acid-base equilibrium practice problems with answers pdf

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04-02-2025

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This guide provides a series of acid-base equilibrium practice problems with detailed solutions. Mastering acid-base chemistry is crucial for success in chemistry, biochemistry, and related fields. These problems cover a range of difficulty, from basic calculations to more complex scenarios involving buffer solutions and titrations. Understanding these concepts is key to comprehending more advanced topics.

Problem 1: Calculating pH from Acid Concentration

Problem: What is the pH of a 0.10 M solution of a strong acid, HCl?

Solution:

Strong acids completely dissociate in water. Therefore, the concentration of H⁺ ions is equal to the initial concentration of the acid.

[H⁺] = 0.10 M

pH = -log₁₀[H⁺] = -log₁₀(0.10) = 1.0

Answer: The pH of the 0.10 M HCl solution is 1.0.

Problem 2: Calculating pH from Base Concentration

Problem: What is the pOH and pH of a 0.05 M solution of a strong base, NaOH?

Solution:

Similar to strong acids, strong bases completely dissociate. The concentration of OH⁻ ions equals the initial base concentration.

[OH⁻] = 0.05 M

pOH = -log₁₀[OH⁻] = -log₁₀(0.05) ≈ 1.3

Since pH + pOH = 14 at 25°C,

pH = 14 - pOH = 14 - 1.3 ≈ 12.7

Answer: The pOH is approximately 1.3, and the pH is approximately 12.7.

Problem 3: Weak Acid Equilibrium Calculation

Problem: Calculate the pH of a 0.10 M solution of acetic acid (CH₃COOH), given its Ka value is 1.8 x 10⁻⁵.

Solution:

This requires using the equilibrium expression for a weak acid:

CH₃COOH ⇌ CH₃COO⁻ + H⁺

Ka = [CH₃COO⁻][H⁺] / [CH₃COOH]

Let x represent the concentration of H⁺ and CH₃COO⁻ at equilibrium. Then:

1.8 x 10⁻⁵ = x² / (0.10 - x)

Since Ka is small, we can approximate 0.10 - x ≈ 0.10:

1.8 x 10⁻⁵ ≈ x² / 0.10

x² ≈ 1.8 x 10⁻⁶

x ≈ 1.34 x 10⁻³ M (This is [H⁺])

pH = -log₁₀(1.34 x 10⁻³) ≈ 2.87

Answer: The pH of the 0.10 M acetic acid solution is approximately 2.87.

Problem 4: Buffer Solution Calculation

Problem: Calculate the pH of a buffer solution containing 0.20 M acetic acid and 0.30 M sodium acetate (CH₃COONa). The Ka of acetic acid is 1.8 x 10⁻⁵. Use the Henderson-Hasselbalch equation.

Solution:

The Henderson-Hasselbalch equation is:

pH = pKa + log₁₀([A⁻]/[HA])

where [A⁻] is the conjugate base concentration and [HA] is the weak acid concentration.

pKa = -log₁₀(1.8 x 10⁻⁵) ≈ 4.74

pH = 4.74 + log₁₀(0.30/0.20) = 4.74 + log₁₀(1.5) ≈ 4.74 + 0.18 ≈ 4.92

Answer: The pH of the buffer solution is approximately 4.92.

Problem 5: Titration Calculation (at the equivalence point)

Problem: What is the pH at the equivalence point of a titration of 25.0 mL of 0.10 M HCl with 0.10 M NaOH?

Solution:

At the equivalence point, the moles of acid equal the moles of base. The resulting solution contains only salt (NaCl) and water. Since NaCl is a neutral salt, the pH will be 7.0.

Answer: The pH at the equivalence point is 7.0.

This is just a small sample of the many types of acid-base equilibrium problems you might encounter. Remember to always consider the strength of the acid or base and whether you're dealing with a buffer solution or titration. Further practice with more diverse problems will solidify your understanding. Consider consulting a textbook or online resources for additional problems and explanations.