Acid-base Equilibria (Edexcel International A Level (IAL) Chemistry): Flashcards

The pH scale is logarithmic. This means each pH unit represents a 10-fold change in hydrogen ion concentration.

pH = -log(1.60 × 10^-4) = 3.80. pH values are given to 2 decimal places.

[H⁺] = 10^-3.10 = 7.94 × 10^-4 mol dm^-3. Use [H⁺] = 10^-pH to find the concentration.

True.

The pH scale is logarithmic with base 10, so a decrease of one pH unit corresponds to a 10-fold increase in [H⁺].

The volume increases by a factor of 100, reducing [H⁺] by a factor of 10^-2. The pH increases by 2 units, giving a final pH of 3.

pH values are typically given to 2 decimal places.

A strong acid is an acid that dissociates almost completely in aqueous solution. The equilibrium position lies so far to the right that the reaction is treated as irreversible.

pK_a = -log K_a. It converts the very small K_a values of weak acids into a more convenient scale, analogous to converting [H⁺] to pH.

A weak acid is an acid that only partially dissociates in aqueous solution. An equilibrium is established with the equilibrium position lying to the left.

False.

A higher K_a value indicates a greater degree of dissociation, meaning the acid is stronger, not weaker.

K_a is the equilibrium constant for the dissociation of a weak acid. A higher K_a value indicates a stronger, more dissociated acid.

When writing the K_a expression for a weak acid, we assume the concentration of H⁺ from the ionisation of water is negligible.

1. Hydrochloric acid, HCl

2. Nitric acid, HNO₃

3. Sulfuric acid, H₂SO₄

False.

Ethanoic acid is a weak acid that only partially dissociates in aqueous solution. Its K_a is approximately 1.74 × 10^-5 mol dm^-3.

Most weak acids have pK_a values in the range of 3 to 7. This reflects the small K_a values associated with partial dissociation.

For a strong acid, [H⁺] equals the acid concentration because it fully dissociates. Substitute directly into pH = -log[H⁺].

[HCl] = [H⁺] = 0.01 mol dm^-3, so pH = -log(0.01) = 2.00.

For a weak acid, [H⁺] can be found using: [H⁺] = √(K_a × [HA])

You need the concentration of the acid [HA] and the acid dissociation constant K_a. These are substituted into [H⁺] = √(K_a × [HA]).

[H⁺] = √(1.74 × 10^-5 × 0.100) = 1.32 × 10^-3 mol dm^-3, so pH = -log(1.32 × 10^-3) = 2.88.

False.

The assumption is that [HA]_eqm ≈ [HA]_initial, because the degree of dissociation is very small. It is [H⁺] and [A^-] that are assumed equal to each other.

1. [H⁺] = [A^-], so K_a × [HA] = [H⁺]²

2. [HA]_eqm ≈ [HA]_initial, because dissociation is negligible

For a strong acid, the [H⁺] from the ionisation of water is very small and can be neglected in pH calculations.

The second ionisation step (HSO₄^- ⇌ SO₄^2- + H⁺) is an equilibrium, not complete. The abundance of H⁺ from the first step suppresses the second, so [H⁺] is less than double the acid concentration.

K_w = [H⁺][OH^-]. At 25 °C, K_w = 1 × 10^-14 mol² dm^-6.

Rearrange K_w = [H⁺][OH^-] to give [H⁺] = K_w ÷ [OH^-], then apply pH = -log[H⁺].

pK_w = -log K_w. It is the negative logarithm of the ionic product of water, analogous to pH.

For a strong base, [OH^-] equals the concentration of the base because the base fully ionises in solution.

[H⁺] = (1 × 10^-14) ÷ 0.15 = 6.67 × 10^-14 mol dm^-3, so pH = -log(6.67 × 10^-14) = 13.18.

True.

The small concentration of H⁺ in strong alkali solutions arises from the ionisation of water, which occurs in all aqueous solutions.

In a neutral solution, [H⁺] = [OH^-] = 1 × 10^-7 mol dm^-3, giving a pH of 7.

In an acidic solution at 25 °C, [H⁺] is greater than [OH^-], and the pH is less than 7.

Measure the pH of equimolar aqueous solutions at the same temperature. The higher the pH, the weaker the acid.

The pH is 7 at 25 °C. Both NaCl and KNO₃ are examples of salts that form neutral solutions.

False.

At such low concentrations, the contribution of H⁺ from the ionisation of water cannot be ignored. The pH would be close to 7, not 8.

NH₄Cl is formed from a strong acid (HCl) and a weak base (NH₃). The NH₄⁺ ion donates a proton to water, producing H⁺ ions and making the solution acidic.

A salt formed from a weak acid and a strong base will be alkaline because the anion hydrolyses water to produce OH^- ions.

Diluting a strong acid by a factor of 10 increases the pH by 1 unit, because [H⁺] decreases by a factor of 10.

Diluting a weak acid by a factor of 10 increases the pH by approximately 0.5 units. This is smaller than for a strong acid because more dissociation occurs as the acid is diluted.

True.

CH₃COONH₄ is formed from a weak acid (CH₃COOH) and a weak base (NH₃) of similar relative strengths, resulting in a neutral solution.

1. Strong acid + strong base

2. Weak acid + strong base

3. Strong acid + weak base

4. Weak acid + weak base

All pH curves have an S-shape, with the midpoint of the inflection being the equivalence point.

The equivalence point is the midpoint of the near-vertical section of a pH curve, where the moles of acid and base have completely neutralised each other.

The equivalence point is above pH 7, because the salt formed (the conjugate base of the weak acid) is alkaline in aqueous solution.

At the half equivalence point in a weak acid-strong base titration, pH = pK_a

As the strong base neutralises the weak acid, a mixture of the weak acid and its conjugate base forms. This buffer mixture resists pH change, producing the gradually rising buffer region.

The half equivalence point is the stage of a titration where exactly half the weak acid has been neutralised, so [HA] = [A^-]. At this point, pH = pK_a.

The pH change at the equivalence point for a weak acid-strong base titration occurs around pH 8–10. Methyl orange changes colour well below this range, so it would change before the end-point is reached.

True.

Phenolphthalein changes colour in the range of approximately pH 8–10, which coincides with the steep near-vertical section of the pH curve for a weak acid-strong base titration.

1. Initial pH of the acid (y-intercept)

2. pH at the equivalence point

3. Volume of base at the equivalence point

4. pH range of the vertical section

A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali are added. It is used to keep pH approximately constant.

A buffer solution works because it contains a large reserve supply of both weak acid and its conjugate base.

An acidic buffer contains a weak acid and its conjugate base (usually supplied by a salt). For example, ethanoic acid and sodium ethanoate.

HCO₃^- (hydrogencarbonate ions) buffer the blood pH between 7.35 and 7.45. CO₂ from respiration dissolves in blood to form H⁺ and HCO₃^-.

The added H⁺ reacts with the conjugate base (e.g. CH₃COO^-) to form more weak acid (CH₃COOH). The equilibrium shifts left, keeping pH approximately constant.

True.

Excess weak acid is partially neutralised by the NaOH, producing a mixture of ethanoic acid and sodium ethanoate. This weak acid-conjugate base mixture acts as a buffer.

The OH^- reacts with H⁺ to form water. The equilibrium shifts right and more weak acid dissociates to replenish H⁺, keeping pH approximately constant.

False.

A buffer resists changes in pH but does not prevent them entirely. Adding a large amount of acid or alkali will eventually overwhelm the buffer and change the pH significantly.

Buffer capacity is a measure of the amount of acid or base required to significantly change the pH of the food. Foods with more protein have a higher buffer capacity.

The Henderson-Hasselbalch equation is: pH = pK_a + log([base]/[acid]). It allows the pH of a buffer solution to be calculated from the pK_a and the ratio of conjugate base to acid.

[H⁺] = K_a × ([acid]/[base]). Rearranging the K_a expression for the weak acid equilibrium gives this relationship.

[H⁺] = (1.74 × 10^-5) × (0.305/0.520) = 1.02 × 10^-5 mol dm^-3, so pH = -log(1.02 × 10^-5) = 4.99.

To make a buffer with pH below 7, you use a mixture of a weak acid and its conjugate base.

A higher [acid]/[base] ratio gives a lower pH (more acidic buffer). A lower ratio gives a higher pH (more alkaline buffer). The K_a of the weak acid determines the midpoint.

True.

At the half equivalence point, [acid] = [base], so log([base]/[acid]) = log(1) = 0. The Henderson-Hasselbalch equation therefore gives pH = pK_a.

[H⁺] = 1.00 × 10^-5 mol dm^-3. Using [H⁺] = K_a × ([acid]/[base]), the ratio = (1.00 × 10^-5) ÷ (1.74 × 10^-5) = 0.575.

In a buffer calculation using the K_a expression, we assume [A^-]_eqm is approximately equal to the concentration of the salt (conjugate base).