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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

A 2.0 L flask holds 0.40 g of helium gas. If the helium is evacuated into a larger container while the temperature is held constant, what will the effect on the entropy of the helium be? (A) It will remain constant because the number of helium molecules does not change. (B) It will decrease because the gas will be more ordered in the larger flask. (C) It will decrease because the molecules will collide with the sides of the larger flask less often than they did in the smaller flask. (D) It will increase because the gas molecules will be more dispersed in the larger flask.

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

Which gas has the strongest IMFs? (A) He (B) Ne (C) NO (D) All gases have identical IMFs.

Directions: Questions 1-3 are long free-response questions that require about 23 minutes each to answer and are worth 10 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. The unbalanced reaction between potassium permanganate and acidified iron (II) sulfate is a redox reaction that proceeds as follows: $$\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-(a q)} \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$ (a) Provide the equations for both half-reactions that occur below: (i) Oxidation half-reaction (ii) Reduction half-reaction (b) What is the balanced net ionic equation? A solution of 0.150 M potassium permanganate is placed in a buret before being titrated into a flask containing 50.00 mL of iron (II) sulfate solution of unknown concentration. The following data describes the colors of the various ions in solution: $$\begin{array}{|c|c|}\hline \text { Ion } & {\text { Color in solution }} \\\ \hline \mathrm{H^{+ }} & {\text { Colorless }} \\ \hline \mathrm{Fe}^{2+} & {\text { Pale Green }} \\ \hline \mathrm{MnO}_{4^{-}} & {\text {Dark Purple }} \\ \hline \mathrm{Mn}^{2+} & {\text { Colorless }} \\ \hline \mathrm{Fe}^{3+} & {\text { Yellow }} \\ \hline \mathrm{K}^{+} & {\text {Colorless }} \\ \hline \mathrm{SO}_{4}^{2-} & {\text { Colorless }} \\\ \hline\end{array}$$ (c) Describe the color of the solution in the flask at the following points: (i) Before titration begins (ii) During titration prior to the endpoint (iii) At the endpoint of the titration (d) (i) If 15.55 mL of permanganate are added to reach the endpoint, what is the initial concentration of the iron (II) sulfate? (ii) The actual concentration of the \(\mathrm{FeSO}_{4}\), is 0.250 \(M\) . Calculate the percent error. (e) Could the following errors have led to the experimental result deviating in the direction that it did? You must justify your answers quantitatively. (i) 55.0 \(\mathrm{mL}\) of \(\mathrm{FeSO}_{4}\) was added to the flask prior to titration instead of 50.0 mL . (ii) The concentration of the potassium permanganate was actually 0.160 \(M\) instead of 0.150 \(M\) .

Which of the following pairs of elements is most likely to create an interstitial alloy? (A) Titanium and copper (B) Aluminum and lead (C) Silver and tin (D) Magnesium and calcium
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