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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 \(111.5 \mathrm{kPa}\) ? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}-\mathrm{kPa} .\)

Short Answer

Expert verified
The pH of the water is approximately 4.41.

Step by step solution

01

Calculate the concentration of dissolved CO2

Use Henry's Law to find the concentration of dissolved CO2 in water. Henry's Law is given by the formula: \[ C = K_H \cdot P_{\text{CO}_2} \]Where - \( C \) is the concentration of dissolved CO2,- \( K_H = 3.1 \times 10^{-4} \text{ mol/L-kPa} \) is the Henry's law constant,- \( P_{\text{CO}_2} = 111.5 \text{ kPa} \) is the partial pressure of CO2.Substituting the given values:\[ C = 3.1 \times 10^{-4} \times 111.5 = 0.034565 \text{ mol/L} \]
02

Use the equilibrium of carbonic acid formation

When CO2 dissolves in water, it reacts to form carbonic acid \( \text{H}_2\text{CO}_3 \). The reaction is:\[ \text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \] Assuming that all the dissolved CO2 forms carbonic acid, \( [\text{H}_2\text{CO}_3] = 0.034565 \text{ mol/L} \).
03

Calculate the hydrogen ion concentration with the dissociation of carbonic acid

\( \text{H}_2\text{CO}_3 \) partially dissociates to form \( \text{H}^+ \) and \( \text{HCO}_3^- \):\[ \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \]The dissociation constant \( K_a \) for carbonic acid is approximately \( 4.3 \times 10^{-7} \). The formulation is:\[ K_a = \frac{[\text{H}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \]Assuming \([\text{H}^+] = [\text{HCO}_3^-] = x\) and \([\text{H}_2\text{CO}_3] \approx 0.034565\):\[ 4.3 \times 10^{-7} = \frac{x^2}{0.034565} \]Solving for \( x \):\[ x^2 = 4.3 \times 10^{-7} \times 0.034565 \]\[ x^2 = 1.486 \times 10^{-8} \]\[ x = 3.855 \times 10^{-5} \text{ mol/L} \]
04

Calculate the pH

The \( \text{pH} \) is defined as:\[ \text{pH} = -\log_{10}([\text{H}^+]) \]Substituting the value of \([\text{H}^+]\):\[ \text{pH} = -\log_{10}(3.855 \times 10^{-5}) \]\[ \text{pH} \approx 4.41 \]

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Key Concepts

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

Henry's Law
Henry's Law describes how gases dissolve in liquids and is used to determine the concentration of dissolved gas within a solution. It's a simple concept that combines a proportional relationship between two variables: the concentration of the gas in the liquid (\( C \)) and the partial pressure (\( P \)) of the gas above the liquid.
  • The formula is given by: \[ C = K_H \cdot P_{\text{CO}_2} \] where \( K_H \) is the Henry's Law constant that changes with temperature and the gas-liquid combination.
  • In our problem, \( K_H \) was provided as \(3.1 \times 10^{-4} \text{ mol/L-kPa}\), effectively showing us how much \( \text{CO}_2 \) will dissolve in water per unit of pressure.
Knowing this helps you start the calculation and understand how environmental changes, like pressure, affect gas solubility.
Carbonic Acid Equilibrium
When carbon dioxide (\( \text{CO}_2 \)) is dissolved in water, it forms carbonic acid (\( \text{H}_2\text{CO}_3 \)), a weak acid. This equilibrium can be described by the chemical reaction:\[ \text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \]
  • Assuming complete conversion of dissolved CO2 to carbonic acid simplifies the calculation, with our earlier concentration giving us \([\text{H}_2\text{CO}_3] \approx 0.034565 \text{ mol/L}\).
  • This equilibrium is crucial as it dictates further reactions, especially the acid dissociation process that follows.
Understanding this equilibrium is essential because it sets the stage for the subsequent dissociation of carbonic acid, which directly affects the \( \text{pH} \) of the solution.
Acid Dissociation Constant
The acid dissociation constant (\( K_a \)) reflects the strength of an acid in a solution and is crucial for understanding how much an acid dissociates into its ions. For carbonic acid (\( \text{H}_2\text{CO}_3 \)), the dissociation into hydrogen ions and bicarbonate ions is given as:\[ \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \]Here,
  • The dissociation constant is: \( K_a = 4.3 \times 10^{-7} \).
  • This value tells us carbonic acid only weakly dissociates, meaning most of it remains intact as \( \text{H}_2\text{CO}_3 \).
You use this to find the concentration of hydrogen ions (\( \text{H}^+ \)) in the solution, which is essential for calculating the \( \text{pH} \). You set up the equilibrium expression and solve for the hydrogen ion concentration, providing the key link between concentration and \( \text{pH} \).
Partial Pressure
Partial pressure is a measurement of the pressure contributed by a single type of gas in a mixture of gases. It is a vital concept when examining how much of a gas can dissolve in a liquid, as seen in Henry's Law.
  • In a closed environment, like the above problem, partial pressure of \( \text{CO}_2 \) is taken as \(111.5 \text{ kPa}\).
  • This value drives the dissolution process by affecting how much \( \text{CO}_2 \) can dissolve, influencing the saturation concentration of dissolved \( \text{CO}_2 \).
Understanding partial pressure helps elucidate why gases behave differently under various environmental conditions and how they interact with liquids like water to form solutions. It provides the initial step in utilizing Henry's Law for calculating concentrations in the mentioned scenario.

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Most popular questions from this chapter

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{\text {ip }}=6.8 \times 10^{-27}\). As discussed in the Chemistry and Life box on page 790 , fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F},\) whose \(K_{y p}=1.0 \times 10^{-60}\) (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

(a) Calculate the percent lonization of \(0.0085 \mathrm{M}\) butanoic acid \(\left(K_{a}=1.5 \times 10^{-5}\right),(\mathbf{b})\) Calculate the percent ionization of \(0.0085 \mathrm{M}\) butanoic acid in a solution containing \(0.075 \mathrm{M}\) sodium butanoate.

(a) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid in a solution containing \(0.050 \mathrm{M}\) sodium lactate.

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating \(0.200 \mathrm{M}\) solutions of each of the following bases with 0.200 M HBr: (a) sodium hydroxide (NaOH), (b) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\)

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

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