Chemistry of Aqueous Solutions – Acid–Base & Solubility Equilibria

Flashcards covering key definitions, equations, qualitative explanations, calculations, buffer theory, salt hydrolysis, titration curves, indicators, and solubility product drawn from the detailed lecture notes on Chemistry of Aqueous Solutions.

According to the Brønsted–Lowry theory, what is an acid and what is a base?

An acid is a proton (H⁺) donor, while a base is a proton (H⁺) acceptor.

How does a strong acid behave in water?

It dissociates completely, producing essentially 100 % H⁺ (or H₃O⁺) ions.

How does a weak acid behave in water?

It dissociates only partially, establishing an equilibrium in which most molecules remain undissociated.

Write the mathematical definition of pH.

pH = –log₁₀[H⁺]

Write the mathematical definition of pOH.

pOH = –log₁₀[OH⁻]

Give the expression for the ionic product of water, K₍w₎.

K_w = [H⁺][OH⁻]

What is the numerical value of K₍w₎ at 25 °C?

1.0 × 10⁻¹⁴ mol² dm⁻⁶

State the relationship linking Kₐ, Kb and Kw for a conjugate acid–base pair.

Ka × Kb = K_w

Express the relationship between pKₐ, pKb and pKw at 25 °C.

pKa + pKb = 14 (because pK_w = 14 at 25 °C)

Define the acid dissociation constant, Kₐ, for a weak acid HA.

K_a = [H⁺][A⁻]/[HA] at equilibrium in aqueous solution.

Define the base dissociation constant, K_b, for a weak base B.

K_b = [BH⁺][OH⁻]/[B] at equilibrium in aqueous solution.

How is pKₐ related to Kₐ?

pKa = –log₁₀ Ka

How is pKb related to Kb?

pKb = –log₁₀ Kb

Write the simplified formula for [H⁺] in a solution of a weak monobasic acid HA of initial concentration C.

[H⁺] ≈ √(K_a × C)

Write the simplified formula for [OH⁻] in a solution of a weak base B of initial concentration C.

[OH⁻] ≈ √(K_b × C)

What is meant by the ‘degree of ionisation’ (α) of an acid or base?

α = (amount ionised) / (initial amount); the fraction of molecules that dissociate into ions.

What range of α values corresponds to a strong acid?

α ≈ 1 (essentially complete ionisation)

What range of α values corresponds to a weak acid?

0 < α < 1 (partial ionisation)

Explain qualitatively why increasing temperature increases K₍w₎.

Self-ionisation of water is endothermic; raising T shifts equilibrium toward more ions, increasing [H⁺] and [OH⁻] and hence K_w.

At 10 °C pure water has pH 7.26. Is the solution acidic, basic or neutral?

Neutral, because [H⁺] = [OH⁻] even though pH is above 7.

Give two common strong acids and two common weak acids.

Strong: HCl, H₂SO₄. Weak: CH₃COOH, H₂CO₃.

Give two common strong bases and two common weak bases.

Strong: NaOH, KOH. Weak: NH₃, CH₃NH₂.

Why must the 1 × 10⁻⁷ mol dm⁻³ contribution of H⁺ from water sometimes be included in pH calculations?

When the acid concentration is ≤ 10⁻⁷ mol dm⁻³, the autoprotolysis of water is no longer negligible compared with added H⁺.

Write the Henderson–Hasselbalch equation for an acidic buffer.

pH = pK_a + log([salt]/[acid])

Write the Henderson–Hasselbalch–type equation for an alkaline buffer.

pOH = pK_b + log([salt]/[base]) (then pH = 14 – pOH)

What two components are required to prepare an acidic buffer?

A weak acid and a salt of that acid (providing its conjugate base).

What two components are required to prepare an alkaline buffer?

A weak base and a salt of that base (providing its conjugate acid).

Explain how an acidic buffer resists pH change upon addition of a small amount of acid.

Added H⁺ is removed by reaction with the conjugate base A⁻ to form HA, shifting equilibrium and minimising [H⁺] increase.

Explain how an acidic buffer resists pH change upon addition of a small amount of base.

Added OH⁻ reacts with HA to form A⁻ and water, removing OH⁻ and restoring [H⁺].

State the condition for maximum buffer capacity for an acidic buffer.

[acid] = [salt] (thus pH = pK_a)

State the condition for maximum buffer capacity for an alkaline buffer.

[base] = [salt] (thus pOH = pK_b)

Describe the role of the H₂CO₃/HCO₃⁻ buffer in blood.

It maintains blood pH near 7.4 by neutralising excess H⁺ with HCO₃⁻ or excess OH⁻ with H₂CO₃; CO₂ exhalation shifts equilibria to restore balance.

What is the ‘common-ion effect’ in solubility equilibria?

The decrease in solubility of an ionic salt when a solution already contains one of its ions, shifting the dissolution equilibrium left.

Define the solubility product, K₍sp₎.

For a sparingly soluble salt, K_sp is the equilibrium constant for its dissolution, equal to the product of the ionic concentrations each raised to the power of their stoichiometric coefficients.

How can K₍sp₎ be used to predict precipitation?

Compare the ionic reaction quotient Q with Ksp; if Q > Ksp, precipitation occurs.

Write the relationship between solubility (s) and K₍sp₎ for AgCl.

K_sp = s² (because [Ag⁺] = [Cl⁻] = s)

When does a salt solution become acidic upon dissolution?

If the cation is the conjugate acid of a weak base (e.g., NH₄⁺) or the anion does not react, producing excess H⁺ via hydrolysis.

Give an example of a basic salt and explain its basicity.

Sodium ethanoate (CH₃COO⁻ Na⁺); acetate ion hydrolyses to form CH₃COOH and OH⁻, raising pH.

Provide the hydrolysis equation for NH₄⁺ in water.

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

Provide the hydrolysis equation for CH₃COO⁻ in water.

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

State the colour change and pH range of methyl orange.

Red → Yellow; pH 3.1–4.4

State the colour change and pH range of phenolphthalein.

Colourless → Pink; pH 8.3–10.0 (approx.)

When titrating a strong acid with a strong base, why may almost any indicator be used?

The pH changes rapidly (≈4–10) near the equivalence point, encompassing most indicator ranges.

Which indicator is suitable for titrating a weak acid with a strong base and why?

Phenolphthalein (or thymolphthalein) because the equivalence pH is >7 (≈8–9), matching its working range.

Which indicator is suitable for titrating a strong acid with a weak base and why?

Methyl orange (or screened methyl orange) because the equivalence pH is <7 (≈4–5), within its working range.

Why is there no suitable indicator for a weak acid–weak base titration?

The pH change at the equivalence point is very gradual; no indicator changes colour over such a narrow, shallow region.

During a strong acid–strong base titration, what is the pH at the equivalence point and why?

pH = 7 because the salt formed does not hydrolyse, and [H⁺] = [OH⁻].

During a weak acid–strong base titration, why is the pH at equivalence > 7?

The anion of the weak acid hydrolyses to produce OH⁻, making the solution basic.

During a strong acid–weak base titration, why is the pH at equivalence < 7?

The cation of the weak base hydrolyses to produce H⁺, making the solution acidic.

Explain what is meant by the ‘buffer region’ on a titration curve.

A relatively flat portion before (or after) the equivalence point where a mixture of weak acid/base and its conjugate forms a buffer, so pH changes slowly.

At half-neutralisation of a weak acid with a strong base, what is the pH of the solution?

pH = pK_a because [acid] = [conjugate base].

What is meant by a ‘polybasic’ (polyprotic) acid?

An acid that can donate more than one proton per molecule, dissociating step-wise with distinct K_a values.

How many equivalence points appear when titrating H₃PO₄ with NaOH?

Three, corresponding to neutralisation of each acidic proton.

At what pH values do maximum buffer capacities occur in the H₃PO₄/NaOH titration?

At pH ≈ pKa₁ (2.15), pKa₂ (7.20) and pK_a₃ (12.35) where successive acid/base pairs are equimolar.

Give one industrial or biological application of buffer solutions other than blood.

Maintaining pH in fermentation media; preparing buffered intravenous injections; controlling pH in electrophoresis; buffering food products.

Why is Kₐ a better indicator of acid strength than pH or degree of ionisation α?

K_a is constant for a given acid at a given temperature and independent of concentration, whereas pH and α vary with concentration.

In calculating the pH of a basic salt solution, why might you need to convert Kₐ to K_b?

If only the Ka of the parent acid is given, Kb for the conjugate base must be obtained via Kb = Kw / K_a to use in [OH⁻] calculations.

What happens to pK_w (and consequently pH + pOH) when temperature increases?

Kw increases, so pKw decreases; therefore pH + pOH = pK_w < 14 at temperatures above 25 °C.

Write the general criterion for precipitation using the reaction quotient Q and K₍sp₎.

If Q > Ksp precipitation occurs; if Q < Ksp the solution is unsaturated; if Q = K_sp the solution is at equilibrium (saturated).

State Le Châtelier’s principle in the context of temperature dependence of K₍w₎.

If an endothermic reaction (like water ionisation) is heated, equilibrium shifts to produce more products, raising K_w.

What is the colour of phenolphthalein in an acidic solution of pH 4?

Colourless (because pH is below its transition range).

What is the colour of methyl orange at pH 9?

Yellow (beyond its transition range, which ends at about 4.4).

Why can pH sometimes be negative for very strong acids?

Because [H⁺] can exceed 1 mol dm⁻³; pH = –log[H⁺] therefore becomes negative.

What physical assumption allows [H₂O] to be omitted in equilibrium expressions such as Kₐ?

Water is the solvent and its molar concentration (~55.5 mol dm⁻³) is effectively constant, so it is incorporated into the equilibrium constant.

Describe the primary chemical event detected by an acid–base indicator during titration.

The indicator’s conjugate acid/base forms interchange dominance; its colour changes when the ratio [In⁻]/[HIn] passes roughly 10:1 or 1:10.

During strong base titration of a weak acid, why does buffering occur before equivalence?

The mixture contains both the unreacted weak acid and the produced conjugate base, forming a buffer that resists pH change.

How can K₍sp₎ be experimentally determined from solubility data?

Measure the molar solubility s in a saturated solution, then substitute ion concentrations into the K_sp expression.

Give the equation for calculating the pH of a solution containing a conjugate acid (of a weak base) at concentration C with Kₐ.

[H⁺] ≈ √(K_a × C); pH = –log[H⁺].