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Chapter 1: Problem 34

A bottle of water is left outside early in the morning. The bottle warms gradually over the course of the day. What will happen to the pH of the water as the bottle warms? (A) Nothing; pure water always has a pH of 7.00. (B) Nothing; the volume would have to change in order for any ion concentration to change. (C) It will increase because the concentration of \(\left[\mathrm{H}^{+}\right]\) is increasing. (D) It will decrease because the auto-ionization of water is an endothermic process.

Short Answer

Expert verified
The pH of the water in the bottle will decrease as it warms because the auto-ionization of water is an endothermic process. Hence, the correct answer is (D)

Step by step solution

01

Understand the Concept of pH

pH is a measure of hydrogen ion concentration, \(\left[\mathrm{H}^{+}\right]\), a measure of the acidity or alkalinity of a solution. Water by nature undergoes auto ionization or self-ionization where one water molecule donates a proton (H鈦�) to another one, becoming \(\mathrm{OH}^-\), and the other molecule becomes \(\mathrm{H}_3\mathrm{O}^+\), creating equal amount of these ions.
02

Identify the Impact of Temperature on pH

Increase in temperature can cause an increase in \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^-\right]\) due to the increase in auto-ionization of water. Since pH is the negative logarithm of hydrogen ion concentration, the pH of pure water decreases as the temperature increases because the increase in \(\left[\mathrm{H}^{+}\right]\) concentration results in an increased acidity and therefore lower pH.
03

Identify the Correct Answer

Looking at the options, the one reflecting our understanding is (D) It will decrease because the auto-ionization of water is an endothermic process. Endothermic because the reaction absorbs heat energy during the process.

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Key Concepts

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

Auto-Ionization of Water
Water is an extraordinary molecule that can ionize on its own, a process known as auto-ionization. This event is essential for understanding how water remains neutral yet dynamic in its composition. During auto-ionization, one water molecule acts as a proton donor and gives a proton (H extsuperscript{+}) to another water molecule, transforming into hydroxide ion (OH extsuperscript{-}). In turn, the receiving water molecule becomes a hydronium ion (H extsubscript{3}O extsuperscript{+}).

This reaction can be represented by the equation:\[ 2 ext{H}_2 ext{O} ightleftharpoons ext{H}_3 ext{O}^+ + ext{OH}^- \]
The auto-ionization of water is crucial because it establishes the basis for water's pH, meaning the neutral point of the pH scale is set by the concentration of these ions. Even though the concentration of hydronium and hydroxide ions in pure water is incredibly low, it is enough to affect changes in the water's acidity or basicity when external factors like temperature come into play.
Hydrogen Ion Concentration
The concentration of hydrogen ions ([H extsuperscript{+}]) in a solution is a direct influencer of the solution's pH level. The concept of pH is derived from these concentrations, calculated as the negative logarithm:\[ ext{pH} = - ext{log} [ ext{H}^+]\]Because of this relationship, even small changes in the concentration of hydrogen ions can lead to significant shifts in pH. In pure water, due to the equal creation of H extsuperscript{+} and OH extsuperscript{-}, the pH remains around 7 at room temperature.

Given the roughly equal balance of hydrogen and hydroxide ions, water is considered neutral. However, changes in circumstances, such as temperature fluctuations, can alter the rate of water's auto-ionization. This alteration directly impacts the concentrations of H extsuperscript{+} and OH extsuperscript{-}, thus affecting the pH.
Endothermic Process
Some chemical reactions absorb heat from their surroundings, designated as endothermic processes. The auto-ionization of water is one such reaction that requires energy input for the dissociation of water molecules into ions.

When temperature increases, more heat is available to drive the reaction forward, leading to an increased concentration of H extsuperscript{+} and OH extsuperscript{-} ions. This causes the pH to fall, indicating an increase in acidity. As water warms, the endothermic nature of the auto-ionization process means that it will engage more readily, thus lowering pH and fostering greater acidity.

This understanding highlights why, despite being an endothermic reaction, the auto-ionization plays a significant part in natural water systems. Increased temperatures accelerate the dissociation rates, making it more noticeable in scenarios like the warming of a bottle of water left outside during the day.

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Most popular questions from this chapter

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. What is the approximate amount of heat released during the reaction? \(\begin{array}{ll}{\text { (A) }} & {1.5 \mathrm{kJ}} \\ {\text { (B) }} & {2.9 \mathrm{kJ}} \\ {\text { (C) }} & {5.9 \mathrm{kJ}} \\ {\text { (D) }} & {11.8 \mathrm{kJ}}\end{array}\)

Which compound, \(\mathrm{CaCl}_{2}\) or \(\mathrm{CaO}\) , would you expect to have a high melting point? Why? (A) \(\mathrm{CaCl}_{2}\) because there are more ions per lattice unit (B) \(\mathrm{CaCl}_{2}\) because a chlorine ion is smaller than an oxygen ion (C) Cao, because the charge of oxygen ion exceeds that of chlorine ion (D) CaO, because the common charges of calcium and oxygen ions are identical in magnitude

The following reaction is found to be at equilibrium at 25掳C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) What is the expression for the equilibrium constant, \(K_{\mathrm{c}} ?\) (A) \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}\) (B) \(\frac{2\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}\) (C) \(\frac{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{SO}_{3}\right]^{2}}\) (D) \(\frac{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}{2\left[\mathrm{SO}_{3}\right]}\)

A 0.1-molar solution of which of the following acids will be the best conductor of electricity? (A) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) (B) \(\mathrm{H}_{2} \mathrm{S}\) (C) \(\mathrm{HF}\) (D) \(\mathrm{HNO}_{3}\)

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$
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