Acids, Bases and Equilibrium Practice

Introduction

Acid–base and solubility questions combine stoichiometry with equilibrium. First account for any strong acid or base added; then analyze what species remain and which equilibrium governs them. The order matters. A formula such as Henderson–Hasselbalch does not replace a mole balance, and a solubility product is not simply “the amount dissolved.” This practice set emphasizes choosing the right model before calculating pH or precipitation.

Core explanation

For dilute introductory aqueous work, pH is often approximated as −log₁₀[H₃O⁺] when concentration is in mol L⁻¹ relative to the standard concentration. Formally, pH uses hydrogen-ion activity. A strong monoprotic acid at low concentration can be treated as fully dissociated to first approximation, but at very low concentrations water's autoionization matters. A weak acid requires an equilibrium calculation, for example Kₐ = a(H⁺)a(A⁻)/a(HA) for HA ⇌ H⁺ + A⁻. If concentrations replace activities, state the dilute-solution approximation. “Strong acid” refers to dissociation tendency, while “concentrated acid” refers to amount per volume.

A buffer contains appreciable amounts of a weak acid and its conjugate base. Before estimating pH after a strong-acid addition, use the essentially complete neutralization step H⁺ + A⁻ → HA to update mole amounts. Then the idealized Henderson–Hasselbalch relation pH ≈ pKₐ + log₁₀(n A⁻/n HA) can be useful when both species remain, volume factors cancel and activities are suitably approximated. If the base component is exhausted, it is no longer a functioning conjugate-pair buffer. Buffer capacity depends on total amounts, not only on initial pH.

For dissolution MₓAᵧ(s) ⇌ xMʸ⁺ + yAˣ⁻ in a simple ionic notation, the solubility product is an activity product of the dissolved ions raised to stoichiometric powers; pure solid activity is taken as one in its standard state. A reaction quotient Q has the same expression using current activities. If Q < Ksp , dissolution is thermodynamically favored toward saturation; if Q > Ksp , precipitation may be favored, though kinetics and complexation matter. Solubility is the equilibrium dissolved amount under specific conditions. A common ion can lower the dissolved amount for a simple system, while complex formation or pH changes can alter it in other directions.

Step-by-step reasoning

1. Write the relevant balanced acid–base or dissolution reaction. 2. Convert supplied volumes and concentrations to moles before any mixing calculation. 3. Complete dominant stoichiometric neutralization steps and identify remaining species. 4. Choose a justified equilibrium expression and state activity approximations. 5. Check whether a buffer pair remains or whether a solid phase is present. 6. Evaluate the numerical result against plausible pH or concentration limits.

Visual explanation

Picture a two-stage flowchart. Stage one has a mole table for strong acid and conjugate base; stage two has an equilibrium balance for the remaining pair. A separate beaker diagram shows dissolved ions above a solid salt, with Q and Ksp compared like two markers on a scale. The layout makes clear that stoichiometric consumption precedes an equilibrium approximation.

Real-world analogy

A reservoir can absorb a modest inflow without a large change in level, but once its capacity is exhausted the response changes sharply. That resembles a buffer's finite capacity. The analogy does not replace the logarithmic pH relation or the specific chemistry of conjugate acid–base pairs.

Real-world example

In a biological assay, a phosphate buffer may keep pH near a desired range while the enzyme produces some acid. If acid production consumes most of the conjugate base, the pH can shift sharply and enzyme activity may change. A protocol should specify buffer species, concentration, starting pH, temperature and anticipated acid production. Reporting only “buffered at pH 7” hides capacity and ionic-medium effects.

Why?

Why not calculate pH directly from the starting buffer ratio after adding HCl? HCl reacts with the buffer base, changing both numerator and denominator. Using the original ratio ignores stoichiometry and can predict no change even when much of the buffer is consumed. Equilibrium calculations must begin from the correct post-reaction composition.

Common misconception

“Neutralization always ends at pH 7.” Weak-acid conjugate bases can shift equivalence pH. “A buffer never changes pH.” It resists finite perturbations. “Ksp is a concentration in mol L⁻¹.” It is a dimensionless equilibrium constant under a defined activity convention. “Q > Ksp guarantees immediate visible precipitate.” Nucleation and complexation can delay or alter observation.

Worked example

Mix 100.0 mL of a buffer containing 0.100 mol L⁻¹ HA and 0.100 mol L⁻¹ A⁻ with 10.0 mL of 0.100 mol L⁻¹ HCl. Initially there are 0.0100 mol of each buffer component; HCl adds 0.00100 mol H⁺. After neutralization, A⁻ is 0.00900 mol and HA is 0.0110 mol. Under a dilute idealized buffer approximation, pH ≈ pKₐ + log₁₀(0.00900/0.0110) = pKₐ − 0.087. Both buffer components remain. If ten times more HCl had been added, A⁻ would be exhausted and this formula would not be valid for the final state.

Quick check

1. Must you update buffer mole amounts after adding strong acid before using a ratio formula? Answer: Yes. Strong acid consumes conjugate base and forms weak acid. 2. Does the pure solid appear in the usual Ksp expression? Answer: No. Its activity is one in the specified standard state.

Exam focus

Show mole tables and equilibrium expressions separately. State whether pH uses an activity or dilute concentration approximation. Check buffer capacity before applying Henderson–Hasselbalch. In precipitation questions, write the dissolution stoichiometry, then compare the correct ion activity product with Ksp .

Advanced insight

At higher ionic strength, activity coefficients can make concentration-based pH and solubility estimates inaccurate. Polyprotic acids have several dissociation steps, and a “single pKa” buffer approximation may fail if multiple steps overlap. Metal complexation can increase total dissolved metal even while free-ion concentration remains constrained by a solubility product. A full speciation calculation may therefore be needed for environmental or analytical samples.

Summary

Acid–base and solubility practice begins with reaction stoichiometry, then equilibrium. pH is activity-based, buffer formulas have capacity and approximation limits, and Ksp refers to a defined activity product. Naming the species and conditions is as important as arithmetic.

Practice questions

1. Approximate pH of dilute 1.0 × 10⁻³ mol L⁻¹ HCl at 25 °C, assuming complete dissociation and ideality. Answer: pH ≈ 3.00. 2. A buffer initially has equal HA and A⁻. If a little strong acid is added, which amount increases? Answer: HA increases as A⁻ consumes H⁺. 3. Write Ksp for AgCl(s) ⇌ Ag⁺ + Cl⁻ using activities. Answer: Ksp = a(Ag⁺)a(Cl⁻) . 4. What does Q < Ksp indicate for a system with solid present under the stated model? Answer: Net dissolution is thermodynamically favored until equilibrium is approached.