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What is meant by the self-ionization of water? Write the expression for \(K_{w} .\) What is its value at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Self-ionization of water generates hydronium and hydroxide ions. The expression for \(K_w\) is \([ ext{H}_3 ext{O}^+][ ext{OH}^-]\), and its value at 25°C is \(1.0 \times 10^{-14}\).

Step by step solution

01

Understanding Self-Ionization of Water

Self-ionization of water refers to the process where water molecules dissociate into hydroxide ions ( ext{OH}^- ext{}) and hydronium ions ( ext{H}_3 ext{O}^+ ext{}). This occurs when two water molecules interact to exchange a proton. The chemical equation for the self-ionization is:\[2 ext{H}_2 ext{O}(l) ightleftharpoons ext{H}_3 ext{O}^+(aq) + ext{OH}^-(aq)\]
02

Writing the Expression for Ion Product of Water, Kw

The equilibrium constant for the self-ionization of water is known as the ion product of water, denoted by \(K_w\). At equilibrium, \(K_w\) is defined by the concentrations of the products, the hydronium ions and the hydroxide ions:\[K_w = [ ext{H}_3 ext{O}^+][ ext{OH}^-]\]In pure water at 25°C, the concentrations of both ions are equal.
03

Value of Kw at 25°C

At a temperature of 25°C, the ion product of water \(K_w\) is a well-established constant with a value of:\[K_w = 1.0 imes 10^{-14}\]This means that at 25°C, the concentration of hydronium ions multiplied by the concentration of hydroxide ions will always equal \(1.0 imes 10^{-14} ext{ M}^2\).

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Key Concepts

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

Ion Product of Water
The ion product of water, often represented as \(K_w\), plays a crucial role in understanding the behavior of water in its self-ionization process. Self-ionization is a rare phenomenon where water molecules undergo a reaction to form equal amounts of hydronium ions (\(\text{H}_3\text{O}^+\)) and hydroxide ions (\(\text{OH}^-\)). This chemical reaction can be represented as:
\[2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)\]
In this equilibrium reaction, two water molecules interact to exchange a proton, resulting in the production of an acid (hydronium ion) and a base (hydroxide ion). The ion product \(K_w\) is expressed mathematically to depict the equilibrium condition:
  • \(K_w = [\text{H}_3\text{O}^+][\text{OH}^-]\)
Here, \(\text{[H}_3\text{O}^+]\) and \(\text{[OH}^-]\) represent the concentrations of the hydronium and hydroxide ions respectively. The concept is vital as it provides insights into the neutrality, acidity or basicity of water depending on how these ion concentrations compare at various conditions.
Equilibrium Constant
An equilibrium constant is a numerical value that describes the ratio of product concentrations to reactant concentrations at equilibrium. In the context of the self-ionization of water, the equilibrium constant \(K_w\) specifically refers to this established balance between the formation and dissociation of hydronium and hydroxide ions in water.

This constant is essential in predicting how chemical reactions progress and maintain balance. For water, defining \(K_w\) as:
  • \(K_w = [\text{H}_3\text{O}^+][\text{OH}^-]\)
allows for an understanding of water's purity and behavior.

The value of the equilibrium constant remains unchanged under set conditions; however, it can vary with changes in temperature or pressure. This constancy at a given condition ensures that chemists can predict the reaction's behavior without performing experiments in every scenario. Understanding \(K_w\) as an equilibrium constant helps shed light on many key reaction dynamics.
Kw at 25°C
The self-ionization of water is markedly influenced by temperature, with 25°C being a standard reference point. At this temperature, the equilibrium constant, or the ion product of water \(K_w\), is fixed at \(1.0 \times 10^{-14}\).

This particular temperature value is significant because it represents the conditions at which water is typically in a near-standard state, where:
  • The concentration of \(\text{H}_3\text{O}^+\) is approximately \(1.0 \times 10^{-7}\, \text{M}\)
  • The concentration of \(\text{OH}^-\) is similarly \(1.0 \times 10^{-7}\, \text{M}\)
Thus, when the concentrations of the two ions are multiplied, they create the constant \(K_w = 1.0 \times 10^{-14} \text{ M}^2\).

This value highlights a perfect neutrality in pure water. Any deviation from this constant would mean either an acidic or basic alteration of water, with higher \([\text{H}_3\text{O}^+]\) indicating acidity and higher \([\text{OH}^-]\) indicating basicity. Understanding \(K_w\) at 25°C is fundamental for calculations involving pH and pOH, thus maintaining its centrality in chemistry and aqueous solutions.

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Most popular questions from this chapter

Aluminum chloride, \(\mathrm{AlCl}_{3}\), reacts with trimethylamine, \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3} .\) What would you guess to be the product of this reaction? Explain why you think so. Describe the reaction in terms of one of the acid-base concepts. Write an appropriate equation to go with this description. Which substance is the acid according to this acid-base concept? Explain.

What is the \(\mathrm{pH}\) of a neutral solution at \(37^{\circ} \mathrm{C}\), where \(K_{w}\) equals \(2.5 \times 10^{-14}\) ?

You make a solution by dissolving \(0.0010\) mol of \(\mathrm{HCl}\) in enough water to make \(1.0 \mathrm{~L}\) of solution. a. Write the chemical equation for the reaction of \(\mathrm{HCl}(a q)\) and water. b. Without performing calculations, give a rough estimate of the \(\mathrm{pH}\) of the \(\mathrm{HCl}\) solution. Justify your answer. c. Calculate the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration and the \(\mathrm{pH}\) of the solution. d. Is there any concentration of the base \(\mathrm{OH}^{-}\) present in this solution of \(\mathrm{HCl}(a q) ?\) If so, where did it come from? e. If you increase the OH \(^{-}\) concentration of the solution by adding \(\mathrm{NaOH}\), does the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration change? If you think it does, explain why this change occurs and whether the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration increases or decreases. f. If you were to measure the \(\mathrm{pH}\) of 10 drops of the original HCl solution, would you expect it to be different from the pH of the entire sample? Explain. g. Explain how two different volumes of your original \(\mathrm{HCl}\) solution can have the same \(\mathrm{pH}\) yet contain different moles of \(\mathrm{H}_{3} \mathrm{O}^{+}\) h. If \(1.0 \mathrm{~L}\) of pure water were added to the \(\mathrm{HCl}\) solution, would this have any impact on the \(\mathrm{pH}\) ? Explain.

Give the conjugate acid to each of the following species regarded as bases. a. HSe b. \(\mathrm{NH}_{2}^{-}\) c. \(\mathrm{ClO}_{2}^{-}\) d. \(\mathrm{N}_{2} \mathrm{H}_{4}\)

The following are solution concentrations. Indicate whether each solution is acidic, basic, or neutral. a. \(2 \times 10^{-11} \mathrm{M} \mathrm{OH}^{-}\) b. \(2 \times 10^{-9} \mathrm{M} \mathrm{H}_{3} \mathrm{O}^{+}\) c. \(6 \times 10^{-5} \mathrm{M} \mathrm{OH}^{-}\) d. \(6 \times 10^{-3} \mathrm{M} \mathrm{H}_{3} \mathrm{O}^{+}\)

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