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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 atm? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{mol} / \mathrm{L}\) -atm.

Short Answer

Expert verified
The pH of water saturated with CO2 at 25°C and a partial pressure of 1.10 atm can be found using Henry's law and the dissociation constant of carbonic acid. First, calculate the concentration of CO2 in water, C, using Henry's law and the given values: \(C = (3.1 \times 10^{-2}\ mol/L\cdot atm) \cdot 1.10\ atm\). Then, set up an ICE table for the dissociation of carbonic acid and use the Ka expression to find the equilibrium concentration of H+ ions, x. As x is very small compared to C, we can simplify the equation to: \(x^2 = (4.3 \times 10^{-7})C\). Finally, plug C into the equation and solve for x, and then calculate the pH using the formula: \(pH = -\log_{10}(x)\).

Step by step solution

01

Calculate the concentration of dissolved CO2 using Henry's law

Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas: \[C = k_HP\] where C is the concentration of the dissolved gas, kH is the Henry's law constant, and P is the partial pressure of the gas. Here, we are given kH = \(3.1 \times 10^{-2}\ mol/L\cdot atm\) and P = 1.10 atm. Plug these values into the formula and calculate C: \[C = (3.1 \times 10^{-2}\ mol/L\cdot atm) \cdot 1.10\ atm\]
02

Write the chemical equation for the reaction of CO2 and water

Carbon dioxide dissolves in water to form carbonic acid (H2CO3), which is a weak acid: \[\mathrm{CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq)}\]
03

Calculate the equilibrium concentrations using the dissociation constant of H2CO3

Carbonic acid is a weak acid and will dissociate in water as follow: \[\mathrm{H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)}\] The dissociation constant, Ka, of carbonic acid can be found in reference materials and is approximately \(4.3 \times 10^{-7}\). To calculate the equilibrium concentrations, set up an ICE table (Initial, Change and Equilibrium) for the reaction: | | H2CO3 | H+ | HCO3- | |---------------|-------|------|-------| | Initial(mol) | C | 0 | 0 | | Change(mol) | -x | +x | +x | | Equilibrium | C-x | x | x | Apply the Ka expression for the dissociation of H2CO3: \[K_a = \frac{[\mathrm{H^+}][\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]}\] Plug in the equilibrium concentrations and Ka value: \[(4.3 \times 10^{-7}) = \frac{x^2}{C-x}\]
04

Calculate the pH using the equilibrium concentration of H+ ions

Solve the equation for x: \[x^2 + (4.3 \times 10^{-7})x - (4.3 \times 10^{-7})C = 0\] We will assume that x is very small compared to C, then C - x ≈ C, and we can simplify the equation to: \[x^2 = (4.3 \times 10^{-7})C\] Now, as derived in step 1, C = (3.1 × 10^(-2) mol/L • atm) • 1.10 atm. Plug C into the equation and solve for x: \[x^2 = (4.3 \times 10^{-7})(3.1 \times 10^{-2} mol/L\cdot atm \cdot 1.10\ atm)\] Finally, calculate the pH using the formula: \[pH = -\log_{10}[\mathrm{H^+}]\] Since x is equal to the equilibrium concentration of H+ ions: \[pH = -\log_{10}(x)\] Once you calculate the value of x in step 4, plug it into the pH formula to find the pH of the water saturated with CO2 at 25°C and a partial pressure of 1.10 atm.

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Key Concepts

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

Henry's Law
When we talk about the solubility of gases in liquids, Henry's law is a fundamental concept. This law states that at a constant temperature, the concentration of a dissolved gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The formula for Henry's law is expressed as:
\[C = k_H P\]
where \( C \) is the concentration of the dissolved gas, \( k_H \) is Henry's law constant specific to the gas, and \( P \) is the partial pressure of the gas.In our exercise, we use Henry's law to calculate the concentration of carbon dioxide (\(CO_2\)) dissolved in water at a given partial pressure. Understanding this step is crucial for further determining the behaviour of \(CO_2\) in the water, including its conversion to carbonic acid.
Equilibrium Concentration
The equilibrium concentration is the amount of a substance present in a solution at equilibrium. Equilibrium is the state where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products over time. In the context of dissolved \(CO_2\), once it reaches equilibrium with water, we have a constant concentration of carbonic acid formed from \(CO_2\).In our problem, we first calculate the initial concentration of \(CO_2\) using Henry's law. As the \(CO_2\) reacts with water to form carbonic acid (\(H_2CO_3\)), the concentrations of the reactants and products change until they reach an equilibrium. The concentrations at this point are essential to calculate the pH of the solution, as they dictate the number of hydrogen ions (\(H^+\)) in the solution.
Dissociation Constant
The dissociation constant, represented by \(K_a\), is a measure of the strength of an acid in solution. It reflects the degree to which an acid dissociates into its ions - in our case, carbonic acid (\(H_2CO_3\)) dissociating into hydrogen ions (\(H^+\)) and bicarbonate ions (\(HCO_3^-\)). A low \(K_a\) value indicates a weak acid, which only partially dissociates in solution.When we approach the exercise, we use the known \(K_a\) value of carbonic acid to set up an ICE (Initial, Change, Equilibrium) table and determine the equilibrium concentrations of the ions. This step is important because the concentration of \(H^+\) directly determines the acidity or the pH of our solution. For a weak acid like carbonic acid, we often assume that the change in concentration due to dissociation (\(x\)) is small compared to the initial concentration, which simplifies our calculations.
Carbonic Acid

Carbonic acid (\(H_2CO_3\)) is a weak acid formed when carbon dioxide (\(CO_2\)) dissolves in water. It plays a significant role in the body's buffering system and in environmental chemistry, contributing to the acidity of natural waters. In chemical equilibrium, carbonic acid can dissociate into hydrogen ions (\(H^+\)) and bicarbonate ions (\(HCO_3^-\)), which is a reversible reaction:


\[\(CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)\)\]

The extent of this reaction depends on the dissociation constant \(K_a\) and the starting concentration of dissolved \(CO_2\), which we calculate using Henry's law. As carbonic acid dissociates into ions, it impacts the pH of the solution, causing the water to become more acidic. Thus, understanding the properties of carbonic acid is key to predicting the pH of CO2-saturated water.

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Most popular questions from this chapter

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, \((\mathbf{b}) \mathrm{Pbl}_{2,}(\mathbf{c}) \mathrm{AuCl}_{3},(\mathbf{d}) \mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) (e) CuBr.

A 1.00 -L. solution saturated at \(25^{\circ} \mathrm{C}\) with lead(lI) iodide contains 0.54 \(\mathrm{g}\) of \(\mathrm{Pbl}_{2}\) . Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\) .

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnCO \(_{3}\) \(\mathbf{b} ) \mathrm{ZnS},(\mathbf{c}) \mathrm{BiI}_{3},(\mathbf{d}) \mathrm{AgCN},(\mathbf{e}) \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\)

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

Consider the equilibrium $$\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ Suppose that a salt of \(\mathrm{HB}^{+}(a q)\) is added to a solution of \(\mathrm{B}(a q)\) at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of \(\mathrm{B}(a q)\) increase, decrease, or stay the same? (c) Will the \(\mathrm{pH}\) of the solution increase, decrease, or stay the same?

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