Acids & Bases (Cambridge (CIE) A Level Chemistry): Flashcards

pK_a = −log₁₀ K_a

A less positive pK_a (or higher K_a) indicates a stronger acid, because the equilibrium lies further to the right and the acid is more extensively ionised.

True.

K_w = [H⁺][OH^-] = 1.00 × 10^-14 mol² dm^-6 at 298 K. This value is used to convert between [H⁺] and [OH^-] in aqueous solutions.

pH is calculated using pH = −log₁₀ [H⁺], and [H⁺] can be found from the pH using [H⁺] = 10^-pH.

1. The concentration of H⁺ from the ionisation of water is negligible.

2. The dissociation of the weak acid is so small that the initial concentration of HA can be used as the equilibrium concentration.

False.

A high K_a value means the equilibrium lies to the right and the acid is almost completely ionised, making it strongly acidic. A low K_a indicates a weakly acidic species.

K_a = [H⁺]² / [CH₃COOH]

K_a = (1.32 × 10^-3)² / 0.100

K_a = 1.74 × 10^-5 mol dm^-3

An ionic product of water (K_w) is the equilibrium constant for the ionisation of water: K_w = [H⁺][OH^-] = 1.00 × 10^-14 mol² dm^-6 at 298 K.

For a 0.100 mol dm^-3 solution of ethanoic acid (K_a = 1.74 × 10^-5 mol dm^-3), [H⁺] = √(1.74 × 10^-5 × 0.100) and the pH = −log₁₀[H⁺].

pH = −log₁₀ [H⁺]

pH = −log₁₀ (1.6 × 10^-4)

pH = 3.80

A strong acid is an acid that completely ionises in aqueous solution, so [H⁺] equals the concentration of the acid. Examples include HCl, HNO₃ and H₂SO₄.

False.

For a strong acid, the contribution of H⁺ from water ionisation is negligible relative to [H⁺] from the acid and can be ignored. [H⁺] ≈ [acid].

[H⁺] = 10^-12.3 = 5.01 × 10^-13 mol dm^-3

[OH^-] = K_w / [H⁺] = (1.00 × 10^-14) / (5.01 × 10^-13)

[OH^-] = 0.0199 mol dm^-3

For a strong alkali, [H⁺] is found using [H⁺] = K_w / [OH^-] and then pH = −log₁₀ [H⁺].

True.

For a weak acid, [H⁺] = [A^-] and the simplified expression K_a = [H⁺]² / [HA] rearranges to [H⁺] = √(K_a × [HA]).

[H⁺] = √(K_a × [CH₃COOH])

[H⁺] = √(1.74 × 10^-5 × 0.100)

[H⁺] = 1.32 × 10^-3 mol dm^-3

A pH scale is a measure of acidity or basicity of a solution, running from 0 to 14. Acids have pH below 7, pure water is neutral at pH 7 (at 25°C), and bases/alkalis have pH above 7.

A solution has pH 3.1. Its [H⁺] = 10^-3.1 = 7.9 × 10^-4 mol dm^-3.

A buffer solution is a solution whose pH does not change significantly when small amounts of acid or alkali are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid).

The added H⁺ ions react with the reserve supply of CH₃COO^- ions from the sodium ethanoate:

CH₃COO^- + H⁺ → CH₃COOH

The equilibrium shifts left, consuming the added H⁺. Because there is a large reservoir of CH₃COO^-, the pH does not change significantly.

False.

A buffer can only maintain approximately constant pH when small amounts of acid or alkali are added. Excessive addition overwhelms the buffer and the pH changes significantly.

Sodium ethanoate is a salt that fully ionises in solution:

CH₃COONa → Na⁺ + CH₃COO^-

This provides a high concentration of CH₃COO^- ions to neutralise any added acid.

When OH^- ions are added to a buffer, they react with H⁺ to form water, causing the equilibrium to shift right as more weak acid ionises.

True.

CO₂ from respiration dissolves in blood: CO₂ + H₂O ⇌ H⁺ + HCO₃^-. This equilibrium buffers the blood pH. If H⁺ rises, the equilibrium shifts left, and if H⁺ falls, it shifts right.

Acidosis is the condition in which there is too much acid in the blood (pH falls below 7.35). It can cause body malfunctioning, organ damage and eventually coma if the blood pH is not regulated.

A conjugate base in a buffer solution is the species formed when the weak acid donates a proton. It acts as a reservoir to neutralise added acid by accepting H⁺ ions.

A buffer resists pH change because it contains reserve supplies of both a weak acid and its conjugate base to neutralise added acid or alkali respectively.

[H⁺] = K_a × [acid] / [salt]

• K_a = acid dissociation constant

• [acid] = equilibrium concentration of the weak acid

• [salt] = equilibrium concentration of the conjugate base

A Henderson-Hasselbalch equation is: pH = pK_a + log₁₀([salt] / [acid]). It is used to calculate the pH of a buffer solution from the pK_a and the concentrations of the weak acid and its conjugate base.

True.

Because the weak acid is only slightly ionised, its concentration barely changes on reaching equilibrium. The initial concentration is used as an approximation for [acid] in the K_a expression.

The K_a expression for a buffer containing CH₃COOH and CH₃COO^- is: K_a = [CH₃COO^-][H⁺] / [CH₃COOH] , rearranged to [H⁺] = K_a × [CH₃COOH] / [CH₃COO^-].

[H⁺] = K_a × [acid] / [salt]

[H⁺] = 1.43 × 10^-5 × (0.305 / 0.520)

[H⁺] = 8.39 × 10^-6 mol dm^-3

pH = −log(8.39 × 10^-6) = 5.08

False.

Increasing [salt] / [acid] increases log₁₀([salt] / [acid]) in the Henderson-Hasselbalch equation, which increases the pH, making the buffer more basic.

The salt (e.g. sodium ethanoate) fully ionises to provide a high concentration of the conjugate base (CH₃COO^-). The contribution of conjugate base from the partial ionisation of the weak acid is negligible by comparison.

Using logarithms, the buffer pH equation becomes: pH = pK_a + log₁₀( [salt] / [acid] ).

A buffer capacity is the ability of a buffer to resist changes in pH. It depends on the concentrations of the weak acid and conjugate base. Higher concentrations provide greater capacity to neutralise added acid or alkali.

A solubility product (K_sp) is the product of the concentrations of each ion in a saturated solution of a sparingly soluble salt at 298 K, each raised to the power of its stoichiometric coefficient.

Ca(OH)₂ (s) ⇌ Ca²⁺ (aq) + 2OH^- (aq)

K_sp = [Ca²⁺(aq)][OH^-(aq)]²

Units: mol dm^-3 × (mol dm^-3)² = mol³ dm^-9

False.

A smaller K_sp value indicates a less soluble (more sparingly soluble) salt. The lower the K_sp, the lower the concentration of ions in a saturated solution.

For MgCl₂ (s) ⇌ Mg²⁺ (aq) + 2Cl^- (aq), the K_sp expression is: K_sp = [Mg²⁺(aq)] × [Cl^-(aq)]²

K_sp applies only to sparingly soluble salts. For highly soluble salts such as Group 1 salts and all nitrates, the ions are at very high concentrations and the concept of a saturated solution equilibrium does not apply in the same way.

False.

The undissolved solid is not included in the K_sp expression. Only the concentrations of the aqueous ions are used, as the concentration of a pure solid is constant.

Fe₂O₃ (s) ⇌ 2Fe³⁺ (aq) + 3O^2- (aq)

K_sp = [Fe³⁺(aq)]²[O^2-(aq)]³

Units: (mol dm^-3)² × (mol dm^-3)³ = mol⁵ dm^-15

A saturated solution is a solution that contains the maximum amount of dissolved salt at a given temperature. Any additional solute remains undissolved and a dynamic equilibrium exists between dissolved ions and the solid.

For SnCO₃ (s) ⇌ Sn²⁺ (aq) + CO₃^2- (aq), the K_sp expression is K_sp = [Sn²⁺(aq)][CO₃^2-(aq)] and the units are mol² dm^-6.

PbBr₂ ⇌ Pb²⁺ + 2Br^-

[Pb²⁺] = 1.39 × 10^-3 mol dm^-3

[Br^-] = 2 × 1.39 × 10^-3 = 2.78 × 10^-3 mol dm^-3

K_sp = (1.39 × 10^-3)(2.78 × 10^-3)² = 1.07 × 10^-8 mol³ dm^-9

A solubility (molar) is the concentration of a saturated solution of a salt, expressed in mol dm^-3. It represents the number of moles of solute that dissolve per dm³ of solvent to form a saturated solution.

False.

For AB₂ ⇌ A²⁺ + 2B^-, [B^-] = 2*s (not s*), because two moles of B^- are produced for every mole of AB₂ that dissolves.

To find solubility from K_sp for CuO (s) ⇌ Cu²⁺ (aq) + O^2- (aq):

K_sp = [Cu²⁺][O^2-] so [Cu²⁺] = √*K~sp*~.

CuO ⇌ Cu²⁺ + O^2-, so K_sp = [Cu²⁺][O^2-]

[Cu²⁺] = [O^2-] = s, so K_sp = s²

s = √(5.9 × 10^-36)

Solubility = 2.4 × 10^-18 mol dm^-3

False.

K_sp only applies to sparingly soluble salts. Soluble salts such as NaCl do not establish a meaningful equilibrium between solid and dissolved ions under normal conditions.

The balanced equation shows that one mole of PbBr₂ produces two moles of Br^- ions. So the concentration of Br^- at equilibrium is double the molar solubility of PbBr₂.

For PbBr₂ with solubility 1.39 × 10^-3 mol dm^-3:

K_sp = (1.39 × 10^-3) × (2.78 × 10^-3)² = 1.07 × 10^-8 mol³ dm^-9

A solubility product calculation is a method that uses the K_sp expression and stoichiometric ratios from the dissolution equation to convert between molar solubility and the product of ion concentrations in a saturated solution.

A common ion effect is the reduction in solubility of a sparingly soluble salt when a soluble compound containing a common ion is added to its saturated solution, shifting the equilibrium towards the solid.

KCl introduces extra Cl^- ions. These are common to the AgCl equilibrium: AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq).

The increased [Cl^-] means [Ag⁺][Cl^-] exceeds K_sp, so the equilibrium shifts left and AgCl precipitates.

True.

If [A^x+]^a[B^y-]^b > K_sp, the solution is supersaturated and the excess ionic compound precipitates until the product of ion concentrations returns to K_sp.

When KCl is added to saturated AgCl, [Cl^-] increases, the equilibrium shifts to the left and AgCl precipitates.

After mixing (equal volumes), [Ca²⁺] = 5.0 × 10^-4 mol dm^-3 and [SO₄^2-] = 1.0 × 10^-3 mol dm^-3.

Product = (5.0 × 10^-4)(1.0 × 10^-3) = 5.0 × 10^-7 mol² dm^-6

5.0 × 10^-7 < K_sp (2.0 × 10^-5), so no precipitate forms.

False.

The common ion effect decreases the solubility of a sparingly soluble salt. Adding a common ion shifts equilibrium towards the solid, reducing how much dissolves.

Calculate the product of the relevant ion concentrations after mixing. If this value is greater than K_sp, a precipitate forms. If it is less than or equal to K_sp, no precipitate forms.

A common ion is an ion that is present in two different ionic compounds dissolved in the same solution, such that adding one compound introduces extra ions already contributed by the other.

In AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq), if a solution of AgNO₃ is added, the Ag⁺ ion is common. Solubility of AgCl decreases.