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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 (H鈧侰O鈧), bicarbonate ion (HCO鈧冣伝), and carbonate ion (CO鈧兟测伝) are 8.31 脳 10鈦烩伓 M, 1.67 脳 10鈦烩伓 M, and 2.00 脳 10鈦烩伖 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鈦 ions using the given pH

We are given the pH of rainwater as 5.60. Using the formula for pH: pH = -log([H鈦篯) We can find the concentration of H鈦 ions: [H鈦篯 = 10^(-pH) = 10^(-5.60) = 2.51 脳 10鈦烩伓 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: [H鈧侰O鈧僝 + [HCO鈧冣伝] + [CO鈧兟测伝] = 1.0 脳 10鈦烩伒 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鈦篯 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 H鈧侰O鈧) = 4.45 脳 10鈦烩伔 Ka2 (for HCO鈧冣伝) = 4.69 脳 10鈦宦孤
06

6. Solve the system of equations to find the concentrations

Now we have three equations and three unknowns: 1. [H鈧侰O鈧僝 + [HCO鈧冣伝] + [CO鈧兟测伝] = 1.0 脳 10鈦烩伒 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: [H鈧侰O鈧僝 = 8.31 脳 10鈦烩伓 M [HCO鈧冣伝] = 1.67 脳 10鈦烩伓 M [CO鈧兟测伝] = 2.00 脳 10鈦烩伖 M These are the concentrations of carbonic acid, bicarbonate ion, and carbonate ion, respectively, in a raindrop with pH 5.60.

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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.

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Most popular questions from this chapter

How many microliters of 1.000\(M\) NaOH solution must be added to 25.00 \(\mathrm{mL}\) of a 0.1000 \(\mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\) or \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce a buffer with \(\mathrm{pH}=3.75 ?\)

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