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What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(1.10 \mathrm{~atm}\) ? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{~mol} / \mathrm{L}-\mathrm{atm}\).

Short Answer

Expert verified
The pH of the water saturated with CO鈧 at 25掳C and a partial pressure of 1.10 atm is approximately 1.77.

Step by step solution

01

Determine the concentration of dissolved CO鈧 using Henry's Law

Henry's Law states that the concentration of a gas in a liquid is proportional to the partial pressure of the gas above the liquid. The Henry's Law constant (KH) relates the concentration (C) of the dissolved gas to its partial pressure (P) in the following manner: \[C = K_H \cdot P\] The partial pressure of CO鈧 (P) is given as 1.10 atm and the Henry's Law constant (KH) for CO鈧 at 25掳C is \(3.1 \times 10^{-2} \frac{mol}{L \cdot atm}\). We can now use this information to determine the concentration of dissolved CO鈧.
02

Calculate the dissolved CO鈧 concentration

Using Henry's Law equation from Step 1, we can calculate the dissolved CO鈧 concentration (C) in the water: \[C = K_H \cdot P = (3.1 \times 10^{-2} \frac{mol}{L \cdot atm}) \cdot (1.10 \, atm) = 3.41 \times 10^{-2} \frac{mol}{L}\] The concentration of dissolved CO鈧 in the water is \(3.41 \times 10^{-2} mol/L\).
03

Calculate the concentration of H鈦 ions

When CO鈧 dissolves in water, it reacts with water to form carbonic acid (H鈧侰O鈧), which then dissociates predominantly into bicarbonate ions (HCO鈧冣伝) and H鈦 ions: \[CO鈧(aq) + H鈧侽(l) \rightleftharpoons H鈧侰O鈧(aq) \rightleftharpoons HCO鈧冣伝(aq) + H鈦(aq)\] The concentration of H鈦 ions is equal to the concentration of HCO鈧冣伝 ions: \[[H鈦篯 = [HCO鈧冣伝] = \frac{1}{2}[CO鈧(aq)]\] Let's calculate the concentration of H鈦 ions in the water: \[[H鈦篯 = \frac{1}{2}(3.41 \times 10^{-2} \frac{mol}{L}) = 1.705 \times 10^{-2} \frac{mol}{L}\] The concentration of H鈦 ions in the water is \(1.705 \times 10^{-2} mol/L\).
04

Calculate the pH of the solution

Finally, we can calculate the pH using the definition of pH, which is the negative base-10 logarithm of the concentration of hydrogen ions: \[pH = -\log_{10} [H鈦篯\] \[pH = -\log_{10} (1.705 \times 10^{-2}) = 1.77\] The pH of the water saturated with CO鈧 at 25掳C and a partial pressure of 1.10 atm is approximately 1.77.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henry's Law
Understanding Henry's Law is crucial for grasping how gases dissolve in liquids, particularly in our discussion about pH calculation involving carbon dioxide in water. The law states that at a constant temperature, the concentration of a dissolved gas is directly proportional to its partial pressure in the gas phase.

In mathematical terms, the law is given as: \[C = K_H \times P\]where C is the concentration of the gas in the liquid, K_H is Henry's Law constant for the particular gas at a given temperature, and P is the partial pressure of the gas.

This relationship allows us to calculate how much CO鈧 is dissolved in water at a specific partial pressure, which is the starting point for determining the pH level of carbonated water. The constant (K_H) varies for different gases and depends on the temperature, enabling us to use it as a unique factor for a given gas, such as CO鈧, at a given condition.
Dissociation of Carbonic Acid
When carbon dioxide dissolves in water, it forms carbonic acid (\(H_2CO_3\)), a reaction key to understanding the acidity of solutions such as carbonated water. The dissociation of carbonic acid into bicarbonate (\(HCO_3^-\)) and hydrogen (\(H^+\)) ions can be represented by the equilibrium:\[CO_2(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) \rightleftharpoons HCO_3^-(aq) + H^+(aq)\]
The degree of this dissociation affects the concentration of hydrogen ions, hence affecting the pH. While carbonic acid is considered a weak acid because it dissociates partially in water, the process significantly contributes to the acidity of solutions like soda water. Understanding the balance between carbonic acid and its dissociated ions is pivotal for accurately predicting the pH of such solutions.
Concentration of Hydrogen Ions
The concentration of hydrogen ions (\([H^+]\)) is directly tied to the pH level of a solution. As per the calculation from the provided exercise, the dissociation of carbonic acid results in the production of hydrogen ions, which determines the acidity.

The reaction between dissolved CO鈧 and water produces carbonic acid, which then dissociates into hydrogen and bicarbonate ions. The concentration of hydrogen ions is derived from the half reaction of carbonic acid dissociation:\[[H^+] = \frac{1}{2}[CO_2(aq)]\]Since the concentration of hydrogen ions is available, you can calculate the pH level of the solution, illustrating the importance of this concept in understanding the chemistry of acids and bases.
Acid-base Equilibrium
The acid-base equilibrium in a solution refers to the reversible chemical reactions involving the transfer of hydrogen ions between reactants. In the context of the dissolution of CO鈧 in water, carbonic acid (\(H_2CO_3\)) forms and slightly dissociates to give bicarbonate (\(HCO_3^-\)) and hydrogen (\(H^+\)) ions.

This equilibrium can be expressed as:\[H_2CO_3(aq) \rightleftharpoons HCO_3^-(aq) + H^+(aq)\]The position of this equilibrium is influenced by various factors, including the concentration of dissolved CO鈧, the temperature, and the presence of other ions in the solution. Understanding this equilibrium is essential for predicting the pH and the behavior of carbonated drinks, ocean acidification, and physiological processes in biology.

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Most popular questions from this chapter

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\), (b) \(\mathrm{Cr}(\mathrm{OH})_{2}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{4}\), (d) \(\mathrm{Ag}^{*}\) and \(\mathrm{Mn}^{2+}\), Suggest how each mixture might be separated.

The solubility of \(\mathrm{CaCO}_{3}\) is \(\mathrm{pH}\) dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{a p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction (c) If we assume that the only sources of \(\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-}\), and \(\mathrm{OH}^{-}\)ions are from the dissolution of \(\mathrm{CaCO}_{3}\), what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the equilibrium expression from part (b)? (d) What is the molar solubility of \(\mathrm{CaCO}_{3}\) at the \(\mathrm{pH}\) of the ocean (8.3)? (e) If the pH is buffered at \(7.5\), what is the molar solubility of \(\mathrm{CaCO}_{3}\) ?

You are asked to prepare a pH \(=4.00\) buffer starting from \(1.50 \mathrm{~L}\) of \(0.0200 \mathrm{M}\) solution of benzoic acid \(\left(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{COOH}\right)\) and any amount you need of sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa}\right)\). (a) What is the \(\mathrm{pH}\) of the benzoic acid solution prior to adding sodium benzoate? (b) How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Rainwater is acidic because \(\mathrm{CO}_{2}(\mathrm{~g})\) dissolves in the water, creating carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{5}\). If the rainwater is toe acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, \(\mathrm{CaCO}_{3}\) ). Calculate the concentrations of carbonic acid, bicarbonate ion \(\left(\mathrm{HCO}_{3}{ }^{-}\right)\)and carbonate ion \(\left(\mathrm{CO}_{3}^{2-}\right)\) that are in a raindrop that has a \(\mathrm{pH}\) of \(5.60\), assuming that the sum of all three species in the raindrop is \(1.0 \times 10^{-5} \mathrm{M} .\)

A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\). of water and titrated with \(0.0950 \mathrm{M}\) \(\mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be 6.50. What is the \(K_{a}\) for the unknown acid?

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