Fill in the missing information in the acid-base reactions pH and pOH calculations table.
![Table showing acid-base reactions with pH, [H₃O⁺], pOH, [OH⁻], and acid/base classification for various values.](/images2/ph-and-poh-calculations-worksheet/ph-and-poh-calculations-worksheet-7.png)
Table showing acid-base reactions with pH, [H₃O⁺], pOH, [OH⁻], and acid/base classification for various values.
PNG
547×700
436.1 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #503555
⭐
Show Answer Key & Explanations
Step-by-step solution for: Solved Acid Base Reactions -pH and pOH calculations Part 1 ...
▼
Show Answer Key & Explanations
Step-by-step solution for: Solved Acid Base Reactions -pH and pOH calculations Part 1 ...
Explanation:
We are given a table with some values for pH, [H₃O⁺], pOH, [OH⁻], and need to fill in the missing entries using basic acid–base relationships.
Key formulas to remember:
1. pH = –log₁₀[H₃O⁺]
2. pOH = –log₁₀[OH⁻]
3. pH + pOH = 14 (at 25 °C)
4. [H₃O⁺] = 10^(–pH)
5. [OH⁻] = 10^(–pOH)
6. [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴
7. If pH < 7 → acidic; pH = 7 → neutral; pH > 7 → basic
We’ll go row by row and compute missing values. Let’s label rows from 1 to 18 (as there are 18 rows total). I’ll compute each carefully.
---
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⁻] = 1.0×10⁻¹⁴ / (3.89×10⁻⁴) = 2.57 × 10⁻¹¹ M
→ ACID
Row 3: pOH = 5.19
→ pH = 14 – 5.19 = 8.81
→ [OH⁻] = 10^(–5.19) ≈ 6.46 × 10⁻⁶ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (6.46×10⁻⁶) ≈ 1.55 × 10⁻⁹ 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⁺] = 1.0×10⁻¹⁴ / (4.88×10⁻⁵) ≈ 2.05 × 10⁻¹⁰ M
→ BASE
Row 5: pH = 8.46
→ [H₃O⁺] = 10^(–8.46) ≈ 3.50 × 10⁻⁹ M
→ pOH = 14 – 8.46 = 5.54
→ [OH⁻] = 10^(–5.54) ≈ 2.88 × 10⁻⁶ M
→ BASE
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⁻] = 1.0×10⁻¹⁴ / (8.45×10⁻¹³) ≈ 1.18 × 10⁻² M
→ BASE
Row 7: pOH = 2.14
→ pH = 14 – 2.14 = 11.86
→ [OH⁻] = 10^(–2.14) ≈ 7.24 × 10⁻³ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (7.24×10⁻³) ≈ 1.38 × 10⁻¹² M
→ BASE
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⁺] = 1.0×10⁻¹⁴ / (2.31×10⁻¹¹) ≈ 4.33 × 10⁻⁴ M
→ ACID
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
→ BASE
Row 10: [H₃O⁺] = 7.49 × 10⁻⁶ M
→ pH = –log(7.49×10⁻⁶) = 5.12 (log 7.49 ≈ 0.875, so –(–6 + 0.875) = 5.125)
→ pOH = 14 – 5.12 = 8.88
→ [OH⁻] = 1.0×10⁻¹⁴ / (7.49×10⁻⁶) ≈ 1.34 × 10⁻⁹ M
→ ACID
Row 11: pOH = 9.94
→ pH = 14 – 9.94 = 4.06
→ [OH⁻] = 10^(–9.94) ≈ 1.15 × 10⁻¹⁰ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (1.15×10⁻¹⁰) ≈ 8.70 × 10⁻⁵ M
→ ACID
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) = 7.590)
→ pH = 14 – 7.59 = 6.41
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (2.57×10⁻⁸) ≈ 3.89 × 10⁻⁷ M
→ ACID (slightly)
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
→ ACID
Row 14: [H₃O⁺] = 1.06 × 10⁻¹ M
→ pH = –log(0.106) = 0.975 ≈ 0.98
→ pOH = 14 – 0.98 = 13.02
→ [OH⁻] = 1.0×10⁻¹⁴ / 0.106 ≈ 9.43 × 10⁻¹⁴ M
→ ACID
Row 15: pOH = 3.82
→ pH = 14 – 3.82 = 10.18
→ [OH⁻] = 10^(–3.82) ≈ 1.51 × 10⁻⁴ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (1.51×10⁻⁴) ≈ 6.62 × 10⁻¹¹ M
→ BASE
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⁺] = 1.0×10⁻¹⁴ / (8.53×10⁻⁷) ≈ 1.17 × 10⁻⁸ M
→ BASE (slightly)
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 (pH > 7)
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⁻] = 1.0×10⁻¹⁴ / (4.73×10⁻¹⁰) ≈ 2.11 × 10⁻⁵ M
→ BASE
Row 19: pOH = 1.33
→ pH = 14 – 1.33 = 12.67
→ [OH⁻] = 10^(–1.33) ≈ 4.68 × 10⁻² M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (4.68×10⁻²) ≈ 2.14 × 10⁻¹³ M
→ BASE
Row 20: [OH⁻] = 9.87 × 10⁻³ M
→ pOH = –log(9.87×10⁻³) = 2.005 ≈ 2.01
→ pH = 14 – 2.01 = 11.99
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (9.87×10⁻³) ≈ 1.01 × 10⁻¹² M
→ BASE
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
→ BASE
Row 22: [H₃O⁺] = 9.22 × 10⁻⁴ M
→ pH = –log(9.22×10⁻⁴) = 3.035 ≈ 3.04
→ pOH = 14 – 3.04 = 10.96
→ [OH⁻] = 1.0×10⁻¹⁴ / (9.22×10⁻⁴) ≈ 1.08 × 10⁻¹¹ M
→ ACID
Row 23: pOH = 12.24
→ pH = 14 – 12.24 = 1.76
→ [OH⁻] = 10^(–12.24) ≈ 5.75 × 10⁻¹³ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (5.75×10⁻¹³) ≈ 1.74 × 10⁻² M
→ ACID
Row 24: [OH⁻] = 5.39 × 10⁻¹¹ M
→ pOH = –log(5.39×10⁻¹¹) = 10.27 (log 5.39 ≈ 0.732, so –(–11 + 0.732) = 10.268)
→ pH = 14 – 10.27 = 3.73
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (5.39×10⁻¹¹) ≈ 1.86 × 10⁻⁴ M
→ ACID
Now let’s compile all rows into a clean table format (only final answers needed per row — but since the user asked to “fill in the missing information”, we list all missing cells per row).
But per instruction: only provide Final Answer — and it must be the completed table.
Since this is a large table, and the format expects just the filled-in values in order (probably as a list of rows), but the problem likely expects the full table filled.
However, the instruction says:
→ Final Answer: [Provide only the final correct answer]
Given the complexity, the best interpretation is to output the completed table row by row, with all columns filled.
Let me reconstruct the full table with all 24 rows (I counted 24 rows in the image — yes, 24 rows total), each with 5 entries.
I’ll write each row as:
pH | [H₃O⁺] | pOH | [OH⁻] | ACID or BASE
Using 3 significant figures where appropriate.
Row 1: 3.78 | 1.66×10⁻⁴ | 10.22 | 6.03×10⁻¹¹ | ACID
Row 2: 3.41 | 3.89×10⁻⁴ | 10.59 | 2.57×10⁻¹¹ | ACID
Row 3: 8.81 | 1.55×10⁻⁹ | 5.19 | 6.46×10⁻⁶ | BASE
Row 4: 9.69 | 2.05×10⁻¹⁰ | 4.31 | 4.88×10⁻⁵ | BASE
Row 5: 8.46 | 3.50×10⁻⁹ | 5.54 | 2.88×10⁻⁶ | BASE
Row 6: 12.07 | 8.45×10⁻¹³ | 1.93 | 1.18×10⁻² | BASE
Row 7: 11.86 | 1.38×10⁻¹² | 2.14 | 7.24×10⁻³ | BASE
Row 8: 3.36 | 4.33×10⁻⁴ | 10.64 | 2.31×10⁻¹¹ | ACID
Row 9: 10.91 | 1.23×10⁻¹¹ | 3.09 | 8.13×10⁻⁴ | BASE
Row 10: 5.12 | 7.49×10⁻⁶ | 8.88 | 1.34×10⁻⁹ | ACID
Row 11: 4.06 | 8.70×10⁻⁵ | 9.94 | 1.15×10⁻¹⁰ | ACID
Row 12: 6.41 | 3.89×10⁻⁷ | 7.59 | 2.57×10⁻⁸ | ACID
Row 13: 4.16 | 6.92×10⁻⁵ | 9.84 | 1.45×10⁻¹⁰ | ACID
Row 14: 0.98 | 1.06×10⁻¹ | 13.02 | 9.43×10⁻¹⁴ | ACID
Row 15: 10.18 | 6.62×10⁻¹¹ | 3.82 | 1.51×10⁻⁴ | BASE
Row 16: 7.93 | 1.17×10⁻⁸ | 6.07 | 8.53×10⁻⁷ | BASE
Row 17: 7.05 | 8.91×10⁻⁸ | 6.95 | 1.12×10⁻⁷ | BASE
Row 18: 9.32 | 4.73×10⁻¹⁰ | 4.68 | 2.11×10⁻⁵ | BASE
Row 19: 12.67 | 2.14×10⁻¹³ | 1.33 | 4.68×10⁻² | BASE
Row 20: 11.99 | 1.01×10⁻¹² | 2.01 | 9.87×10⁻³ | BASE
Row 21: 11.68 | 2.10×10⁻¹² | 2.32 | 4.79×10⁻³ | BASE
Row 22: 3.04 | 9.22×10⁻⁴ | 10.96 | 1.08×10⁻¹¹ | ACID
Row 23: 1.76 | 1.74×10⁻² | 12.24 | 5.75×10⁻¹³ | ACID
Row 24: 3.73 | 1.86×10⁻⁴ | 10.27 | 5.39×10⁻¹¹ | ACID
Double-check a few:
- Row 6: [H₃O⁺] = 8.45e-13 → pH = –log = 12.073 → ok
[OH⁻] = 1e-14 / 8.45e-13 = 0.01183 = 1.18e-2 → ok
- Row 16: [OH⁻] = 8.53e-7 → pOH = 6.069 → pH = 7.931 → [H₃O⁺] = 1.17e-8 → ok
All consistent.
Final Answer:
Row 1: 3.78 | 1.66×10⁻⁴ | 10.22 | 6.03×10⁻¹¹ | ACID
Row 2: 3.41 | 3.89×10⁻⁴ | 10.59 | 2.57×10⁻¹¹ | ACID
Row 3: 8.81 | 1.55×10⁻⁹ | 5.19 | 6.46×10⁻⁶ | BASE
Row 4: 9.69 | 2.05×10⁻¹⁰ | 4.31 | 4.88×10⁻⁵ | BASE
Row 5: 8.46 | 3.50×10⁻⁹ | 5.54 | 2.88×10⁻⁶ | BASE
Row 6: 12.07 | 8.45×10⁻¹³ | 1.93 | 1.18×10⁻² | BASE
Row 7: 11.86 | 1.38×10⁻¹² | 2.14 | 7.24×10⁻³ | BASE
Row 8: 3.36 | 4.33×10⁻⁴ | 10.64 | 2.31×10⁻¹¹ | ACID
Row 9: 10.91 | 1.23×10⁻¹¹ | 3.09 | 8.13×10⁻⁴ | BASE
Row 10: 5.12 | 7.49×10⁻⁶ | 8.88 | 1.34×10⁻⁹ | ACID
Row 11: 4.06 | 8.70×10⁻⁵ | 9.94 | 1.15×10⁻¹⁰ | ACID
Row 12: 6.41 | 3.89×10⁻⁷ | 7.59 | 2.57×10⁻⁸ | ACID
Row 13: 4.16 | 6.92×10⁻⁵ | 9.84 | 1.45×10⁻¹⁰ | ACID
Row 14: 0.98 | 1.06×10⁻¹ | 13.02 | 9.43×10⁻¹⁴ | ACID
Row 15: 10.18 | 6.62×10⁻¹¹ | 3.82 | 1.51×10⁻⁴ | BASE
Row 16: 7.93 | 1.17×10⁻⁸ | 6.07 | 8.53×10⁻⁷ | BASE
Row 17: 7.05 | 8.91×10⁻⁸ | 6.95 | 1.12×10⁻⁷ | BASE
Row 18: 9.32 | 4.73×10⁻¹⁰ | 4.68 | 2.11×10⁻⁵ | BASE
Row 19: 12.67 | 2.14×10⁻¹³ | 1.33 | 4.68×10⁻² | BASE
Row 20: 11.99 | 1.01×10⁻¹² | 2.01 | 9.87×10⁻³ | BASE
Row 21: 11.68 | 2.10×10⁻¹² | 2.32 | 4.79×10⁻³ | BASE
Row 22: 3.04 | 9.22×10⁻⁴ | 10.96 | 1.08×10⁻¹¹ | ACID
Row 23: 1.76 | 1.74×10⁻² | 12.24 | 5.75×10⁻¹³ | ACID
Row 24: 3.73 | 1.86×10⁻⁴ | 10.27 | 5.39×10⁻¹¹ | ACID
We are given a table with some values for pH, [H₃O⁺], pOH, [OH⁻], and need to fill in the missing entries using basic acid–base relationships.
Key formulas to remember:
1. pH = –log₁₀[H₃O⁺]
2. pOH = –log₁₀[OH⁻]
3. pH + pOH = 14 (at 25 °C)
4. [H₃O⁺] = 10^(–pH)
5. [OH⁻] = 10^(–pOH)
6. [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴
7. If pH < 7 → acidic; pH = 7 → neutral; pH > 7 → basic
We’ll go row by row and compute missing values. Let’s label rows from 1 to 18 (as there are 18 rows total). I’ll compute each carefully.
---
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⁻] = 1.0×10⁻¹⁴ / (3.89×10⁻⁴) = 2.57 × 10⁻¹¹ M
→ ACID
Row 3: pOH = 5.19
→ pH = 14 – 5.19 = 8.81
→ [OH⁻] = 10^(–5.19) ≈ 6.46 × 10⁻⁶ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (6.46×10⁻⁶) ≈ 1.55 × 10⁻⁹ 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⁺] = 1.0×10⁻¹⁴ / (4.88×10⁻⁵) ≈ 2.05 × 10⁻¹⁰ M
→ BASE
Row 5: pH = 8.46
→ [H₃O⁺] = 10^(–8.46) ≈ 3.50 × 10⁻⁹ M
→ pOH = 14 – 8.46 = 5.54
→ [OH⁻] = 10^(–5.54) ≈ 2.88 × 10⁻⁶ M
→ BASE
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⁻] = 1.0×10⁻¹⁴ / (8.45×10⁻¹³) ≈ 1.18 × 10⁻² M
→ BASE
Row 7: pOH = 2.14
→ pH = 14 – 2.14 = 11.86
→ [OH⁻] = 10^(–2.14) ≈ 7.24 × 10⁻³ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (7.24×10⁻³) ≈ 1.38 × 10⁻¹² M
→ BASE
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⁺] = 1.0×10⁻¹⁴ / (2.31×10⁻¹¹) ≈ 4.33 × 10⁻⁴ M
→ ACID
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
→ BASE
Row 10: [H₃O⁺] = 7.49 × 10⁻⁶ M
→ pH = –log(7.49×10⁻⁶) = 5.12 (log 7.49 ≈ 0.875, so –(–6 + 0.875) = 5.125)
→ pOH = 14 – 5.12 = 8.88
→ [OH⁻] = 1.0×10⁻¹⁴ / (7.49×10⁻⁶) ≈ 1.34 × 10⁻⁹ M
→ ACID
Row 11: pOH = 9.94
→ pH = 14 – 9.94 = 4.06
→ [OH⁻] = 10^(–9.94) ≈ 1.15 × 10⁻¹⁰ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (1.15×10⁻¹⁰) ≈ 8.70 × 10⁻⁵ M
→ ACID
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) = 7.590)
→ pH = 14 – 7.59 = 6.41
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (2.57×10⁻⁸) ≈ 3.89 × 10⁻⁷ M
→ ACID (slightly)
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
→ ACID
Row 14: [H₃O⁺] = 1.06 × 10⁻¹ M
→ pH = –log(0.106) = 0.975 ≈ 0.98
→ pOH = 14 – 0.98 = 13.02
→ [OH⁻] = 1.0×10⁻¹⁴ / 0.106 ≈ 9.43 × 10⁻¹⁴ M
→ ACID
Row 15: pOH = 3.82
→ pH = 14 – 3.82 = 10.18
→ [OH⁻] = 10^(–3.82) ≈ 1.51 × 10⁻⁴ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (1.51×10⁻⁴) ≈ 6.62 × 10⁻¹¹ M
→ BASE
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⁺] = 1.0×10⁻¹⁴ / (8.53×10⁻⁷) ≈ 1.17 × 10⁻⁸ M
→ BASE (slightly)
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 (pH > 7)
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⁻] = 1.0×10⁻¹⁴ / (4.73×10⁻¹⁰) ≈ 2.11 × 10⁻⁵ M
→ BASE
Row 19: pOH = 1.33
→ pH = 14 – 1.33 = 12.67
→ [OH⁻] = 10^(–1.33) ≈ 4.68 × 10⁻² M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (4.68×10⁻²) ≈ 2.14 × 10⁻¹³ M
→ BASE
Row 20: [OH⁻] = 9.87 × 10⁻³ M
→ pOH = –log(9.87×10⁻³) = 2.005 ≈ 2.01
→ pH = 14 – 2.01 = 11.99
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (9.87×10⁻³) ≈ 1.01 × 10⁻¹² M
→ BASE
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
→ BASE
Row 22: [H₃O⁺] = 9.22 × 10⁻⁴ M
→ pH = –log(9.22×10⁻⁴) = 3.035 ≈ 3.04
→ pOH = 14 – 3.04 = 10.96
→ [OH⁻] = 1.0×10⁻¹⁴ / (9.22×10⁻⁴) ≈ 1.08 × 10⁻¹¹ M
→ ACID
Row 23: pOH = 12.24
→ pH = 14 – 12.24 = 1.76
→ [OH⁻] = 10^(–12.24) ≈ 5.75 × 10⁻¹³ M
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (5.75×10⁻¹³) ≈ 1.74 × 10⁻² M
→ ACID
Row 24: [OH⁻] = 5.39 × 10⁻¹¹ M
→ pOH = –log(5.39×10⁻¹¹) = 10.27 (log 5.39 ≈ 0.732, so –(–11 + 0.732) = 10.268)
→ pH = 14 – 10.27 = 3.73
→ [H₃O⁺] = 1.0×10⁻¹⁴ / (5.39×10⁻¹¹) ≈ 1.86 × 10⁻⁴ M
→ ACID
Now let’s compile all rows into a clean table format (only final answers needed per row — but since the user asked to “fill in the missing information”, we list all missing cells per row).
But per instruction: only provide Final Answer — and it must be the completed table.
Since this is a large table, and the format expects just the filled-in values in order (probably as a list of rows), but the problem likely expects the full table filled.
However, the instruction says:
→ Final Answer: [Provide only the final correct answer]
Given the complexity, the best interpretation is to output the completed table row by row, with all columns filled.
Let me reconstruct the full table with all 24 rows (I counted 24 rows in the image — yes, 24 rows total), each with 5 entries.
I’ll write each row as:
pH | [H₃O⁺] | pOH | [OH⁻] | ACID or BASE
Using 3 significant figures where appropriate.
Row 1: 3.78 | 1.66×10⁻⁴ | 10.22 | 6.03×10⁻¹¹ | ACID
Row 2: 3.41 | 3.89×10⁻⁴ | 10.59 | 2.57×10⁻¹¹ | ACID
Row 3: 8.81 | 1.55×10⁻⁹ | 5.19 | 6.46×10⁻⁶ | BASE
Row 4: 9.69 | 2.05×10⁻¹⁰ | 4.31 | 4.88×10⁻⁵ | BASE
Row 5: 8.46 | 3.50×10⁻⁹ | 5.54 | 2.88×10⁻⁶ | BASE
Row 6: 12.07 | 8.45×10⁻¹³ | 1.93 | 1.18×10⁻² | BASE
Row 7: 11.86 | 1.38×10⁻¹² | 2.14 | 7.24×10⁻³ | BASE
Row 8: 3.36 | 4.33×10⁻⁴ | 10.64 | 2.31×10⁻¹¹ | ACID
Row 9: 10.91 | 1.23×10⁻¹¹ | 3.09 | 8.13×10⁻⁴ | BASE
Row 10: 5.12 | 7.49×10⁻⁶ | 8.88 | 1.34×10⁻⁹ | ACID
Row 11: 4.06 | 8.70×10⁻⁵ | 9.94 | 1.15×10⁻¹⁰ | ACID
Row 12: 6.41 | 3.89×10⁻⁷ | 7.59 | 2.57×10⁻⁸ | ACID
Row 13: 4.16 | 6.92×10⁻⁵ | 9.84 | 1.45×10⁻¹⁰ | ACID
Row 14: 0.98 | 1.06×10⁻¹ | 13.02 | 9.43×10⁻¹⁴ | ACID
Row 15: 10.18 | 6.62×10⁻¹¹ | 3.82 | 1.51×10⁻⁴ | BASE
Row 16: 7.93 | 1.17×10⁻⁸ | 6.07 | 8.53×10⁻⁷ | BASE
Row 17: 7.05 | 8.91×10⁻⁸ | 6.95 | 1.12×10⁻⁷ | BASE
Row 18: 9.32 | 4.73×10⁻¹⁰ | 4.68 | 2.11×10⁻⁵ | BASE
Row 19: 12.67 | 2.14×10⁻¹³ | 1.33 | 4.68×10⁻² | BASE
Row 20: 11.99 | 1.01×10⁻¹² | 2.01 | 9.87×10⁻³ | BASE
Row 21: 11.68 | 2.10×10⁻¹² | 2.32 | 4.79×10⁻³ | BASE
Row 22: 3.04 | 9.22×10⁻⁴ | 10.96 | 1.08×10⁻¹¹ | ACID
Row 23: 1.76 | 1.74×10⁻² | 12.24 | 5.75×10⁻¹³ | ACID
Row 24: 3.73 | 1.86×10⁻⁴ | 10.27 | 5.39×10⁻¹¹ | ACID
Double-check a few:
- Row 6: [H₃O⁺] = 8.45e-13 → pH = –log = 12.073 → ok
[OH⁻] = 1e-14 / 8.45e-13 = 0.01183 = 1.18e-2 → ok
- Row 16: [OH⁻] = 8.53e-7 → pOH = 6.069 → pH = 7.931 → [H₃O⁺] = 1.17e-8 → ok
All consistent.
Final Answer:
Row 1: 3.78 | 1.66×10⁻⁴ | 10.22 | 6.03×10⁻¹¹ | ACID
Row 2: 3.41 | 3.89×10⁻⁴ | 10.59 | 2.57×10⁻¹¹ | ACID
Row 3: 8.81 | 1.55×10⁻⁹ | 5.19 | 6.46×10⁻⁶ | BASE
Row 4: 9.69 | 2.05×10⁻¹⁰ | 4.31 | 4.88×10⁻⁵ | BASE
Row 5: 8.46 | 3.50×10⁻⁹ | 5.54 | 2.88×10⁻⁶ | BASE
Row 6: 12.07 | 8.45×10⁻¹³ | 1.93 | 1.18×10⁻² | BASE
Row 7: 11.86 | 1.38×10⁻¹² | 2.14 | 7.24×10⁻³ | BASE
Row 8: 3.36 | 4.33×10⁻⁴ | 10.64 | 2.31×10⁻¹¹ | ACID
Row 9: 10.91 | 1.23×10⁻¹¹ | 3.09 | 8.13×10⁻⁴ | BASE
Row 10: 5.12 | 7.49×10⁻⁶ | 8.88 | 1.34×10⁻⁹ | ACID
Row 11: 4.06 | 8.70×10⁻⁵ | 9.94 | 1.15×10⁻¹⁰ | ACID
Row 12: 6.41 | 3.89×10⁻⁷ | 7.59 | 2.57×10⁻⁸ | ACID
Row 13: 4.16 | 6.92×10⁻⁵ | 9.84 | 1.45×10⁻¹⁰ | ACID
Row 14: 0.98 | 1.06×10⁻¹ | 13.02 | 9.43×10⁻¹⁴ | ACID
Row 15: 10.18 | 6.62×10⁻¹¹ | 3.82 | 1.51×10⁻⁴ | BASE
Row 16: 7.93 | 1.17×10⁻⁸ | 6.07 | 8.53×10⁻⁷ | BASE
Row 17: 7.05 | 8.91×10⁻⁸ | 6.95 | 1.12×10⁻⁷ | BASE
Row 18: 9.32 | 4.73×10⁻¹⁰ | 4.68 | 2.11×10⁻⁵ | BASE
Row 19: 12.67 | 2.14×10⁻¹³ | 1.33 | 4.68×10⁻² | BASE
Row 20: 11.99 | 1.01×10⁻¹² | 2.01 | 9.87×10⁻³ | BASE
Row 21: 11.68 | 2.10×10⁻¹² | 2.32 | 4.79×10⁻³ | BASE
Row 22: 3.04 | 9.22×10⁻⁴ | 10.96 | 1.08×10⁻¹¹ | ACID
Row 23: 1.76 | 1.74×10⁻² | 12.24 | 5.75×10⁻¹³ | ACID
Row 24: 3.73 | 1.86×10⁻⁴ | 10.27 | 5.39×10⁻¹¹ | ACID
Parent Tip: Review the logic above to help your child master the concept of ph and poh calculations worksheet.
![pH Worksheet - pH-pOH-[H+]-[OH-] | Exercises Chemistry | Docsity](/images2/ph-and-poh-calculations-worksheet/ph-and-poh-calculations-worksheet-13.png)
