AP Chemistry worksheet on acid-base equilibria, including a pH scale diagram and practice problems on calculating pH, pOH, and ion concentrations.

Here are the solutions to the problems on the worksheet, worked out step-by-step.

1. Write the pH of each solution above the [H⁺]s.
*Formula:* $pH = -\log[H^+]$

* Box 1: $[H^+] = 8.8 \times 10^{-4}$
* Calculation: $-\log(8.8 \times 10^{-4}) \approx 3.056$
* Answer: 3.06 (rounded to 2 decimal places)
* Box 2: $[H^+] = 2 \times 10^{-4}$
* Calculation: $-\log(2 \times 10^{-4}) \approx 3.699$
* Answer: 3.70
* Box 3: $[H^+] = 1 \times 10^{-7}$
* Calculation: $-\log(1 \times 10^{-7}) = 7$
* Answer: 7.00
* Box 4: $[H^+] = 5 \times 10^{-9}$
* Calculation: $-\log(5 \times 10^{-9}) \approx 8.301$
* Answer: 8.30
* Box 5: $[H^+] = 3.6 \times 10^{-11}$
* Calculation: $-\log(3.6 \times 10^{-11}) \approx 10.444$
* Answer: 10.44

2. Label the "Z" diagram as "Acidic," "Basic," and "Neutral."
* Top Left (Low pH): Acidic
* Middle (pH 7): Neutral
* Bottom Right (High pH): Basic

3. Fill in the [OH⁻] for each of the five solutions.
*Rule:* $[H^+] \times [OH^-] = 1 \times 10^{-14}$. Therefore, $[OH^-] = \frac{1 \times 10^{-14}}{[H^+]}$.

* Box 1: $\frac{1 \times 10^{-14}}{8.8 \times 10^{-4}} \approx 1.1 \times 10^{-11}$
* Box 2: $\frac{1 \times 10^{-14}}{2 \times 10^{-4}} = 5 \times 10^{-11}$
* Box 3: $\frac{1 \times 10^{-14}}{1 \times 10^{-7}} = 1 \times 10^{-7}$
* Box 4: $\frac{1 \times 10^{-14}}{5 \times 10^{-9}} = 2 \times 10^{-6}$
* Box 5: $\frac{1 \times 10^{-14}}{3.6 \times 10^{-11}} \approx 2.8 \times 10^{-4}$

4. Write the pOH of each solution below the [OH⁻]s.
*Formula:* $pOH = -\log[OH^-]$

* Box 1: $-\log(1.1 \times 10^{-11}) \approx 10.94$
* Box 2: $-\log(5 \times 10^{-11}) \approx 10.30$
* Box 3: $-\log(1 \times 10^{-7}) = 7.00$
* Box 4: $-\log(2 \times 10^{-6}) \approx 5.70$
* Box 5: $-\log(2.8 \times 10^{-4}) \approx 3.56$

5. pH + pOH always equals...
* Answer: 14 (at 25°C)

6. A solution of acid has $[H^+] = 3.0 \times 10^{-3}$ M.
* a. Calculate the [OH⁻]:
* $\frac{1 \times 10^{-14}}{3.0 \times 10^{-3}} = 3.33... \times 10^{-12}$
* Answer: $3.3 \times 10^{-12}$ M
* b. Calculate the pH:
* $-\log(3.0 \times 10^{-3}) \approx 2.522$
* Answer: 2.52
* c. Calculate the pOH:
* $14 - 2.52 = 11.48$ (or $-\log(3.3 \times 10^{-12}) \approx 11.48$)
* Answer: 11.48

7. A solution of base has an $[OH^-] = 4.25 \times 10^{-5}$ M.
* a. Calculate the [H⁺]:
* $\frac{1 \times 10^{-14}}{4.25 \times 10^{-5}} \approx 2.35 \times 10^{-10}$
* Answer: $2.35 \times 10^{-10}$ M
* b. Calculate the pH:
* First find pOH: $-\log(4.25 \times 10^{-5}) \approx 4.37$
* Then pH: $14 - 4.37 = 9.63$
* Answer: 9.63
* c. Calculate the pOH:
* Answer: 4.37

8. Calculate the pH's of the following solutions:
*(Note: HCl is a strong acid, so $[H^+]$ equals the concentration of the acid).*

* $2.53 \times 10^{-2}$ M HCl:
* $pH = -\log(2.53 \times 10^{-2}) \approx 1.597$
* Answer: 1.60
* $2.53 \times 10^{-3}$ M HCl:
* $pH = -\log(2.53 \times 10^{-3}) \approx 2.597$
* Answer: 2.60
* $2.53 \times 10^{-4}$ M HCl:
* $pH = -\log(2.53 \times 10^{-4}) \approx 3.597$
* Answer: 3.60

Final Sentence Completion:
A pH with 3 significant figures is written with 2 numbers after the decimal place.
*(Reasoning: In logarithms like pH, the number of significant figures in the original concentration determines the number of decimal places in the answer. Since $2.53$ has 3 sig figs, the pH must have 3 total digits, which means 1 before the decimal and 2 after).*

──────────────────────────────────────

Final Answer:
1. pH Values (Left to Right): 3.06, 3.70, 7.00, 8.30, 10.44
2. Labels: Top-Left: Acidic; Middle: Neutral; Bottom-Right: Basic
3. [OH⁻] Values (Left to Right): $1.1 \times 10^{-11}$, $5 \times 10^{-11}$, $1 \times 10^{-7}$, $2 \times 10^{-6}$, $2.8 \times 10^{-4}$
4. pOH Values (Left to Right): 10.94, 10.30, 7.00, 5.70, 3.56
5. 14
6a. $3.3 \times 10^{-12}$ M
6b. 2.52
6c. 11.48
7a. $2.35 \times 10^{-10}$ M
7b. 9.63
7c. 4.37
8. 1.60, 2.60, 3.60
Sentence: 2