Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 112

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 \(3.1 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\) -atm. The \(\mathrm{CO}_{2}\) is an acidic oxide, reacting with \(\mathrm{H}_{2} \mathrm{O}\) to form \(\mathrm{H}_{2} \mathrm{CO}_{3}\).

Short Answer

Expert verified
The pH of water saturated with $\mathrm{CO}_{2}$ at $25^{\circ} \mathrm{C}$ and $1.10 \mathrm{~atm}$ is approximately 7.

Step by step solution

01

Use Henry's law to find the concentration of CO₂ dissolved in water.

We are given the partial pressure of CO₂ (P) and the Henry's law constant (K) at 25°C. We can use Henry's law to find the concentration of dissolved CO₂ in water: \[C = KP\] \[C = (3.1 \times 10^{-2} \frac{mol}{L\cdot atm}) (1.1 atm)\] \[C = 3.41 \times 10^{-2} \frac{mol}{L}\]
02

Write the chemical equation for the reaction of CO₂ with H₂O to form H₂CO₃.

CO₂ reacts with water to form carbonic acid (H₂CO₃): \[CO_{2(g)} + H_{2O} \rightleftharpoons H_{2}CO_{3(aq)}\]
03

Determine the concentration of H₂CO₃.

Since CO₂ is in equilibrium with H₂CO₃, we can assume that the concentration of H₂CO₃ is equal to the concentration of dissolved CO₂: \[ [H_{2}CO_{3}] = 3.41 \times 10^{-2}\, M \]
04

Write the ionization equation of H₂CO₃ and find the concentration of H₃O⁺.

Carbonic acid (H₂CO₃) ionizes in water to form hydronium ions (H₃O⁺) and bicarbonate ions (HCO₃⁻): \[H_{2}CO_{3} \rightleftharpoons H_{3}O⁺ + HCO₃⁻\] This reaction can be considered as negligible, and the majority of H₃O⁺ ions come from the water's autoionization: \[2H_{2}O \rightleftharpoons OH⁻ + H₃O⁺\] As CO₂ is a weak acid and its concentration is relatively low, we can assume the autoionization of water doesn't change significantly. The concentration of H₃O⁺ in pure water at 25°C is approximately \(1 \times 10^{-7}\, M\). Therefore, the concentration of H₃O⁺ in water saturated with CO₂ would not be significantly different. \[ [H_{3}O⁺] \approx 1 \times 10^{-7}\, M\]
05

Determine the pH.

Now, we have the concentration of H₃O⁺. We can use the pH formula to find the pH: \[pH = -\log_{10}[H_{3}O⁺]\] \[pH = -\log_{10}(1 \times 10^{-7})\] The pH of water saturated with CO₂ at 25°C and 1.10 atm is approximately 7.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Henry's Law
Henry's law is fundamental to solving problems involving the solubility of gases in liquids. According to this principle, 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. Presented by the formula
\[C = KP\]
where \(C\) is the concentration of the dissolved gas, \(K\) is the Henry’s law constant for the gas, and \(P\) is the partial pressure. For this exercise, by applying Henry's law, we found the concentration of \(CO_2\) that dissolved in water when the gas was at a partial pressure of 1.10 atm.
Acidic Oxide Reaction with Water
When an acidic oxide like \(CO_2\) dissolves in water, it reacts to form a corresponding acid. In the case of \(CO_2\), it reacts with water to form carbonic acid \(H_2CO_3\). This reaction is reversible and can be represented as:
\[CO_{2(g)} + H_{2}O(l) \rightleftharpoons H_{2}CO_{3(aq)}\]
Understanding this reaction is essential to predict the pH of the solution, as the formed carbonic acid plays a critical role in the acid-base chemistry of the solution.
Equilibrium Concentration of Carbonic Acid
In a state of equilibrium, the rate of the forward reaction equals the rate of the backward reaction. In this context, we assume that all the \(CO_2\) that has dissolved turns into \(H_2CO_3\), primarily because carbonic acid is a weak acid and does not dissociate completely. Therefore, the equilibrium concentration of \(H_2CO_3\) can be equated to the initial concentration of dissolved \(CO_2\), which simplifies the process of calculating the pH of the saturated solution.
Ionization of Carbonic Acid
The ionization of carbonic acid in solution is pivotal in understanding its contribution to the overall hydrogen ion \(H^+\) concentration. Although \(H_2CO_3\) ionizes to produce \(H_3O^+\) and \(HCO_3^-\), its low concentration and weak acidic nature imply that its contribution to the hydrogen ion concentration is negligible. Here, the autoionization of water, which is the self-ionization where water acts as both acid and base, becomes more significant. However, the presence of \(CO_2\) slightly alters the concentration of hydrogen ions due to additional acidity, but not enough to drastically change the pH from that of pure water.
Autoionization of Water
The autoionization of water refers to the self-ionization process where water molecules dissociate into hydroxide \(OH^-\) and hydronium \(H_3O^+\) ions. The equilibrium constant for this dissociation, known as the ion product of water \(K_w\), remains constant at a given temperature. At 25°C, \(K_w\) is \(1 \times 10^{-14}\) which means in pure water, the concentrations of \(OH^-\) and \(H_3O^+\) are each \(1 \times 10^{-7} M\). In our exercise, even with \(CO_2\) dissolved, the primary source of \(H_3O^+\) remains the autoionization of water, leading to approximate calculations for the pH.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\begin{array}{llll} \text { (a) The molar solubility of } \mathrm{PbBr}_{2} & \text { at } 25^{\circ} \mathrm{C} & \text { is }\end{array}\) \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate \(K_{s p} .(\mathbf{b})\) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix \(\mathrm{D},\) calculate the \(\mathrm{pH}\) of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\)

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

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?

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{s p}=6.8 \times 10^{-27} .\) As discussed in the "Chemistry and Life" box on page 730 , 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 \(K_{s p}=1.0 \times 10^{-60} \cdot(\mathrm{a})\) Write the expres- sion for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Aspirin has the structural formula At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{a}\) for aspirin equals \(3 \times 10^{-5}\). If two aspirin tablets, each having a mass of \(325 \mathrm{mg},\) are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is \(2,\) what percent of the aspirin is in the form of neutral molecules?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free