Acid Strength, Conjugate Pairs & Ka — Print · Grade 11 Science
Chapter: Acids, Bases & TitrationAcid Strength, Conjugate Pairs & Ka
Learning goals
- I can identify conjugate pairs and distinguish strength from concentration.
- I can estimate weak-acid ionization and check a simplifying assumption.
Readings for this unit
Before this unit
Acids, equilibrium, logarithms, and scientific notation.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
When is the square-root weak-acid shortcut justified?
Illustrative evidence case. A selected acid’s predicted ionization is compared with its starting concentration.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
When is the square-root weak-acid shortcut justified?
Your first model
Write the equilibrium before simplifying and retain the discarded term for a check.
Build the language from the idea
- conjugate pair
- Species that differ by one proton.
- Ka
- Acid-ionization equilibrium constant for a specified temperature.
- pKa
- Negative base-ten logarithm of Ka.
A Brønsted–Lowry acid donates a proton and a base accepts one. After donating, an acid becomes its conjugate base; after accepting, a base becomes its conjugate acid. Members of a conjugate pair differ by one proton. Water can play either role depending on its reaction partner.
Strength describes the extent of ionization in water. Concentration describes amount per solution volume. A dilute strong acid can have lower hydrogen-ion concentration than a concentrated weak acid, or higher, depending on numbers. Neither label alone determines pH; use actual concentration and ionization.
For HA + H₂O ⇌ H₃O⁺ + A⁻, Ka=[H₃O⁺][A⁻]/[HA] in the introductory concentration approximation. Water is omitted because it is the solvent. Larger Ka generally means more extensive ionization at comparable concentration and temperature. pKa=−log 10 Ka provides a convenient scale.
For a weak monoprotic acid initially at concentration C, let x be the concentration ionized. Then Ka≈x²/(C−x), assuming no important added ions and negligible initial hydronium. If x is small relative to C, x≈√(KaC). Check the approximation by calculating x/C; a rough 5% criterion is common.
Strong-acid pH calculations often use [H₃O⁺]≈C for a fully ionized monoprotic acid. This approximation needs care at extremely low concentrations where water’s own ionization matters. Neutral pH is 7 at about 25°C, not a temperature-independent law.
Use a reaction and numbers to explain an acid claim. Do not treat a weak acid as harmless: concentration, exposure, and other chemical properties matter. These calculations are for supplied data, with household observations limited to safe, labeled food materials and no tasting or mixing cleaners.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Write the equilibrium before the constant
The acid dissociation constant Ka describes a specified proton-transfer equilibrium at a stated temperature. A thermodynamic expression uses activities; the familiar bracket expression approximates those activities using suitably normalized dilute concentrations. Pure liquid water is omitted from this chosen expression. A larger Ka means a greater product-to-reactant equilibrium tendency for the specified step, while pKa=−log₁₀Ka reverses the numerical direction. The constant does not give a reaction rate or make every initial acid molecule ionized.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Does a smaller pKa indicate a larger or smaller Ka for the same convention?
Visual reasoning — after your attempt
A larger Ka, because taking a negative logarithm reverses the numerical ordering.
For HA+H₂O⇌H₃O⁺+A⁻, a dilute-solution approximation writes Ka≈[H₃O⁺][A⁻]/[HA], with water activity treated as approximately constant. Larger Ka under comparable conditions indicates greater acid ionization tendency. It is not the same as a larger initial acid concentration.
Use the supplied acid identity and temperature. Comparing two Ka values from different conditions needs care, and a concentration label alone cannot identify acid strength.
Pause and explain
Why is water omitted from the simple Ka expression?
Check your reasoning after answering
Its solvent activity is treated as effectively constant in the dilute aqueous approximation and included in the constant.
Part 2 · Use an initial-change-equilibrium table
An ICE table records initial concentrations, their stoichiometric changes and the equilibrium values. For the selected monoprotic acid starting with negligible conjugate base, a change x lowers HA by x and raises hydronium and A⁻ by x. Substituting into Ka gives x²/(C−x). This simple table neglects water autoionization and assumes no added common ion; those assumptions must be checked for very dilute or unusual conditions. ICE is bookkeeping, not a new law independent of conservation and equilibrium.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why is the equilibrium HA concentration C−x rather than C+x?
Visual reasoning — after your attempt
HA is consumed as the forward proton-transfer amount x forms its products. The sign follows the chosen reaction direction and stoichiometry.
For a simple initially un-ionized monoprotic acid concentration C, write equilibrium concentrations as x,x,C−x when water’s contribution is negligible. Then Ka=x²/(C−x). The table links stoichiometry to equilibrium and prevents substituting initial acid concentration as its ionized amount.
A weak acid does not generally give [H₃O⁺]=C. Solve for x or use a justified approximation.
Pause and explain
What does x represent in this simple table?
Check your reasoning after answering
It is the equilibrium amount per volume ionized, also approximating hydronium and conjugate-base concentrations under the specified initial and water-neglect assumptions.
Part 3 · Validate the small-ionization approximation
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.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What calculation tests the assumption C−x≈C in this example?
Visual reasoning — after your attempt
Compute x/C, approximately 0.00100/0.100=0.0100, or one percent, and compare with the desired accuracy.
If x is much smaller than C, replace C−x by C and estimate x≈√(KaC). Check x/C; a conventional roughly 5% check helps decide whether the simplification is adequate for the task. Extremely dilute cases may also require water autoionization.
For Ka=1.0×10⁻⁵,C=0.100 M, x≈0.00100 M and x/C≈1%, supporting the approximation.
Pause and explain
What should happen if the calculated ionization fraction is too large?
Check your reasoning after answering
Use the full equilibrium equation or another suitable method rather than keep an unjustified C−x≈C approximation.
Part 4 · Relate conjugate strengths with conditions
For a corresponding conjugate acid/base pair in water, consistent equilibrium definitions give KaKb=Kw. A stronger acid therefore has a weaker conjugate base under the same conditions. This does not mean equal solution concentrations or equal pH values, and comparing constants from different temperatures or solvents requires care. A salt of a weak acid can produce basic behavior through its conjugate base’s reaction with water; charge alone does not determine whether every ion solution is acidic or basic.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Which conjugate base is weaker in the displayed comparison, A⁻ or B⁻?
Visual reasoning — after your attempt
B⁻ is weaker because its parent HB has the larger Ka, giving the smaller corresponding Kb at the same Kw.
For a conjugate pair at one temperature, KaKb=Kw under consistent definitions. A stronger acid has a weaker conjugate base. pKa=−log Ka is another scale: lower pKa corresponds to larger Ka. This relationship does not make a weak-acid solution neutral.
At 25°C with Kw=1.0×10⁻¹⁴, Ka=1.0×10⁻⁵ gives Kb=1.0×10⁻⁹ for its conjugate base.
Pause and explain
Why does a smaller pKa mean a stronger acid?
Check your reasoning after answering
The negative logarithm reverses the order: smaller pKa corresponds to larger Ka and greater ionization tendency under comparable conditions.
Apply the whole unit · Estimate a weak-acid pH and check the approximation
Illustrative monoprotic acid: C=0.100 M,Ka=1.0×10⁻⁵ at 25°C; initially negligible conjugate base. Use Kw=1.0×10⁻¹⁴ and neglect water contribution if justified.
Your task
Write the equilibrium table and equation, estimate hydronium and pH, check percent ionization, and find conjugate-base Kb. Explain the distinction between weak and dilute.
What your response should show
- Choose the appropriate chemical model or relationship and state its conditions.
- Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
- Use supplied evidence to justify the conclusion; distinguish observation from inference.
- Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
Equilibrium: HA=C−x,H₃O⁺≈x,A⁻≈x.
Estimate x=√(KaC)=0.00100 M.
Ionization fraction=0.00100/0.100=1.0%, supporting the small-x approximation.
pH≈3.00; Kb=Kw/Ka=1.0×10⁻⁹.
The supplied 0.100 M acid is weak by its equilibrium behavior but not described as weak merely because of its concentration. The approximation is internally consistent here and hydronium is far above the water-only scale. A different concentration could require the full equation or a water contribution; the shortcut is conditional.
Scientific references and source-model foundations
Parent / teacher teaching notes
Check the equilibrium table before the square-root calculation. Ask students to verify the approximation and explain how concentration and strength differ. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.
The key idea
Strength and concentration are independent descriptors; acid equilibria determine hydronium concentration.
Worked examples
Example 1
- x≈√(10⁻⁵×.10)=.0010M.
- x/C=.01, so the small-x assumption is reasonable.
- pH=−log(.0010).
Example 2 · Worked application
- Larger K_a indicates more favorable ionization in water.
- HA has larger K_a, so it is stronger as an acid under these conditions.
- Equal concentration separates strength from amount present.
A smaller pK_a corresponds to larger K_a.
Example 3
- NH₃ accepts H⁺ to become NH₄⁺.
- H₂O donates H⁺ to become OH⁻.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: A smaller pK_a corresponds to larger K_a.
- Strength concerns equilibrium ionization, while concentration concerns amount per solution volume.
A smaller pK_a corresponds to larger K_a.
Example 5 · Reason from the visual model
- 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.
Original selected calculation with a hypothetical Ka. The quadratic value is computed from x²/(C−x)=Ka under the same simplified species assumptions.
Avoid these mistakes
- Do not infer actual pH from strength alone.
Check your understanding
What distinguishes a conjugate base from a different unrelated base?
Discuss the answer
It is the species remaining after that particular acid donates one proton.
Return to the phenomenon
Return to the initial case. When is the square-root weak-acid shortcut justified? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
The shortcut is conditional on small relative ionization and other neglected contributions being small. Validate it with x/C and use the fuller model when the assumptions fail.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. When is the square-root weak-acid shortcut justified? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the chemical mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep species identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
The shortcut is conditional on small relative ionization and other neglected contributions being small. Validate it with x/C and use the fuller model when the assumptions fail.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
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.
Success 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.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
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.
Words to know
- conjugate pair
- Species that differ by one proton.
- Ka
- Acid-ionization equilibrium constant for a specified temperature.
- pKa
- Negative base-ten logarithm of Ka.
In the real world
Use the chapter investigation to connect this unit to measured evidence and a practical decision.
Exit check
What distinguishes a conjugate base from a different unrelated base?
Check your explanation after answering
It is the species remaining after that particular acid donates one proton.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
6
−log 10(10⁻⁶)=6.
Problem 2
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
Hints
List the given quantities, select a relationship, and check units before substituting.
Answer and explanation
4
−log 10(10⁻⁴)=4.
Problem 4
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
2
−log(.010)=2.
Problem 5
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
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
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
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.