Acids & Bases Worksheet with eight problems on calculating pH, pOH, and Ka values for various chemical solutions.

Let's solve each problem step by step from the Acids & Bases Worksheet. I'll provide clear explanations and calculations for each.

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1. Determine the pH and pOH of 1.5 mol/L solution of NH₃OH (Kb = 8.7 × 10⁻⁹)

Wait — NH₃OH is not a standard compound. This might be a typo. Likely, it should be NH₃ (ammonia), which is a weak base with Kb = 1.8 × 10⁻⁵. But here, it says Kb = 8.7 × 10⁻⁹, which is unusually low for ammonia. Alternatively, it could be hydroxylamine, NH₂OH, which has Kb ≈ 1.1 × 10⁻⁸, close to this value.

But the formula given is NH₃OH, which is not correct. The correct formula for hydroxylamine is NH₂OH, and its Kb is approximately 1.1 × 10⁻⁸, but the problem gives Kb = 8.7 × 10⁻⁹. So we will proceed assuming it’s a weak base with Kb = 8.7 × 10⁻⁹ and concentration = 1.5 mol/L.

Assume:
- Base: B (e.g., NH₂OH)
- Kb = 8.7 × 10⁻⁹
- [B]₀ = 1.5 M

Reaction:
B + H₂O ⇌ BH⁺ + OH⁻

Set up ICE table:

| | B | BH⁺ | OH⁻ |
|----------|-------|--------|--------|
| Initial | 1.5 | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 1.5 - x | x | x |

Kb = [BH⁺][OH⁻] / [B] = x² / (1.5 - x) ≈ x² / 1.5
(We’ll check approximation later)

So:
\[
\frac{x^2}{1.5} = 8.7 \times 10^{-9}
\Rightarrow x^2 = 1.5 \times 8.7 \times 10^{-9} = 1.305 \times 10^{-8}
\Rightarrow x = \sqrt{1.305 \times 10^{-8}} \approx 1.142 \times 10^{-4}
\]

So:
- [OH⁻] = 1.142 × 10⁻⁴ M
- pOH = -log(1.142 × 10⁻⁴) ≈ 3.94
- pH = 14 - pOH = 14 - 3.94 = 10.06

✔ Answer:
- pH ≈ 10.06
- pOH ≈ 3.94

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2. A 1.0×10⁻³ mol/L solution of a weak acid HA is 20.0% dissociated. What is the pH and Ka?

Given:
- [HA]₀ = 1.0 × 10⁻³ M
- % dissociation = 20.0% → so [H⁺] = 20.0% of 1.0 × 10⁻³ = 0.20 × 10⁻³ = 2.0 × 10⁻⁴ M

So:
- [H⁺] = [A⁻] = 2.0 × 10⁻⁴ M
- [HA] remaining = 1.0 × 10⁻³ - 2.0 × 10⁻⁴ = 8.0 × 10⁻⁴ M

Now use Ka expression:
\[
Ka = \frac{[H^+][A^-]}{[HA]} = \frac{(2.0 \times 10^{-4})(2.0 \times 10^{-4})}{8.0 \times 10^{-4}} = \frac{4.0 \times 10^{-8}}{8.0 \times 10^{-4}} = 5.0 \times 10^{-5}
\]

pH = -log[H⁺] = -log(2.0 × 10⁻⁴) = 3.70

✔ Answer:
- pH = 3.70
- Ka = 5.0 × 10⁻⁵

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3. Calculate [H₃O⁺], [OH⁻], pH, pOH of 1.56×10⁻⁴ mol/L Ba(OH)₂ solution

Ba(OH)₂ is a strong base and dissociates completely:
\[
Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-
\]
So [OH⁻] = 2 × [Ba(OH)₂] = 2 × 1.56 × 10⁻⁴ = 3.12 × 10⁻⁴ M

Now:
- [OH⁻] = 3.12 × 10⁻⁴ M
- pOH = -log(3.12 × 10⁻⁴) ≈ 3.50
- pH = 14 - pOH = 14 - 3.50 = 10.50
- [H₃O⁺] = 10⁻¹⁴ / [OH⁻] = 10⁻¹⁴ / 3.12 × 10⁻⁴ ≈ 3.20 × 10⁻¹¹ M

✔ Answer:
- [H₃O⁺] = 3.20 × 10⁻¹¹ M
- [OH⁻] = 3.12 × 10⁻⁴ M
- pH = 10.50
- pOH = 3.50

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4. Citric acid, a monoprotic acid, has pKa = 7.94 × 10⁻⁴. Wait — that can't be right.

Wait — pKa = 7.94 × 10⁻⁴? That would mean Ka = 10⁻⁷.⁹⁴ × 10⁻⁴ = no — wait.

Actually, pKa = 7.94 × 10⁻⁴ is not possible because pKa is a logarithmic value, typically between 0–14. But 7.94 × 10⁻⁴ is a very small number, like 0.000794, which is not a pKa.

Likely, the problem meant:
> "Citric acid has a pKa of 7.94" or "Ka = 7.94 × 10⁻⁴"

But it says: "pKa of 7.94×10⁻⁴" — that must be a typo.

If pKa = 7.94, then Ka = 10⁻⁷.⁹⁴ ≈ 1.15 × 10⁻⁸ → very weak acid.

But citric acid is actually triprotic, and its first pKa is around 3.1, so 7.94 is too high.

Alternatively, if Ka = 7.94 × 10⁻⁴, then pKa = -log(7.94 × 10⁻⁴) ≈ 3.10

That makes more sense.

Let me assume the intended meaning is:
> Ka = 7.94 × 10⁻⁴, and concentration = 0.0034 mol/L

But the problem says: "pKa of 7.94×10⁻⁴", which is incorrect syntax.

Possibility: It meant Ka = 7.94 × 10⁻⁴, so pKa = -log(7.94 × 10⁻⁴) ≈ 3.10

Let’s proceed with:

- Ka = 7.94 × 10⁻⁴
- [HA]₀ = 0.0034 M = 3.4 × 10⁻³ M

Since Ka = 7.94 × 10⁻⁴ and [HA]₀ = 3.4 × 10⁻³, they are comparable, so we cannot assume x << [HA].

Use quadratic.

HA ⇌ H⁺ + A⁻

ICE:
| | HA | H⁺ | A⁻ |
|-------|-----------|--------|--------|
| I | 3.4×10⁻³ | 0 | 0 |
| C | -x | +x | +x |
| E | 3.4×10⁻³ - x | x | x |

Ka = x² / (3.4×10⁻³ - x) = 7.94×10⁻⁴

Multiply both sides:
x² = 7.94×10⁻⁴ (3.4×10⁻³ - x)
x² = (7.94×10⁻⁴)(3.4×10⁻³) - (7.94×10⁻⁴)x
x² = 2.70 × 10⁻⁶ - 7.94×10⁻⁴ x

Bring all terms to one side:
x² + 7.94×10⁻⁴ x - 2.70×10⁻⁶ = 0

Use quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a=1, b=7.94\times10^{-4}, c=-2.70\times10^{-6}
\]

Discriminant:
D = (7.94×10⁻⁴)² + 4×2.70×10⁻⁶ = 6.29×10⁻⁷ + 1.08×10⁻⁵ = 1.143×10⁻⁵

√D ≈ √(1.143×10⁻⁵) ≈ 3.38×10⁻³

x = [-7.94×10⁻⁴ + 3.38×10⁻³]/2 ≈ (2.586×10⁻³)/2 ≈ 1.293×10⁻³

Check: x ≈ 1.29×10⁻³, but initial [HA] = 3.4×10⁻³ → OK, but let's see if it fits.

Wait: x² = (1.29×10⁻³)² ≈ 1.66×10⁻⁶
Denominator = 3.4×10⁻³ - 1.29×10⁻³ = 2.11×10⁻³
Ka = 1.66×10⁻⁶ / 2.11×10⁻³ ≈ 7.87×10⁻⁴ → close enough.

So [H⁺] = 1.29×10⁻³ M
pH = -log(1.29×10⁻³) ≈ 2.93

✔ Answer: pH ≈ 2.93

But note: If the original was pKa = 7.94, then Ka = 10⁻⁷.⁹⁴ ≈ 1.15×10⁻⁸, which is much smaller.

Then:
Ka = 1.15×10⁻⁸
[HA] = 0.0034 M
x² / (0.0034 - x) ≈ x² / 0.0034 = 1.15×10⁻⁸
→ x² = 4.01×10⁻¹¹ → x ≈ 6.33×10⁻⁶
pH = -log(6.33×10⁻⁶) ≈ 5.20

But that doesn’t make sense for citric acid.

Given confusion, I believe the intended meaning was: Ka = 7.94 × 10⁻⁴, so pKa = 3.10, and pH ≈ 2.93

✔ Final Answer: pH ≈ 2.93

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5. Calculate a) [H₃O⁺], b) pH, c) pOH, d) [OH⁻] of 2.06×10⁻³ mol/L acetic acid

Acetic acid: CH₃COOH, Ka = 1.8 × 10⁻⁵

[HA]₀ = 2.06 × 10⁻³ M

HA ⇌ H⁺ + A⁻

Ka = x² / (2.06×10⁻³ - x) ≈ x² / 2.06×10⁻³ = 1.8×10⁻⁵

x² = 1.8×10⁻⁵ × 2.06×10⁻³ = 3.708×10⁻⁸
x = √(3.708×10⁻⁸) ≈ 1.925×10⁻⁴

Check: x / [HA] = 1.925×10⁻⁴ / 2.06×10⁻³ ≈ 9.3%, so approximation okay (but not great; better to use quadratic)

Quadratic:
x² = 1.8×10⁻⁵ (2.06×10⁻³ - x)
x² = 3.708×10⁻⁸ - 1.8×10⁻⁵ x
x² + 1.8×10⁻⁵ x - 3.708×10⁻⁸ = 0

a=1, b=1.8×10⁻⁵, c=-3.708×10⁻⁸

D = (1.8×10⁻⁵)² + 4×3.708×10⁻⁸ = 3.24×10⁻¹⁰ + 1.483×10⁻⁷ = 1.483×10⁻⁷
√D ≈ 3.85×10⁻⁴

x = [-1.8×10⁻⁵ + 3.85×10⁻⁴]/2 = 3.67×10⁻⁴ / 2 = 1.835×10⁻⁴

So [H₃O⁺] = 1.835 × 10⁻⁴ M

- a) [H₃O⁺] = 1.84 × 10⁻⁴ M
- b) pH = -log(1.84×10⁻⁴) ≈ 3.71
- c) pOH = 14 - 3.71 = 10.29
- d) [OH⁻] = 10⁻¹⁴ / 1.84×10⁻⁴ = 5.43 × 10⁻¹¹ M

✔ Answer:
a) [H₃O⁺] = 1.84 × 10⁻⁴ M
b) pH = 3.71
c) pOH = 10.29
d) [OH⁻] = 5.43 × 10⁻¹¹ M

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6. A 2.0×10⁻³ mol/L solution of a weak acid HA is 40.0% dissociated. Find pH and Ka

Given:
- [HA]₀ = 2.0 × 10⁻³ M
- % dissociation = 40.0% → [H⁺] = 0.40 × 2.0×10⁻³ = 8.0×10⁻⁴ M

So:
- [H⁺] = [A⁻] = 8.0×10⁻⁴ M
- [HA] = 2.0×10⁻³ - 8.0×10⁻⁴ = 1.2×10⁻³ M

Ka = (8.0×10⁻⁴)² / (1.2×10⁻³) = 6.4×10⁻⁷ / 1.2×10⁻³ = 5.33 × 10⁻⁴

pH = -log(8.0×10⁻⁴) = 3.09

✔ Answer:
- pH = 3.09
- Ka = 5.33 × 10⁻⁴

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7. Determine a) [OH⁻], b) pH, c) pOH of 1.5 mol/L C₅H₅N (pyridine)

C₅H₅N is pyridine, a weak base. Kb = 1.7 × 10⁻⁹ (standard value)

So:
- [B]₀ = 1.5 M
- Kb = 1.7 × 10⁻⁹

B + H₂O ⇌ BH⁺ + OH⁻

Kb = x² / (1.5 - x) ≈ x² / 1.5 = 1.7×10⁻⁹
x² = 1.5 × 1.7×10⁻⁹ = 2.55×10⁻⁹
x = √(2.55×10⁻⁹) ≈ 5.05×10⁻⁵

So:
- a) [OH⁻] = 5.05 × 10⁻⁵ M
- b) pOH = -log(5.05×10⁻⁵) ≈ 4.30
- c) pH = 14 - 4.30 = 9.70

✔ Answer:
a) [OH⁻] = 5.05 × 10⁻⁵ M
b) pH = 9.70
c) pOH = 4.30

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8. Determine the pH of a 1.50 mol/L solution of HN₃ (hydrazoic acid)

HN₃ is a weak acid. Ka = 1.9 × 10⁻⁵ (standard value)

[HA]₀ = 1.50 M

HA ⇌ H⁺ + A⁻

Ka = x² / (1.50 - x) ≈ x² / 1.50 = 1.9×10⁻⁵
x² = 1.50 × 1.9×10⁻⁵ = 2.85×10⁻⁵
x = √(2.85×10⁻⁵) ≈ 5.34×10⁻³

Check: x / 1.50 = 0.356% → OK for approximation

[H⁺] = 5.34×10⁻³ M
pH = -log(5.34×10⁻³) ≈ 2.27

✔ Answer: pH ≈ 2.27

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## ✔ Final Answers Summary:

1. pH = 10.06, pOH = 3.94
2. pH = 3.70, Ka = 5.0 × 10⁻⁵
3. [H₃O⁺] = 3.20 × 10⁻¹¹ M, [OH⁻] = 3.12 × 10⁻⁴ M, pH = 10.50, pOH = 3.50
4. pH ≈ 2.93 (assuming Ka = 7.94×10⁻⁴)
5. a) [H₃O⁺] = 1.84 × 10⁻⁴ M, b) pH = 3.71, c) pOH = 10.29, d) [OH⁻] = 5.43 × 10⁻¹¹ M
6. pH = 3.09, Ka = 5.33 × 10⁻⁴
7. a) [OH⁻] = 5.05 × 10⁻⁵ M, b) pH = 9.70, c) pOH = 4.30
8. pH = 2.27

Let me know if you'd like these in a formatted table or need corrections based on different values!