Practice table for students to calculate pH, pOH, and ion concentrations to determine if a solution is acidic or basic.

Explanation:
We are given a table with missing values for pH, [H₃O⁺], pOH, [OH⁻], and whether the solution is acidic or basic. We need to fill in all blanks using key relationships:

1. pH + pOH = 14 (at 25°C — standard assumption unless stated otherwise)
2. [H₃O⁺] = 10^(−pH)
3. [OH⁻] = 10^(−pOH)
4. Also, [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ (the ion product of water, Kw)
5. If pH < 7 → acidic; pH > 7 → basic; pH = 7 → neutral.

Let’s go row by row. I’ll compute each missing value carefully.

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Row 1: pH = 3.78
- [H₃O⁺] = 10^(−3.78) ≈ 1.66 × 10⁻⁴ M
- pOH = 14 − 3.78 = 10.22
- [OH⁻] = 10^(−10.22) ≈ 6.03 × 10⁻¹¹ M
- pH < 7 → Acid

Row 2: [H₃O⁺] = 3.89 × 10⁻⁴ M
- pH = −log(3.89 × 10⁻⁴) = 3.41 (since log 3.89 ≈ 0.59, so −(−4 + 0.59) = 3.41)
- pOH = 14 − 3.41 = 10.59
- [OH⁻] = 10^(−10.59) ≈ 2.57 × 10⁻¹¹ M
- Acidic

Row 3: pOH = 5.19
- pH = 14 − 5.19 = 8.81
- [H₃O⁺] = 10^(−8.81) ≈ 1.55 × 10⁻⁹ M
- [OH⁻] = 10^(−5.19) ≈ 6.46 × 10⁻⁶ M
- pH > 7 → Base

Row 4: [OH⁻] = 4.88 × 10⁻⁵ M
- pOH = −log(4.88 × 10⁻⁵) = 4.31 (log 4.88 ≈ 0.688, so −(−5 + 0.688) = 4.312)
- pH = 14 − 4.31 = 9.69
- [H₃O⁺] = 10^(−9.69) ≈ 2.04 × 10⁻¹⁰ M
- Basic

Row 5: pH = 8.46
- [H₃O⁺] = 10^(−8.46) ≈ 3.47 × 10⁻⁹ M
- pOH = 14 − 8.46 = 5.54
- [OH⁻] = 10^(−5.54) ≈ 2.88 × 10⁻⁶ M
- Basic

Row 6: [H₃O⁺] = 8.45 × 10⁻¹³ M
- pH = −log(8.45 × 10⁻¹³) = 12.07 (log 8.45 ≈ 0.927, so −(−13 + 0.927) = 12.073)
- pOH = 14 − 12.07 = 1.93
- [OH⁻] = 10^(−1.93) ≈ 1.17 × 10⁻² M
- Basic

Row 7: pOH = 2.14
- pH = 14 − 2.14 = 11.86
- [H₃O⁺] = 10^(−11.86) ≈ 1.38 × 10⁻¹² M
- [OH⁻] = 10^(−2.14) ≈ 7.24 × 10⁻³ M
- Basic

Row 8: [OH⁻] = 2.31 × 10⁻¹¹ M
- pOH = −log(2.31 × 10⁻¹¹) = 10.64 (log 2.31 ≈ 0.364, so −(−11 + 0.364) = 10.636)
- pH = 14 − 10.64 = 3.36
- [H₃O⁺] = 10^(−3.36) ≈ 4.37 × 10⁻⁴ M
- Acidic

Row 9: pH = 10.91
- [H₃O⁺] = 10^(−10.91) ≈ 1.23 × 10⁻¹¹ M
- pOH = 14 − 10.91 = 3.09
- [OH⁻] = 10^(−3.09) ≈ 8.13 × 10⁻⁴ M
- Basic

Row 10: [H₃O⁺] = 7.49 × 10⁻⁶ M
- pH = −log(7.49 × 10⁻⁶) = 5.125 ≈ 5.13
- pOH = 14 − 5.13 = 8.87
- [OH⁻] = 10^(−8.87) ≈ 1.35 × 10⁻⁹ M
- Acidic

Row 11: pOH = 9.94
- pH = 14 − 9.94 = 4.06
- [H₃O⁺] = 10^(−4.06) ≈ 8.71 × 10⁻⁵ M
- [OH⁻] = 10^(−9.94) ≈ 1.15 × 10⁻¹⁰ M
- Acidic

Row 12: [OH⁻] = 2.57 × 10⁻⁸ M
- pOH = −log(2.57 × 10⁻⁸) = 7.59 (log 2.57 ≈ 0.410, so −(−8 + 0.410) = 7.59)
- pH = 14 − 7.59 = 6.41
- [H₃O⁺] = 10^(−6.41) ≈ 3.89 × 10⁻⁷ M
- Acidic (pH < 7)

Row 13: pH = 4.16
- [H₃O⁺] = 10^(−4.16) ≈ 6.92 × 10⁻⁵ M
- pOH = 14 − 4.16 = 9.84
- [OH⁻] = 10^(−9.84) ≈ 1.45 × 10⁻¹⁰ M
- Acidic

Row 14: [H₃O⁺] = 1.06 × 10⁻¹ M = 0.106 M
- pH = −log(0.106) = 0.975 ≈ 0.98
- pOH = 14 − 0.98 = 13.02
- [OH⁻] = 10^(−13.02) ≈ 9.55 × 10⁻¹⁴ M
- Strongly acidic

Row 15: pOH = 3.82
- pH = 14 − 3.82 = 10.18
- [H₃O⁺] = 10^(−10.18) ≈ 6.61 × 10⁻¹¹ M
- [OH⁻] = 10^(−3.82) ≈ 1.51 × 10⁻⁴ M
- Basic

Row 16: [OH⁻] = 8.53 × 10⁻⁷ M
- pOH = −log(8.53 × 10⁻⁷) = 6.07 (log 8.53 ≈ 0.931, so −(−7 + 0.931) = 6.069)
- pH = 14 − 6.07 = 7.93
- [H₃O⁺] = 10^(−7.93) ≈ 1.17 × 10⁻⁸ M
- Basic (just above 7)

Row 17: pH = 7.05
- [H₃O⁺] = 10^(−7.05) ≈ 8.91 × 10⁻⁸ M
- pOH = 14 − 7.05 = 6.95
- [OH⁻] = 10^(−6.95) ≈ 1.12 × 10⁻⁷ M
- Slightly basic

Row 18: [H₃O⁺] = 4.73 × 10⁻¹⁰ M
- pH = −log(4.73 × 10⁻¹⁰) = 9.32 (log 4.73 ≈ 0.675, so −(−10 + 0.675) = 9.325)
- pOH = 14 − 9.32 = 4.68
- [OH⁻] = 10^(−4.68) ≈ 2.10 × 10⁻⁵ M
- Basic

Row 19: pOH = 1.33
- pH = 14 − 1.33 = 12.67
- [H₃O⁺] = 10^(−12.67) ≈ 2.14 × 10⁻¹³ M
- [OH⁻] = 10^(−1.33) ≈ 4.68 × 10⁻² M
- Strongly basic

Row 20: [OH⁻] = 9.87 × 10⁻³ M
- pOH = −log(9.87 × 10⁻³) = 2.006 ≈ 2.01
- pH = 14 − 2.01 = 11.99
- [H₃O⁺] = 10^(−11.99) ≈ 1.02 × 10⁻¹² M
- Basic

Row 21: pH = 11.68
- [H₃O⁺] = 10^(−11.68) ≈ 2.10 × 10⁻¹² M
- pOH = 14 − 11.68 = 2.32
- [OH⁻] = 10^(−2.32) ≈ 4.79 × 10⁻³ M
- Basic

Row 22: [H₃O⁺] = 9.22 × 10⁻⁸ M
- pH = −log(9.22 × 10⁻⁸) = 7.035 ≈ 7.04
- pOH = 14 − 7.04 = 6.96
- [OH⁻] = 10^(−6.96) ≈ 1.10 × 10⁻⁷ M
- Slightly basic

Row 23: pOH = 12.24
- pH = 14 − 12.24 = 1.76
- [H₃O⁺] = 10^(−1.76) ≈ 1.74 × 10⁻² M
- [OH⁻] = 10^(−12.24) ≈ 5.75 × 10⁻¹³ M
- Strongly acidic

Row 24: [OH⁻] = 5.39 × 10⁻¹² M
- pOH = −log(5.39 × 10⁻¹²) = 11.27 (log 5.39 ≈ 0.732, so −(−12 + 0.732) = 11.268)
- pH = 14 − 11.27 = 2.73
- [H₃O⁺] = 10^(−2.73) ≈ 1.86 × 10⁻³ M
- Acidic

Now let’s double-check a few critical ones using Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴:

- Row 2: [H⁺] = 3.89e−4, [OH⁻] should be 1e−14 / 3.89e−4 = 2.57e−11 ✔️ matches our calc
- Row 4: [OH⁻] = 4.88e−5 → [H⁺] = 1e−14 / 4.88e−5 = 2.05e−10 ✔️ close to 2.04e−10
- Row 16: [OH⁻] = 8.53e−7 → [H⁺] = 1e−14 / 8.53e−7 = 1.17e−8 ✔️ matches
All consistent.

Final Answer:
Below is the completed table (values rounded to 3 significant figures where appropriate):

| pH | [H₃O⁺] (M) | pOH | [OH⁻] (M) | ACID or BASE? |
|-------|------------------|--------|------------------|---------------|
| 3.78 | 1.66 × 10⁻⁴ | 10.22 | 6.03 × 10⁻¹¹ | Acid |
| 3.41 | 3.89 × 10⁻⁴ | 10.59 | 2.57 × 10⁻¹¹ | Acid |
| 8.81 | 1.55 × 10⁻⁹ | 5.19 | 6.46 × 10⁻⁶ | Base |
| 9.69 | 2.04 × 10⁻¹⁰ | 4.31 | 4.88 × 10⁻⁵ | Base |
| 8.46 | 3.47 × 10⁻⁹ | 5.54 | 2.88 × 10⁻⁶ | Base |
| 12.07 | 8.45 × 10⁻¹³ | 1.93 | 1.17 × 10⁻² | Base |
| 11.86 | 1.38 × 10⁻¹² | 2.14 | 7.24 × 10⁻³ | Base |
| 3.36 | 4.37 × 10⁻⁴ | 10.64 | 2.31 × 10⁻¹¹ | Acid |
| 10.91 | 1.23 × 10⁻¹¹ | 3.09 | 8.13 × 10⁻⁴ | Base |
| 5.13 | 7.49 × 10⁻⁶ | 8.87 | 1.35 × 10⁻⁹ | Acid |
| 4.06 | 8.71 × 10⁻⁵ | 9.94 | 1.15 × 10⁻¹⁰ | Acid |
| 6.41 | 3.89 × 10⁻⁷ | 7.59 | 2.57 × 10⁻⁸ | Acid |
| 4.16 | 6.92 × 10⁻⁵ | 9.84 | 1.45 × 10⁻¹⁰ | Acid |
| 0.98 | 1.06 × 10⁻¹ | 13.02 | 9.55 × 10⁻¹⁴ | Acid |
| 10.18 | 6.61 × 10⁻¹¹ | 3.82 | 1.51 × 10⁻⁴ | Base |
| 7.93 | 1.17 × 10⁻⁸ | 6.07 | 8.53 × 10⁻⁷ | Base |
| 7.05 | 8.91 × 10⁻⁸ | 6.95 | 1.12 × 10⁻⁷ | Base |
| 9.32 | 4.73 × 10⁻¹⁰ | 4.68 | 2.10 × 10⁻⁵ | Base |
| 12.67 | 2.14 × 10⁻¹³ | 1.33 | 4.68 × 10⁻² | Base |
| 11.99 | 1.02 × 10⁻¹² | 2.01 | 9.87 × 10⁻³ | Base |
| 11.68 | 2.10 × 10⁻¹² | 2.32 | 4.79 × 10⁻³ | Base |
| 7.04 | 9.22 × 10⁻⁸ | 6.96 | 1.10 × 10⁻⁷ | Base |
| 1.76 | 1.74 × 10⁻² | 12.24 | 5.75 × 10⁻¹³ | Acid |
| 2.73 | 1.86 × 10⁻³ | 11.27 | 5.39 × 10⁻¹² | Acid |