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When carbon dioxide dissolves in water, it is in equilibrium with carbonic acid \(\mathrm{H}_{2} \mathrm{CO}_{3},\) which is a weak electrolyte. What solutes are present in aqueous solution of this compound? Write the chemical equation for the ionization of \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

Short Answer

Expert verified
The solutes present in the aqueous solution of carbonic acid (H₂CO₃) are H⁺ ions and HCO₃⁻ ions. The chemical equation for the ionization of carbonic acid is: \[H_2CO_3 \rightleftharpoons H^+ + HCO_3^-\]

Step by step solution

01

Identify the solutes present in the aqueous solution of carbonic acid.

Carbonic acid is a weak acid, and when it dissolves in water, it partially ionizes, meaning it splits into its constituent ions. In the case of H₂CO₃, the solutes present in an aqueous solution will be H⁺ ions and HCO₃⁻ ions.
02

Write the chemical equation for the ionization of carbonic acid.

To write the chemical equation, we will begin with the chemical formula of carbonic acid (H₂CO₃) and show it ionizing into its constituent ions: \[H_2CO_3 \rightleftharpoons H^+ + HCO_3^-\] This equation represents the ionization of carbonic acid when it dissolves in water, forming H⁺ and HCO₃⁻ ions. The double arrow (\(\rightleftharpoons\)) indicates that the reaction is in equilibrium, meaning it's both forward (ionization) and reverse (recombination) reactions occur simultaneously.

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Key Concepts

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

Weak Acid Equilibrium
Carbonic acid (\(\mathrm{H}_2\mathrm{CO}_3\)) is a great example of a weak acid in chemistry. Weak acids differ from strong acids because they do not fully ionize in solution.
This means only a small fraction of carbonic acid molecules in water will release hydrogen ions (\(\mathrm{H}^+\)). The rest remain intact as \(\mathrm{H}_2\mathrm{CO}_3\).
Because the dissociation (ionization) is incomplete, the solution reaches a dynamic equilibrium. This is represented by the double arrow symbol \((\rightleftharpoons)\) in the chemical equation that shows both forward and reverse reactions occurring simultaneously.
  • In the forward reaction, some \(\mathrm{H}_2\mathrm{CO}_3\) molecules split into \(\mathrm{H}^+\) and \(\mathrm{HCO}_3^-\) ions.
  • In the reverse reaction, these ions recombine to form \(\mathrm{H}_2\mathrm{CO}_3\) again.
Understanding weak acid equilibrium is crucial because it controls the pH of solutions where weak acids are present, including natural processes like carbonic acid's behavior in the ocean and bloodstream.
H2CO3 Chemical Equation
The ionization of carbonic acid involves writing its chemical equation. Specifically, the equation illustrates how \(\mathrm{H}_2\mathrm{CO}_3\) ionizes to form \(\mathrm{H}^+\) and \(\mathrm{HCO}_3^-\) ions.
Here's the chemical equation:\[\mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^-\] This equation highlights the weak acid nature of \(\mathrm{H}_2\mathrm{CO}_3\) by showing how it splits into ions in water.
The equation uses the double arrow \((\rightleftharpoons)\) to indicate equilibrium, meaning it's a reversible process. While forward reactions increase \(\mathrm{H}^+\) and \(\mathrm{HCO}_3^-\) concentration, the reverse reaction lets them recombine into \(\mathrm{H}_2\mathrm{CO}_3\), maintaining balance.
This balance means that at equilibrium, the concentration of \(\mathrm{H}_2\mathrm{CO}_3 \) and its ions stays relatively constant. Understanding this balance helps chemists predict how the solution's pH will behave when carbonic acid is dissolved in water.
Solutes in Aqueous Solution
In aqueous solutions, solutes are the substances dissolved in a solvent like water. When carbon dioxide dissolves, it forms carbonic acid \(\mathrm{H}_2\mathrm{CO}_3\) in water, which partially ionizes providing specific solutes.
These solutes are important to understand:
  • Hydrogen ions \((\mathrm{H}^+)\): These ions make the solution acidic, lowering its pH.
  • Bicarbonate ions \((\mathrm{HCO}_3^-)\): They are a common component in buffering systems, stabilizing pH by absorbing excess hydrogen ions.
In equilibrium, both these ions interact dynamically with undissociated carbonic acid molecules, harmonizing the solution.
This dynamic equilibrium is essential for many biological and environmental processes. For instance, in the human body, bicarbonate ions help maintain the pH of blood, while in oceans, they contribute to the carbon balance influencing marine life. Understanding the nature of solutes in aqueous solutions helps in predicting reactions and outcomes in different chemical environments.

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Most popular questions from this chapter

(a) Which will have the highest concentration of sodium ions: \(0.25 \mathrm{MNaCl}, 0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3},\) or \(0.075 \mathrm{MNa}_{3} \mathrm{PO}_{4} ?(\mathbf{b})\) Which will contain the greater number of moles of sodium ion: \(20.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{NaHCO}_{3}\) or \(15.0 \mathrm{~mL}\) of \(0.04 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S} ?\)

Ritalin is the trade name of a drug, methylphenidate, used to treat attention- deficit/hyperactivity disorder in young adults. The chemical structure of methylphenidate is (a) Is Ritalin an acid or a base? An electrolyte or a nonelectrolyte? (b) A tablet contains a \(10.0-\mathrm{mg}\) dose of Ritalin. Assuming all the drug ends up in the bloodstream, and the average man has a total blood volume of \(5.0 \mathrm{~L}\), calculate the initial molarity of Ritalin in a man's bloodstream. (c) Ritalin has a half-life of 3 hours in the blood, which means that after 3 hours the concentration in the blood has decreased by half of its initial value. For the man in part (b), what is the concentration of Ritalin in his blood after 6 hours?

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other platinum. You have three solutions at your disposal: \(1 \mathrm{M}\) sodium nitrate, \(1 \mathrm{M}\) nitric acid, and \(1 \mathrm{M}\) nickel nitrate. How could you use these solutions to determine the identities of each metal powder? [Section 4.4]

A solid sample of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is added to \(0.500 \mathrm{~L}\) of \(0.250 \mathrm{M}\) aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The solution that remains is still acidic. It is then titrated with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution, and it takes \(12.5 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. What mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) was added to the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution?

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{MHCl} .\) The \(\mathrm{NH}_{3}\) reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

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