Acid Strength, Conjugate Pairs & Ka — Practice · Grade 11 Science

Chapter: Acids, Bases & Titration

What this practice covers

By the end of Acid Strength, Conjugate Pairs & Ka you should be able to:

8 questions, numeric and written answer, each with a worked explanation.

Vocabulary used: conjugate pair, Ka, pKa.

Problem 1

Ka=10⁻⁶. Find pKa.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

6

−log 10(10⁻⁶)=6.

Problem 2

Does strong mean concentrated?
Answer and explanation

No. Strong concerns fraction ionized; concentrated concerns amount per volume.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

Find pKa for Ka=10⁻⁴.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

4

−log 10(10⁻⁴)=4.

Problem 4

A monoprotic strong acid has concentration .010 M. Estimate pH.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

2

−log(.010)=2.

Problem 5

Which is stronger at the same temperature: acid with pK_a=3 or pK_a=5? Show the reasoning, not just the conclusion.
Hints

A smaller pK_a corresponds to larger K_a.

Answer and explanation

The pK_a=3 acid, since its K_a is 100 times larger.

Problem 6

Using x≈√(K_aC), estimate [H⁺] for K_a=10⁻⁵ and C=0.10 M, then check approximation.
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

x≈√10⁻⁶=0.0010 M; x/C=1%, so small-ionization approximation is reasonable and pH≈3.

Problem 7

A useful approximation earns its place through a consistency checkA useful approximation earns its place through a consistency checkC = 0.100 M; Ka = 1.0×10⁻⁵Approximation x ≈ √(KaC) = 0.00100 MFraction ionized ≈ 0.00100/0.100 = 1.00%Exact quadratic x ≈ 0.000995 M; pH ≈ 3.00Check x ≪ C instead of assuming it automatically.
Original selected calculation with a hypothetical Ka. The quadratic value is computed from x²/(C−x)=Ka under the same simplified species assumptions.
Interpret the supplied visual. What calculation tests the assumption C−x≈C in this example?
Answer and explanation

Compute x/C, approximately 0.00100/0.100=0.0100, or one percent, and compare with the desired accuracy.

If x is small relative to initial acid concentration C, replace C−x by C to obtain x≈√(KaC). Then check the inferred x/C against a stated tolerance; a classroom five-percent rule is a heuristic, not a universal error guarantee. The selected numbers give about one-percent ionization, so the approximation agrees closely with the quadratic result. When ionization is substantial, added conjugate base is present or water matters, solve the appropriate fuller equilibrium rather than forcing the shortcut.

Compute x/C, approximately 0.00100/0.100=0.0100, or one percent, and compare with the desired accuracy.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 8

For a hypothetical monoprotic acid with C=0.200 M and Ka=2.0×10⁻⁵, estimate hydronium and pH, then test the small-ionization approximation.
Answer and explanation

x≈√(KaC)=√(4.0×10⁻⁶)=0.0020 M. pH≈2.70. x/C≈0.010=1%, supporting the approximation for ordinary classroom accuracy under the stated species model.

x≈√(KaC)=√(4.0×10⁻⁶)=0.0020 M. pH≈2.70. x/C≈0.010=1%, supporting the approximation for ordinary classroom accuracy under the stated species model.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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