g}\) solute?
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