Investigating The Ph Scale Answer Key

Investigating the pH Scale: An Answer Key Guide

When students tackle chemistry problems involving the pH scale, they often face a maze of numbers, equations, and conceptual twists. This answer key guide is designed to walk you through the most common pH-related questions, explaining each step, the underlying science, and how to arrive at the correct answer. Whether you’re a high‑school student preparing for a test or a teacher looking for a quick reference, this resource breaks down the process into clear, manageable parts.

This changes depending on context. Keep that in mind.

Introduction

The pH scale is a logarithmic measure of hydrogen ion concentration, [H⁺], in a solution. On the flip side, it ranges from 0 (highly acidic) to 14 (highly basic), with 7 being neutral. Calculating pH involves understanding the relationship between [H⁺], [OH⁻], and the ion product of water, Kₐw.

1. Determine the concentration of [H⁺] or [OH⁻].
2. Convert that concentration to pH or pOH.
3. Apply equilibrium concepts for weak acids/bases.
4. Use stoichiometry for neutralization reactions.

Below is a step‑by‑step answer key covering typical scenarios.

1. Direct pH Calculation from a Strong Acid or Base

Problem

A 0.025 M solution of hydrochloric acid (HCl) is prepared. What is its pH?

Solution

1. Identify the acid as strong: HCl fully dissociates → [H⁺] = 0.025 M.
2. Apply the pH formula:
   [ \text{pH} = -\log[H⁺] = -\log(0.025) ]
3. Compute:
   (-\log(0.025) = 1.60).

Answer: pH = 1.60

2. pOH and Conversion to pH

Problem

A 0.010 M sodium hydroxide (NaOH) solution has a pOH of what value? What is its pH?

Solution

1. Strong base fully dissociates → [OH⁻] = 0.010 M.
2. Calculate pOH:
   [ \text{pOH} = -\log[OH⁻] = -\log(0.010) = 2.00 ]
3. Convert to pH:
   [ \text{pH} = 14 - \text{pOH} = 14 - 2.00 = 12.00 ]

Answer: pOH = 2.00; pH = 12.00

3. Weak Acid Equilibrium (Using Ka)

Problem

Determine the pH of a 0.050 M solution of acetic acid (CH₃COOH) with (K_a = 1.8 \times 10^{-5}).

Solution

1. Set up the equilibrium expression:
   [ K_a = \frac{[H⁺][CH₃COO⁻]}{[CH₃COOH]} ]
2. Assume x = [H⁺] = [CH₃COO⁻] (since each acid molecule yields one H⁺ and one conjugate base).
   Initial [CH₃COOH] = 0.050 M; change = –x; equilibrium = 0.050 – x.
3. Write the equation:
   [ 1.8 \times 10^{-5} = \frac{x^2}{0.050 - x} ]
4. Approximate: Since (K_a) is small, (x \ll 0.050); thus (0.050 - x \approx 0.050).
   [ x^2 \approx 1.8 \times 10^{-5} \times 0.050 = 9.0 \times 10^{-7} ] [ x \approx \sqrt{9.0 \times 10^{-7}} = 9.49 \times 10^{-4},\text{M} ]
5. Calculate pH:
   [ \text{pH} = -\log(9.49 \times 10^{-4}) \approx 3.02 ]

Answer: pH ≈ 3.02

4. Weak Base Equilibrium (Using Kb)

Problem

What is the pH of a 0.020 M solution of ammonia (NH₃) with (K_b = 1.8 \times 10^{-5})?

Solution

1. Set up the base equilibrium:
   [ NH₃ + H₂O \rightleftharpoons NH₄⁺ + OH⁻ ] [ K_b = \frac{[NH₄⁺][OH⁻]}{[NH₃]} ]
2. Assume y = [OH⁻] = [NH₄⁺]; initial [NH₃] = 0.020 M; equilibrium = 0.020 – y.
3. Equation:
   [ 1.8 \times 10^{-5} = \frac{y^2}{0.020 - y} ]
4. Approximate: (y \ll 0.020).
   [ y^2 \approx 1.8 \times 10^{-5} \times 0.020 = 3.6 \times 10^{-7} ] [ y \approx \sqrt{3.6 \times 10^{-7}} = 6.0 \times 10^{-4},\text{M} ]
5. Find pOH:
   [ \text{pOH} = -\log(6.0 \times 10^{-4}) \approx 3.22 ]
6. Convert to pH:
   [ \text{pH} = 14 - 3.22 = 10.78 ]

Answer: pH ≈ 10.78

5. Neutralization Reaction (Stoichiometry)

Problem

Mix 25 mL of 0.100 M H₂SO₄ with 30 mL of 0.200 M NaOH. What is the final pH?

Solution

1. Determine moles of acid and base:
   H₂SO₄ is diprotic: 1 mol H₂SO₄ → 2 mol H⁺.
   [ n_{\text{H⁺}} = 0.025,\text{L} \times 0.100,\text{M} \times 2 = 0.0050,\text{mol} ] [ n_{\text{NaOH}} = 0.030,\text{L} \times 0.200,\text{M} = 0.0060,\text{mol} ]
2. Find limiting reactant: NaOH (0.006 mol) > H⁺ (0.005 mol). Excess NaOH = 0.006 – 0.005 = 0.001 mol.
3. Total volume: 25 mL + 30 mL = 55 mL = 0.055 L.
4. Calculate [OH⁻]:
   [ [OH⁻] = \frac{0.001,\text{mol}}{0.055,\text{L}} \approx 0.0182,\text{M} ]
5. Find pOH:
   [ \text{pOH} = -\log(0.0182) \approx 1.74 ]
6. Convert to pH:
   [ \text{pH} = 14 - 1.74 = 12.26 ]

Answer: pH ≈ 12.26

6. Buffer Capacity (pH of a Buffer Solution)

Problem

A buffer contains 0.150 M acetic acid and 0.300 M sodium acetate. What is its pH? (Use (K_a = 1.8 \times 10^{-5})).

Solution

1. Use the Henderson–Hasselbalch equation:
   [ \text{pH} = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right) ]
2. Calculate (pK_a):
   [ pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74 ]
3. Insert concentrations:
   [ \text{pH} = 4.74 + \log\left(\frac{0.300}{0.150}\right) = 4.74 + \log(2) ] [ \log(2) \approx 0.301 ]
4. Final pH:
   [ \text{pH} = 4.74 + 0.301 = 5.04 ]

Answer: pH ≈ 5.04

7. Titration Endpoint (Half‑Equivalence Point)

Problem

During the titration of 25 mL of 0.050 M HCl with 0.100 M NaOH, what is the pH when 12.5 mL of NaOH has been added?

Solution

1. At half‑equivalence: moles of NaOH added = half the moles of HCl.
   Moles HCl = 0.025 L × 0.050 M = 0.00125 mol.
   Half = 0.000625 mol → volume NaOH = 0.000625 mol / 0.100 M = 6.25 mL.
   Since 12.5 mL is twice this, the solution is beyond the half‑equivalence point; we must calculate the exact pH.
2. Determine excess base:
   Moles NaOH added = 0.0125 L × 0.100 M = 0.00125 mol.
   Moles HCl = 0.00125 mol → neutralized completely, leaving 0.00125 mol NaOH excess.
3. Total volume: 25 mL + 12.5 mL = 37.5 mL = 0.0375 L.
   [ [OH⁻] = \frac{0.00125}{0.0375} = 0.0333,\text{M} ]
4. Find pOH:
   [ \text{pOH} = -\log(0.0333) \approx 1.48 ]
5. Convert to pH:
   [ \text{pH} = 14 - 1.48 = 12.52 ]

Answer: pH ≈ 12.52

8. Common Mistakes to Avoid

• Forgetting the diprotic nature of sulfuric acid: Always double the moles of H⁺.
• Using linear approximations for weak acids/bases when Ka or Kb is large: The assumption (x \ll C) may fail.
• Misapplying the Henderson–Hasselbalch equation: It is valid only when both acid and conjugate base are present in appreciable amounts.
• Ignoring activity coefficients in highly concentrated solutions: In most school problems, this can be neglected, but be aware of the limitation.

FAQ

Conclusion

Mastering pH calculations requires a solid grasp of acid–base equilibria, stoichiometry, and logarithmic relationships. On top of that, by following the systematic approach illustrated in these examples—identifying the species involved, setting up the correct equilibrium expression, applying approximations judiciously, and converting between pH and pOH—you can confidently solve a wide array of pH‑related problems. Keep this answer key handy as a quick reference, and use it to reinforce the concepts that form the backbone of analytical chemistry And that's really what it comes down to..