Chemistry worksheet for calculating pH and pOH values of different solutions.

You're working on Worksheet 4-1: Calculating pH and pOH Values, which requires you to fill in missing values for various solutions using the relationships between:

- `[H₃O⁺]` (hydronium ion concentration)
- `[OH⁻]` (hydroxide ion concentration)
- `pH = -log[H₃O⁺]`
- `pOH = -log[OH⁻]`
- `pH + pOH = 14` (at 25°C)

Also, remember:
- For strong acids (like HCl, HNO₃, H₂SO₄), `[H₃O⁺]` ≈ initial acid concentration (assuming complete dissociation).
- For strong bases (like NaOH, Ca(OH)₂), `[OH⁻]` depends on the number of OH⁻ ions per formula unit.
- Water is neutral: `[H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M`

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Let’s go row by row and fill in the blanks.

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1. Water

Given:
`[H₃O⁺] = 1.0 × 10⁻⁷ M`

→ `[OH⁻] = 1.0 × 10⁻⁷ M` (since water is neutral)
→ `pH = -log(1.0 × 10⁻⁷) = 7.00`
→ `pOH = 14.00 - 7.00 = 7.00`

✔ Filled:
- `[OH⁻] = 1.0 × 10⁻⁷ M`
- `pH = 7.00`
- `pOH = 7.00`

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2. 0.10 M HCl

HCl is a strong acid → fully dissociates → `[H₃O⁺] = 0.10 M = 1.0 × 10⁻¹ M`

→ `pH = -log(1.0 × 10⁻¹) = 1.00` (given — matches)
→ `pOH = 14.00 - 1.00 = 13.00`
→ `[OH⁻] = 10^(-pOH) = 10^(-13.00) = 1.0 × 10⁻¹³ M`

✔ Filled:
- `[OH⁻] = 1.0 × 10⁻¹³ M`
- `pOH = 13.00`

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3. 0.010 M NaOH

NaOH is a strong base → fully dissociates → `[OH⁻] = 0.010 M = 1.0 × 10⁻² M`

→ `pOH = -log(1.0 × 10⁻²) = 2.00`
→ `pH = 14.00 - 2.00 = 12.00` (given — matches)
→ `[H₃O⁺] = 10^(-pH) = 10^(-12.00) = 1.0 × 10⁻¹² M`

✔ Filled:
- `[H₃O⁺] = 1.0 × 10⁻¹² M`
- `pOH = 2.00`

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4. 1.0 M HNO₃

Strong acid → `[H₃O⁺] = 1.0 M = 1.0 × 10⁰ M`

→ `pH = -log(1.0) = 0.00`
→ `pOH = 14.00 - 0.00 = 14.00`
→ `[OH⁻] = 10^(-14.00) = 1.0 × 10⁻¹⁴ M` (given — matches)

✔ Filled:
- `[H₃O⁺] = 1.0 M`
- `pH = 0.00`
- `pOH = 14.00`

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5. 1.0 M H₂SO₄

Sulfuric acid is diprotic. The first proton is strong, second is weak, but for concentrated solutions like 1.0 M, we often approximate as fully dissociating both protons → `[H₃O⁺] ≈ 2.0 M`

But let’s check the given value: `[H₃O⁺] = 2.00 × 10⁰ M` → that’s 2.00 M

→ `pH = -log(2.00) ≈ -0.301 → rounded to -0.30`
→ `pOH = 14.00 - (-0.30) = 14.30` (given — matches)
→ `[OH⁻] = 10^(-14.30) = 5.01 × 10⁻¹⁵ M` → usually written as `5.0 × 10⁻¹⁵ M`

✔ Filled:
- `[OH⁻] = 5.0 × 10⁻¹⁵ M`
- `pH = -0.30`

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6. Blood

Given:
`[H₃O⁺] = 2.00 × 10⁻⁸ M`

→ `pH = -log(2.00 × 10⁻⁸) = 7.699 → 7.70`
→ `pOH = 14.00 - 7.70 = 6.30`
→ `[OH⁻] = 10^(-6.30) = 5.01 × 10⁻⁷ M → 5.0 × 10⁻⁷ M`

✔ Filled:
- `[OH⁻] = 5.0 × 10⁻⁷ M`
- `pH = 7.70`
- `pOH = 6.30`

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7. 10 M HCl

Strong acid → `[H₃O⁺] = 10 M = 1.0 × 10¹ M`

→ `pH = -log(10) = -1.00`
→ `pOH = 14.00 - (-1.00) = 15.00`
→ `[OH⁻] = 10^(-15.00) = 1.0 × 10⁻¹⁵ M` (given — matches)

✔ Filled:
- `[H₃O⁺] = 10 M`
- `pH = -1.00`
- `pOH = 15.00`

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8. 0.50 M NaOH

Strong base → `[OH⁻] = 0.50 M = 5.0 × 10⁻¹ M`

→ `pOH = -log(0.50) = 0.301 → 0.30`
→ `pH = 14.00 - 0.30 = 13.70`
→ `[H₃O⁺] = 10^(-13.70) = 2.0 × 10⁻¹⁴ M` (given — matches)

✔ Filled:
- `[H₃O⁺] = 2.0 × 10⁻¹⁴ M`
- `pH = 13.70`
- `pOH = 0.30`

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9. 0.10 M Na₂O

Na₂O reacts with water:
`Na₂O + H₂O → 2NaOH` → so 0.10 M Na₂O produces 0.20 M NaOH

→ `[OH⁻] = 0.20 M = 2.0 × 10⁻¹ M`

→ `pOH = -log(0.20) = 0.699 → 0.70`
→ `pH = 14.00 - 0.70 = 13.30`
→ `[H₃O⁺] = 10^(-13.30) = 5.01 × 10⁻¹⁴ M → 5.0 × 10⁻¹⁴ M`

✔ Filled:
- `[H₃O⁺] = 5.0 × 10⁻¹⁴ M`
- `pH = 13.30`
- `pOH = 0.70`

---

10. 0.010 M Ca(OH)₂

Ca(OH)₂ is a strong base → each mole gives 2 OH⁻ → `[OH⁻] = 2 × 0.010 = 0.020 M = 2.0 × 10⁻² M`

→ `pOH = -log(0.020) = 1.699 → 1.70`
→ `pH = 14.00 - 1.70 = 12.30`
→ `[H₃O⁺] = 10^(-12.30) = 5.01 × 10⁻¹³ M → 5.0 × 10⁻¹³ M`

✔ Filled:
- `[H₃O⁺] = 5.0 × 10⁻¹³ M`
- `pH = 12.30`
- `pOH = 1.70`

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11. 2.0 M NaOH

Strong base → `[OH⁻] = 2.0 M = 2.0 × 10⁰ M`

→ `pOH = -log(2.0) = 0.301 → 0.30`
→ `pH = 14.00 - 0.30 = 13.70`
→ `[H₃O⁺] = 10^(-13.70) = 2.0 × 10⁻¹⁴ M`

✔ Filled:
- `[H₃O⁺] = 2.0 × 10⁻¹⁴ M`
- `pH = 13.70`
- `pOH = 0.30`

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## ✔ Final Completed Table:

| Solution | [H₃O⁺] (M) | [OH⁻] (M) | pH | pOH |
|--------------------|------------------|------------------|-------|-------|
| water | 1.0 × 10⁻⁷ | 1.0 × 10⁻⁷ | 7.00 | 7.00 |
| 0.10 M HCl | 1.0 × 10⁻¹ | 1.0 × 10⁻¹³ | 1.00 | 13.00 |
| 0.010 M NaOH | 1.0 × 10⁻¹² | 1.0 × 10⁻² | 12.00 | 2.00 |
| 1.0 M HNO₃ | 1.0 | 1.0 × 10⁻¹⁴ | 0.00 | 14.00 |
| 1.0 M H₂SO₄ | 2.00 | 5.0 × 10⁻¹⁵ | -0.30 | 14.30 |
| Blood | 2.00 × 10⁻⁸ | 5.0 × 10⁻⁷ | 7.70 | 6.30 |
| 10 M HCl | 10 | 1.0 × 10⁻¹⁵ | -1.00 | 15.00 |
| 0.50 M NaOH | 2.0 × 10⁻¹⁴ | 5.0 × 10⁻¹ | 13.70 | 0.30 |
| 0.10 M Na₂O | 5.0 × 10⁻¹⁴ | 2.0 × 10⁻¹ | 13.30 | 0.70 |
| 0.010 M Ca(OH)₂ | 5.0 × 10⁻¹³ | 2.0 × 10⁻² | 12.30 | 1.70 |
| 2.0 M NaOH | 2.0 × 10⁻¹⁴ | 2.0 | 13.70 | 0.30 |

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## 💡 Key Takeaways:
- Use `pH = -log[H₃O⁺]`, `pOH = -log[OH⁻]`, and `pH + pOH = 14`.
- For strong acids/bases, assume full dissociation.
- For salts like Na₂O or Ca(OH)₂, account for stoichiometry (how many H⁺ or OH⁻ ions are released).
- Negative pH values are possible for very concentrated strong acids.

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