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Consider the following statements. Write out an example reaction and \(K\) expression that is associated with each statement. a. The autoionization of water. b. An acid reacts with water to produce the conjugate base of the acid and the hydronium ion. c. A base reacts with water to produce the conjugate acid of the base and the hydroxide ion.

Short Answer

Expert verified
a. Autoionization of water: \[2 H_2O \rightleftharpoons H_3O^+ + OH^-\], \(K_w = \frac{[H_3O^+][OH^-]}{[H_2O]^2}\) b. Acid reaction with water: \[HA + H_2O \rightleftharpoons H_3O^+ + A^-\], \(K_a = \frac{[H_3O^+][A^-]}{[HA]}\) c. Base reaction with water: \[B + H_2O \rightleftharpoons HB^+ + OH^-\], \(K_b = \frac{[HB^+][OH^-]}{[B]}\)

Step by step solution

01

a. Autoionization of water

In this case, we consider the autoionization of water, where two water molecules react with each other to produce a hydronium ion (H鈧僌鈦) and a hydroxide ion (OH鈦). The balanced chemical equation for the autoionization of water is: \[2 H_2O \rightleftharpoons H_3O^+ + OH^-\] The equilibrium constant expression (K) for this reaction is given by the product of the concentrations of the products (hydronium and hydroxide ions) divided by the concentration of the reactant (water) raised to the power of its stoichiometric coefficient: \[K_w = \frac{[H_3O^+][OH^-]}{[H_2O]^2}\] Here, \(K_w\) is the ion product of water, which is equal to \(1 \times 10^{-14}\) at \(25 ^\circ C\).
02

b. Acid reaction with water

For this case, consider an acid HA that reacts with water to form its conjugate base A鈦 and hydronium ion H鈧僌鈦. The balanced chemical equation for this reaction is: \[HA + H_2O \rightleftharpoons H_3O^+ + A^-\] The equilibrium constant expression (K) for this reaction is given by the product of the concentrations of the products (hydronium ion and conjugate base) divided by the concentration of the reactant (acid): \[K_a = \frac{[H_3O^+][A^-]}{[HA]}\] In this equation, \(K_a\) represents the acid dissociation constant, which is a measure of the acidity of the compound.
03

c. Base reaction with water

Now, let's consider a base B that reacts with water to form its conjugate acid HB鈦 and a hydroxide ion OH鈦. The balanced chemical equation for this reaction is: \[B + H_2O \rightleftharpoons HB^+ + OH^-\] The equilibrium constant expression (K) for this reaction is given by the product of the concentrations of the products (conjugate acid and hydroxide ion) divided by the concentration of the reactant (base): \[K_b = \frac{[HB^+][OH^-]}{[B]}\] Here, \(K_b\) represents the base dissociation constant, which is a measure of the basicity of the compound.

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Key Concepts

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

Autoionization of Water
The autoionization of water is a fascinating process where water molecules interact to form ions. This reaction is self-sustaining, meaning it involves pure water without the addition of any other substances. In this process, two water molecules (H鈧侽) react with each other to produce a hydronium ion (H鈧僌鈦) and a hydroxide ion (OH鈦).

The chemical equation depicting this autoionization is:
  • \[2 H_2O \rightleftharpoons H_3O^+ + OH^-\]
This reaction is in equilibrium, which means that the rate of the forward reaction (forming ions) equals the rate of the backward reaction (reforming water molecules). This equilibrium is described by an equilibrium constant known as the ion product of water \(K_w\), which at 25掳C is \(1 \times 10^{-14}\).

The equilibrium expression for this is:
  • \[K_w = \frac{[H_3O^+][OH^-]}{[H_2O]^2}\]
However, because water is a pure liquid, its concentration is typically incorporated into the constant, simplifying the expression to:
  • \[K_w = [H_3O^+][OH^-]\]
Acid-Base Reactions
Acid-base reactions are interactions between acids and bases, and they are central to numerous chemical processes in both the lab and nature. An acid (like HA) reacts with water, donating a proton to the water and yielding a hydronium ion (H鈧僌鈦) and a conjugate base (A鈦).

The general chemical equation for this reaction is:
  • \[HA + H_2O \rightleftharpoons H_3O^+ + A^-\]
This reaction reaches equilibrium, and its equilibrium state is described using the acid dissociation constant \(K_a\). This is a measure of how easily an acid donates a proton to water, typically indicating the strength of the acid.

The equilibrium constant expression for this reaction is:
  • \[K_a = \frac{[H_3O^+][A^-]}{[HA]}\]
The value of \(K_a\) helps predict the acidity behavior in solutions. Acids with a higher \(K_a\) are considered strong acids because they dissociate more fully in solution.
Equilibrium Constants
Equilibrium constants are pivotal in understanding chemical reactions, including acid-base interactions. They represent the ratio of the concentrations of products to reactants at equilibrium, providing insights into the position of equilibrium and the nature of the substances involved.

For a base reacting with water, like the base B, the process involves gaining a proton to form its conjugate acid (HB鈦) and producing a hydroxide ion (OH鈦). The balanced equation is:
  • \[B + H_2O \rightleftharpoons HB^+ + OH^-\]
The equilibrium constant for this reaction is the base dissociation constant \(K_b\), which informs us about the base's ability to accept protons.

The equilibrium constant expression for this reaction is:
  • \[K_b = \frac{[HB^+][OH^-]}{[B]}\]
In summary:
  • \(K_a\) and \(K_b\) help us understand and quantify the strength of acids and bases.
  • The higher the value of \(K_b\), the stronger the base.
  • Equilibrium constants allow chemists to predict the behavior of chemical systems in equilibrium.

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Most popular questions from this chapter

A \(0.20-M\) sodium chlorobenzoate \(\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution has a pH of \(8.65 .\) Calculate the \(\mathrm{pH}\) of a \(0.20-M\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.

A typical sample of vinegar has a pH of \(3.0\). Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons{\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)}\) is \(6.0 \times 10^{-3}\). a. Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). b. Will a \(1.0-M\) solution of iron(II) nitrate have a higher or lower \(\mathrm{pH}\) than a \(1.0-M\) solution of iron(III) nitrate? Explain.

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), in a \(0.10-M\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\).

The \(\mathrm{pH}\) of a \(0.063-M\) solution of hypobromous acid \((\mathrm{HOBr}\) but usually written \(\mathrm{HBrO}\) ) is 4.95. Calculate \(K_{\mathrm{a}}\).

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