Why can a salt solution be acidic or basic?
You will be able to: Relate Ka and Kb for a conjugate pair and predict salt hydrolysis.
Why can a salt solution be acidic or basic?
Dissolving sodium acetate separates ions, but the chemistry does not stop there. Acetate can accept a proton from water, making a basic solution.
A useful starting point: How does a weak base make hydroxide? →
Words and symbols before equations
- Conjugate-strength relation
- For a matching pair at one temperature, KaKb=Kw.
- Hydrolysis
- An ion reacts with water as an acid or base.
- Spectator ion
- An ion not participating appreciably in the acid–base equilibrium considered.
- pKa
- −log₁₀Ka; smaller pKa indicates a stronger acid.
What this picture assumes
Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. Only a matching acid/conjugate-base pair satisfies this relation. Temperature is fixed.
Read the picture in three steps
- Read the species and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- pKa=5.00; pKb=9.00. Ka×Kb=1.00e-14=Kw at 25 °C. A larger acid Ka means a smaller conjugate-base Kb.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
For HA/A⁻, acid donation and conjugate-base acceptance combine to water autoionization. The constants therefore satisfy Ka(HA)Kb(A⁻)=Kw at the same temperature.
A stronger acid has a weaker conjugate base. At 25 °C, pKa+pKb=14. This relationship applies to a matching conjugate pair, not arbitrary acids and bases.
Acetate is the conjugate base of a weak acid and hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Ammonium is a weak conjugate acid: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.
Na⁺ and Cl⁻ are treated as spectators in these dilute classroom cases. Salts with both reactive ions require comparing their acid/base equilibria; not every salt is neutral.
A worked example, step by step
For HA with Ka=1.0×10⁻⁵ at 25 °C, find Kb of A⁻ and predict whether sodium A is acidic, basic or neutral.
- Identify the conjugate pair HA/A⁻.
- Kb=Kw/Ka=10⁻¹⁴/10⁻⁵=1.0×10⁻⁹.
- Write A⁻ + H₂O ⇌ HA + OH⁻.
- The anion produces hydroxide, so its sodium salt is basic under these assumptions.
Do not multiply Ka and Kb of unrelated substances or assume that dissolving any salt gives pH 7.
If an acid’s Ka increases tenfold, what happens to its conjugate base’s Kb at fixed temperature?
Compare with an explanation
Kb decreases tenfold because their product stays Kw.
Predict. Change one thing. Explain.
Increase the acid pKa at fixed pKw. Predict whether its conjugate base becomes stronger or weaker, and explain the inverse relationship between the constants.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
pKa=5.00; pKb=9.00. Ka×Kb=1.00e-14=Kw at 25 °C. A larger acid Ka means a smaller conjugate-base Kb.
Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. Only a matching acid/conjugate-base pair satisfies this relation. Temperature is fixed.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using proton transfer, charge and atom conservation, a mole balance or the stated acid–base equilibrium. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA conjugate acid has pKa=9.00 at 25 °C. Find pKb and Kb of its base and explain which constant is larger.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: pKb=14.00−9.00=5.00.
- 1 point: Kb=1.0×10⁻⁵.
- 1 point: Ka=1.0×10⁻⁹, so Kb is larger for this pair.
- 1 point: The comparison uses a matching conjugate pair and the same temperature.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What connects conjugate Ka and Kb?
Their product is Kw.
RECALL 2Which acetate reaction explains basicity?
Acetate accepts H⁺ from water and produces OH⁻.
RECALL 3Is every salt neutral?
No; inspect the acid/base behavior of its ions.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why can a salt solution be acidic or basic?
- Matching pair: KaKb=Kw; pKa+pKb=pKw.
- Weak-acid anions can be basic; weak-base cations can be acidic.
Remember: Do not multiply Ka and Kb of unrelated substances or assume that dissolving any salt gives pH 7.
Conditions: Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. Only a matching acid/conjugate-base pair satisfies this relation. Temperature is fixed.
Refresh Kid · AP Chemistry Unit 8 · Objectives 8.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.3, objective 8.3.A. CED effective Fall 2024 and June 2026 clarifications checked September 17, 2026. Unit 8: Acids and Bases, Topics 8.1–8.11. Focused lesson names, examples, models and assessments are original Refresh Kid teaching materials, not additional official topics or official AP questions. Official corrections.
The model states its assumptions beside the diagram. Dilute ideal-solution concentrations approximate activities; numerical models use 25 °C and Kw=1.00×10⁻¹⁴ unless another pKw is supplied. pH need not be restricted to 0–14 in all real solutions. The optional 3D views show original schematic molecular geometry, not a measured trajectory or a reaction mechanism. Computation of a buffer’s pH change after adding acid/base, derivation of Henderson–Hasselbalch, concentrations of every species in a polyprotic titration, and solubility as a function of pH are excluded from assessed scope. Buffer response and pH-dependent solubility are taught qualitatively. Calculating the pH of a buffer formed by partial neutralization remains in Topic 8.4 scope.
Teaching resources: The Organic Chemistry Tutor video titles/descriptions and topic coverage were checked for optional links; no claim is made to have watched every video. No creator scripts, examples, worksheets or artwork were copied. GitHub’s 3D website collection and its Three.js camera-control example informed the idea of controllable spatial inspection. Scientific diagrams, geometry and interactions here are original. The self-hosted Three.js runtime retains its MIT license. Camera rotation changes the view, not the chemistry.
Independent teacher review and observation of students remain pending. Implementation checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Chemistry questions and scoring guides. This archive contains questions across units; it is not an assignment of every question to this lesson.
The teaching sequence is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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