Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
In a raindrop with pH 5.60, the concentrations of carbonic acid (H₂CO₃), bicarbonate ion (HCO₃⁻), and carbonate ion (CO₃²⁻) are 8.31 × 10⁻⁶ M, 1.67 × 10⁻⁶ M, and 2.00 × 10⁻⁹ M, respectively.

Step by step solution

01

1. Write the equilibrium reactions for carbonic acid dissociation

First, let's write down the two equilibrium reactions involving carbonic acid dissociation: Reaction 1: \(H_{2}CO_{3} \rightleftharpoons H^{+} + HCO_{3}^{-}\) Reaction 2: \(HCO_{3}^{-} \rightleftharpoons H^{+} + CO_{3}^{2-}\)
02

2. Write the equilibrium constants for the reactions

The equilibrium constants for the two reactions involving carbonic acid dissociation are given as follows: Equilibrium Constant 1: \(K_{a1} = \frac{[H^{+}][HCO_{3}^{-}]}{[H_{2}CO_{3}]}\) Equilibrium Constant 2: \(K_{a2} = \frac{[H^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}\)
03

3. Find the concentration of H⁺ ions using the given pH

We are given the pH of rainwater as 5.60. Using the formula for pH: pH = -log([H⁺]) We can find the concentration of H⁺ ions: [H⁺] = 10^(-pH) = 10^(-5.60) = 2.51 × 10⁻⁶ M
04

4. Set up two equations using the total concentration and the equilibrium constants

We know that the sum of the three species is: [H₂CO₃] + [HCO₃⁻] + [CO₃²⁻] = 1.0 × 10⁻⁵ M Using the equilibrium constants from Step 2, we can set up two more equations: \(K_{a1} = \frac{[H^{+}][HCO_{3}^{-}]}{[H_{2}CO_{3}]}\) \(K_{a2} = \frac{[H^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}\) Now, let's plug in the [H⁺] value that we calculated in Step 3: \(K_{a1} = \frac{(2.51 \times 10^{-6})([HCO_{3}^{-}])}{[H_{2}CO_{3}]}\) \(K_{a2} = \frac{(2.51 \times 10^{-6})([CO_{3}^{2-}])}{[HCO_{3}^{-}]}\)
05

5. Find the values for Ka1 and Ka2 for carbonic acid

To solve the system of equations, we need the Ka1 and Ka2 values for carbonic acid. Ka1 (for H₂CO₃) = 4.45 × 10⁻⁷ Ka2 (for HCO₃⁻) = 4.69 × 10⁻¹¹
06

6. Solve the system of equations to find the concentrations

Now we have three equations and three unknowns: 1. [H₂CO₃] + [HCO₃⁻] + [CO₃²⁻] = 1.0 × 10⁻⁵ M 2. \(\frac{(2.51 \times 10^{-6})([HCO_{3}^{-}])}{[H_{2}CO_{3}]} = 4.45 \times 10^{-7}\) 3. \(\frac{(2.51 \times 10^{-6})([CO_{3}^{2-}])}{[HCO_{3}^{-}]} = 4.69 \times 10^{-11}\) Solve this system of equations to get: [H₂CO₃] = 8.31 × 10⁻⁶ M [HCO₃⁻] = 1.67 × 10⁻⁶ M [CO₃²⁻] = 2.00 × 10⁻⁹ M These are the concentrations of carbonic acid, bicarbonate ion, and carbonate ion, respectively, in a raindrop with pH 5.60.

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.

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.

Carbonic Acid Dissociation
Carbonic acid dissociation is a key concept in understanding acid rain. When carbon dioxide (\(\text{CO}_2\)) from the atmosphere dissolves in water, it forms carbonic acid (\(\text{H}_2\text{CO}_3\)). This weak acid can dissociate in a two-step process to release hydrogen ions (\(\text{H}^+\)) and form bicarbonate (\(\text{HCO}_3^-\)) and carbonate ions (\(\text{CO}_3^{2-}\)).

The first dissociation step involves the conversion of carbonic acid to bicarbonate ion and a proton. This reaction can be represented as:
  • \(\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-\)

The second step involves the further dissociation of bicarbonate into carbonate ion:
  • \(\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}\)

Understanding these steps is integral to comprehend the behaviors of acids and buffers in natural systems.
Equilibrium Reactions
Equilibrium reactions are crucial in chemical processes as they define a state where the reactants and products are in balance. This balance allows one to measure the concentrations of different species in a system, like the dissociation of carbonic acid in rainwater.

In the case of carbonic acid, the dissociation reactions are equilibrium processes. For each dissociation of carbonic acid, there is a definitive equilibrium between the forward and backward reactions. At equilibrium, the rates of the forward reaction (dissociation) and the reverse reaction (recombination) are equal. This balance of reactions is represented by reversible arrows (\( \rightleftharpoons \)).

Recognizing the state of equilibrium allows us to calculate the concentrations of involved species, having a wide implication in acid-base chemistry and environmental science.
pH Calculation
Calculating \(\text{pH}\) is a fundamental step in understanding the acidity or basicity of a solution. It is a measure of hydrogen ion concentration and is expressed logarithmically:

\(\text{pH} = -\log([\text{H}^+])\)

For instance, in exercise with rainwater having a \(\text{pH}\) of 5.60, the concentration of hydrogen ions can be calculated using the formula: \([\text{H}^+] = 10^{-\text{pH}}\). Substituting the given \(\text{pH}\) :
  • \([\text{H}^+] = 10^{-5.60} = 2.51 \times 10^{-6}\,\text{M}\)

This concentration is crucial for further calculations of various species in the system, as it influences the equilibrium of the dissociation reactions.
Bicarbonate Ion
The bicarbonate ion (\(\text{HCO}_3^-\)) plays a significant role in the system of carbonic acid dissociation. It acts as both a product and reactant depending on its position in the dissociation process.

In the first dissociation step, bicarbonate is formed when carbonic acid releases a hydrogen ion. It is an intermediary species and can further dissociate into carbonate ion in the second step.

Bicarbonate also functions as a buffer in solutions, maintaining the \(\text{pH}\) within specific ranges, which is vital for natural waters and biological systems. Its concentration can be calculated by understanding the equilibrium reactions it participates in, balancing changes in \(\text{pH}\), and adjusting to added acids or bases.
Carbonate Ion
The carbonate ion (\(\text{CO}_3^{2-}\)) is the final species in the dissociation process of carbonic acid. Formed from the further dissociation of bicarbonate ions, it carries a 2- negative charge.

Carbonate plays an essential role in buffering systems and natural processes, like the formation of calcium carbonate found in limestone and shells.

Understanding its concentration in solutions is crucial, especially in environmental chemistry, where it interacts with calcium ions to form insoluble calcium carbonate. This reaction has significant implications, such as the buffering capacity of oceans and potential effects on marine life.
Equilibrium Constants
Equilibrium constants (\(K_a\)) are a measure of the extent of a reaction at equilibrium for dissociation reactions.

In the scenario of carbonic acid dissociation, two constants are involved:
  • \(K_{a1}\) relates to the first dissociation step (\(\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-\)) and is given by \(K_{a1} = \frac{[\text{H}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\).
  • \(K_{a2}\) for the second step (\(\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}\)) is \(K_{a2} = \frac{[\text{H}^+][\text{CO}_3^{2-}]}{[\text{HCO}_3^-]}\).

These constants reflect how readily the acid dissociates in water. They are vital for calculating the equilibrium concentrations of species in acidic solutions and are commonly used as reference values for reactions involving acids and bases.

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

A concentration of 10–100 parts per billion (by mass) of Ag+ is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the Ag+ can cause adverse health effects. One way to maintain an appropriate concentration of Ag+ is to add a slightly soluble salt to the pool. Using \(K_{s p}\) values from Appendix \(\mathrm{D},\) calculate the the equilibrium concentration of Ag+ in parts per billion that would exist in equilibrium with (a) \(\mathrm{AgCl},(\mathbf{b}) \mathrm{AgBr},(\mathbf{c}) \mathrm{AgI}\)

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?

(a) If the molar solubility of \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(1.24 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) what is \(K_{s p}\) at this temperature? (b) It is found that \(1.1 \times 10^{-2} \mathrm{g} \mathrm{SrF}_{2}\) dissolves per 100 \(\mathrm{mL}\) of aqueous solution at \(25^{\circ} \mathrm{C}\) Calculate the solubility product for \(\mathrm{SrF}_{2 .}(\mathbf{c})\) The \(K_{s p}\) of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-10} .\) What is the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} ?\)

Write the expression for the solubility-product constant for each of the following ionic compounds: AgI, SrSO \(_{4}, \mathrm{Fe}(\mathrm{OH})_{2},\) and \(\mathrm{Hg}_{2} \mathrm{Br}_{2}\) .

Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is required to reach the equivalence point for the strong acid than the weak acid. (b) The pH at the beginning of the titration is lower for the weak acid than the strong acid. (c) The pH at the equivalence point is 7 no matter which acid is titrated.

See all solutions

Recommended explanations on Chemistry Textbooks

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