A particulate model of pure water at 25°C shows that for every $$5.5 \times 10^8$$ water molecules, one hydronium ion and one hydroxide ion are present. This corresponds to concentrations of $$[H_3O^+] = [OH^-] = 1.0 \times 10^{-7} M$$. Which values for pH and pOH are consistent with the quantitative information in this model?

$$pH = 1.0$$ and $$pOH = 13.0$$

$$pH = 14.0$$ and $$pOH = 0.0$$

Two particulate representations of pure water are compared: Model X at 25°C and Model Y at 50°C. Model Y shows a higher density of both hydronium and hydroxide ions than Model X. Which conclusion about Model Y is supported by this observation?

The pH is lower than 7.0, and the solution is neutral

The pH is lower than 7.0, and the solution is acidic

The pH is higher than 7.0, and the solution is basic

The pH is higher than 7.0, and the solution is neutral

A particulate model of pure water at 50°C shows a higher density of $$H_3O^+$$ and $$OH^-$$ ions compared to a model at 25°C. Which claim about the pH and neutrality of the water at 50°C is supported by this evidence?

pH < 7.0 and the solution is neutral due to equal ion concentrations

pH < 7.0 and the solution is acidic due to higher hydronium levels

pH = 7.0 and the solution is neutral due to the constant ion product

pH > 7.0 and the solution is basic due to increased hydroxide levels

The autoionization reaction $$2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$$ produces ions in a 1:1 stoichiometric ratio. Which claim about pure water at any temperature is best supported by this particulate-level stoichiometry?

pH equals pOH because the concentrations of hydronium and hydroxide are identical

pH equals 7.0 because the concentrations of hydronium and hydroxide are constant

pH is less than pOH because hydronium ions are smaller than hydroxide ions

pH plus pOH equals 14.0 because the ion product constant is temperature stable

AP Chemistry Unit 8 Practice Test

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The exam experience, sized to one unit — timed like the real thing

20 Multiple Choice Questions

30 min

1 FRQ – FRQ 1 – Long Answer

15 min

1 FRQ – FRQ 4 – Short Answer

15 min

Sections run back-to-back with the same directions, timers, and scoring as the real exam

Total: 1 hr

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multiple choice questions

free response questions (choose as many as you want)

FRQ 1 – Long Answer · 15 minFRQ 4 – Short Answer · 15 minFRQ 2 – Long Answer · 15 minFRQ 3 – Long Answer · 15 minFRQ 5 – Short Answer · 15 minFRQ 6 – Short Answer · 15 minFRQ 7 – Short Answer · 15 min

Total: 30 min

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AP Chemistry Unit 8 practice questions

Try 10 multiple-choice questions from Unit 8 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

Question 1
A graphical representation of species concentration in pure water at 25°C shows two horizontal lines for $[H_3O^+]$ and $[OH^-]$ intersecting at $1.0 \times 10^{-7}$ M. If the temperature is increased and the intersection point shifts to $2.0 \times 10^{-7}$ M, what is the new pH?
6.7
7.0
7.3
13.4
Show answer
Answer: (A)
Calculating the new pH requires taking the negative logarithm of the new hydronium concentration identified from the intersection. $pH = -\log(2.0 \times 10^{-7}) \approx 6.7$ .
Question 2
For aqueous solutions at 25°C, a plot of $-\log[OH^-]$ (on the y-axis) versus $-\log[H_3O^+]$ (on the x-axis) is a straight line with a slope of -1 and a y-intercept of 14. Using this linear relationship, determine the pOH of a solution where the pH is 4.5.
9.5
4.5
5.5
18.5
Show answer
Answer: (A)
Using the linear relationship defined by the graph ( $y = -x + 14$ ), where y is pOH and x is pH, allows for direct calculation. $pOH = -4.5 + 14 = 9.5$ .
Question 3
The ion-product constant of water, $K_w$ , is $1.0 \times 10^{-15}$ at 0°C, $1.0 \times 10^{-14}$ at 25°C, and $9.6 \times 10^{-14}$ at 60°C. Based on the value at 0°C, what is the pOH of a neutral water sample at freezing?
7.5
6.5
7.0
8.0
Show answer
Answer: (A)
Calculating pOH requires finding the hydroxide concentration from the $K_w$ at 0°C. $[OH^-] = \sqrt{1.0 \times 10^{-15}} = 1.0 \times 10^{-7.5}$ . Therefore, $pOH = 7.5$ .
Question 4
In pure water at 40°C, the pH and pOH are both 6.77. What is the value of $K_w$ at this temperature?
$2.9 \times 10^{-14}$
$1.0 \times 10^{-14}$
$1.7 \times 10^{-7}$
$5.8 \times 10^{-14}$
Show answer
Answer: (A)
Calculating $K_w$ requires using the identified pH/pOH value. Since $pH = pOH = 6.77$ , $[H_3O^+] = [OH^-] = 10^{-6.77}$ . $K_w = (10^{-6.77})^2 = 10^{-13.54} \approx 2.9 \times 10^{-14}$ .
Question 5
A particulate model of pure water at 25°C shows that for every $5.5 \times 10^8$ water molecules, one hydronium ion and one hydroxide ion are present. This corresponds to concentrations of $[H_3O^+] = [OH^-] = 1.0 \times 10^{-7} M$ . Which values for pH and pOH are consistent with the quantitative information in this model?
$pH = 7.0$ and $pOH = 7.0$
$pH = 7.0$ and $pOH = 1.0$
$pH = 1.0$ and $pOH = 13.0$
$pH = 14.0$ and $pOH = 0.0$
Show answer
Answer: (A)
Calculating pH and pOH requires applying the negative logarithm function to the ion concentrations provided in the model. Since both concentrations are $1.0 \times 10^{-7} M$ , the calculation $-\log(1.0 \times 10^{-7})$ yields 7.0 for both values.
Question 6
A researcher develops a particulate model for water at 0°C where the extent of ionization is lower than at room temperature. If the model corresponds to a $pK_w$ of 15.0, what is the pOH of the neutral water represented?
$pOH = 7.5$
$pOH = 7.0$
$pOH = 8.0$
$pOH = 1.0$
Show answer
Answer: (A)
Calculating the pOH involves using the relationship $pK_w = pH + pOH$ for a neutral solution where pH and pOH are equal. Dividing the given $pK_w$ of 15.0 by 2 yields a pOH of 7.5.
Question 7
Two particulate representations of pure water are compared: Model X at 25°C and Model Y at 50°C. Model Y shows a higher density of both hydronium and hydroxide ions than Model X. Which conclusion about Model Y is supported by this observation?
The pH is lower than 7.0, and the solution is neutral
The pH is lower than 7.0, and the solution is acidic
The pH is higher than 7.0, and the solution is basic
The pH is higher than 7.0, and the solution is neutral
Show answer
Answer: (A)
Interpreting the model shows that higher ion concentrations lead to a lower pH ( $pH = -\log[H_3O^+]$ ). However, because the hydronium and hydroxide ions increase by the same amount, their concentrations remain equal, keeping the solution neutral.
Question 8
A particulate model depicts the autoionization reaction $2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$ . If the model shows that $1.0 \times 10^{-4}$ moles of $H_3O^+$ are present in a 1.0 L volume of pure water at a specific temperature, what is the pOH?
4.0
7.0
10.0
14.0
Show answer
Answer: (A)
Analyzing the stoichiometry of the autoionization reaction shown in the model reveals that hydronium and hydroxide ions are produced in a 1:1 ratio. Therefore, $[OH^-] = 1.0 \times 10^{-4} M$ , and $pOH = -\log(10^{-4}) = 4.0$ .
Question 9
A particulate model of pure water at 50°C shows a higher density of $H_3O^+$ and $OH^-$ ions compared to a model at 25°C. Which claim about the pH and neutrality of the water at 50°C is supported by this evidence?
pH < 7.0 and the solution is neutral due to equal ion concentrations
pH < 7.0 and the solution is acidic due to higher hydronium levels
pH = 7.0 and the solution is neutral due to the constant ion product
pH > 7.0 and the solution is basic due to increased hydroxide levels
Show answer
Answer: (A)
Connecting the increased ion concentration to pH calculation reveals that while pH decreases (due to higher $[H_3O^+]$ ), the solution remains neutral because the stoichiometry of autoionization produces equal amounts of hydronium and hydroxide.
Question 10
The autoionization reaction $2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$ produces ions in a 1:1 stoichiometric ratio. Which claim about pure water at any temperature is best supported by this particulate-level stoichiometry?
pH equals pOH because the concentrations of hydronium and hydroxide are identical
pH equals 7.0 because the concentrations of hydronium and hydroxide are constant
pH is less than pOH because hydronium ions are smaller than hydroxide ions
pH plus pOH equals 14.0 because the ion product constant is temperature stable
Show answer
Answer: (A)
Connecting the reaction stoichiometry to the mathematical definitions of pH and pOH confirms that if $[H_3O^+] = [OH^-]$ , their negative logarithms must also be equal.

What's on the AP Chemistry Unit 8 practice test

The standard version takes about 60 minutes and follows the timing and directions of the AP exam, sized to one unit:

20 Multiple Choice Questions (30 minutes)
1 FRQ – FRQ 1 – Long Answer (15 minutes)
1 FRQ – FRQ 4 – Short Answer (15 minutes)

Questions are drawn from 3639 multiple-choice questions and 72 FRQs written for Unit 8, Acids & Bases.

Practice AP Chemistry Unit 8 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

FRQ 1 – Long Answer (4 questions)
FRQ 2 – Long Answer (3 questions)
FRQ 3 – Long Answer (5 questions)
FRQ 4 – Short Answer (15 questions)
FRQ 5 – Short Answer (19 questions)
FRQ 6 – Short Answer (9 questions)
FRQ 7 – Short Answer (17 questions)

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