Define a Brønsted–Lowry acid and a Brønsted–Lowry base .

A Brønsted–Lowry acid is a species that can donate a proton (H + ). A Brønsted–Lowry base is a species that can accept a proton (H + ).

True or False? A conjugate acid-base pair differs by one H + ion.

True. A conjugate acid-base pair is two species related by the transfer of a single proton. The acid loses H + to give the conjugate base; the base gains H + to give the conjugate acid.

In the reaction CH 3 COOH (aq) + H 2 O (l) ⇌ CH 3 COO - (aq) + H 3 O + (aq), the acid is .......... and its conjugate base is .......... . The base is .......... and its conjugate acid is .......... .

In the reaction CH 3 COOH (aq) + H 2 O (l) ⇌ CH 3 COO - (aq) + H 3 O + (aq), the acid is CH 3 COOH and its conjugate base is CH 3 COO - . The base is H 2 O and its conjugate acid is H 3 O + .

Identify the two conjugate acid-base pairs in: HCO 3 - (aq) + H 2 O (l) ⇌ CO 3 2- (aq) + H 3 O + (aq)

Pair 1: HCO 3 - (acid) and CO 3 2- (conjugate base) Pair 2: H 3 O + (conjugate acid) and H 2 O (base)

What does it mean to say water is amphoteric in the Brønsted–Lowry model?

Water is amphoteric because it can act as both a Brønsted–Lowry acid (donating H + ) and a Brønsted–Lowry base (accepting H + ) depending on the other species present in the reaction.

In the reaction H 2 SO 4 (aq) + HNO 3 (aq) ⇌ HSO 4 - (aq) + NO 2 + (aq) + H 2 O (l), H 2 SO 4 acts as the .......... and HNO 3 acts as the .......... .

In the reaction H 2 SO 4 (aq) + HNO 3 (aq) ⇌ HSO 4 - (aq) + NO 2 + (aq) + H 2 O (l), H 2 SO 4 acts as the acid and HNO 3 acts as the base .

Give the equations linking pH and [H + ].

pH = −log[H + ] [H + ] = 10 −pH where [H + ] is in mol dm −3 . pH values are normally given to 2 decimal places.

True or False? The pH scale is linear — a decrease of 1 pH unit corresponds to a decrease of 1 in [H + ].

False. The pH scale is logarithmic (base 10). A decrease of 1 pH unit corresponds to a 10-fold increase in [H + ]. For example, pH 5 is 10 times more acidic than pH 6.

A solution has [H + ] = 1.60 × 10 -4 mol dm -3 . Its pH = −log( .......... ) = .......... . A solution with pH = 3.10 has [H + ] = 10 .......... = .......... mol dm -3 .

A solution has [H + ] = 1.60 × 10 -4 mol dm -3 . Its pH = −log( 1.60 × 10 -4 ) = 3.80 . A solution with pH = 3.10 has [H + ] = 10 -3.10 = 7.94 × 10 -4 mol dm -3 .

Why is the [H + ] from a strong acid assumed to equal the acid concentration when calculating pH?

Strong acids are completely ionised in solution (HA → H + + A - ), so every mole of acid produces one mole of H + . The tiny contribution of H + from water is negligible in comparison.

For 0.01 mol dm -3 HCl: [H + ] = .......... mol dm -3 , so pH = −log( .......... ) = .......... .

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

True or False? 0.1 mol dm -3 sulfuric acid gives [H + ] = 0.2 mol dm -3 because H 2 SO 4 is diprotic and fully ionises in both steps.

False. H 2 SO 4 is a diprotic acid, but the second ionisation (HSO 4 - ⇌ SO 4 2- + H + ) is only partial. This means [H + ] is greater than the acid concentration but less than twice it. At A Level, the pH of dilute H 2 SO 4 is calculated assuming only the first (complete) ionisation contributes.

10.0 cm 3 of a solution at pH 1.0 is diluted to 1000.0 cm 3 with distilled water. What is the pH of the final solution?

The volume increased by a factor of 100, so [H + ] decreased by a factor of 100 (= 10 2 ). Each factor of 10 decrease in [H + ] increases pH by 1, so pH increases by 2 . Final pH = 3 .

A diprotic acid is an acid that can donate two protons (H + ions) per molecule in aqueous solution, such as sulfuric acid (H 2 SO 4 ) or carbonic acid (H 2 CO 3 ).

What is the ionic product of water, K w ?

Ionic product of water is the equilibrium constant for the dissociation of water: K w = [H + ][OH - ], which equals 1 x 10 -14 mol 2 dm -6 at 298 K.

Complete the expression for the ionic product of water and give its value at 298 K. K w = [ .......... ][ .......... ] = .......... mol 2 dm -6

K w = [ H + ][ OH - ] = 1 x 10 -14 mol 2 dm -6

True or False? As temperature increases, the value of K w increases and the pH of pure water decreases.

True. The dissociation of water is endothermic. Increasing temperature shifts the equilibrium right, producing more H + and OH - ions, raising K w and lowering the pH of pure water below 7.

Why is pure water neutral even when its pH falls below 7 at higher temperatures?

Neutrality requires [H + ] = [OH - ], not pH = 7. At higher temperatures K w increases but [H + ] still equals [OH - ], so the solution remains neutral even though pH < 7.

For a strong base fully ionised in solution, the steps to find pH are: [OH - ] = .......... (equal to base concentration) [H + ] = K w ÷ .......... pH = ..........

[OH - ] = [BOH] [H + ] = K w ÷ [OH - ] pH = -log[H + ]

Calculate the pH of 0.15 mol dm -3 NaOH. ( K w = 1 x 10 -14 mol 2 dm -6 )

[H + ] = (1 x 10 -14 ) ÷ 0.15 = 6.67 x 10 -14 mol dm -3 pH = -log(6.67 x 10 -14 ) = 13.18

True or False? The concentration of OH - ions in a strong base solution is always much greater than [H + ], but H + ions are still present due to the ionisation of water.

True. Even in strongly alkaline solutions, water ionises to produce a small concentration of H + ions. This is why K w can be used to find [H + ] from [OH - ].

A titration curve is a graph of pH against volume of titrant added during an acid–base titration, used to determine the equivalence point and choose a suitable indicator.

A LevelChemistryAQAFlashcardsAdvanced Physical ChemistryFundamentals of Acids & Bases

Fundamentals of Acids & Bases (AQA A Level Chemistry): Flashcards

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FrontBrønsted–Lowry Acid & Bases
Define a Brønsted–Lowry acid and a Brønsted–Lowry base.

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Define a Brønsted–Lowry acid and a Brønsted–Lowry base.
A Brønsted–Lowry acid is a species that can donate a proton (H⁺). A Brønsted–Lowry base is a species that can accept a proton (H⁺).
True or False?
A conjugate acid-base pair differs by one H⁺ ion.
True.
A conjugate acid-base pair is two species related by the transfer of a single proton. The acid loses H⁺ to give the conjugate base; the base gains H⁺ to give the conjugate acid.
In the reaction CH₃COOH (aq) + H₂O (l) ⇌ CH₃COO^- (aq) + H₃O⁺ (aq), the acid is .......... and its conjugate base is .......... . The base is .......... and its conjugate acid is .......... .
In the reaction CH₃COOH (aq) + H₂O (l) ⇌ CH₃COO^- (aq) + H₃O⁺ (aq), the acid is CH₃COOH and its conjugate base is CH₃COO^-. The base is H₂O and its conjugate acid is H₃O⁺.
Identify the two conjugate acid-base pairs in: HCO₃^- (aq) + H₂O (l) ⇌ CO₃^2- (aq) + H₃O⁺ (aq)
Pair 1: HCO₃^- (acid) and CO₃^2- (conjugate base)
Pair 2: H₃O⁺ (conjugate acid) and H₂O (base)
What does it mean to say water is amphoteric in the Brønsted–Lowry model?
Water is amphoteric because it can act as both a Brønsted–Lowry acid (donating H⁺) and a Brønsted–Lowry base (accepting H⁺) depending on the other species present in the reaction.
In the reaction H₂SO₄ (aq) + HNO₃ (aq) ⇌ HSO₄^- (aq) + NO₂⁺ (aq) + H₂O (l), H₂SO₄ acts as the .......... and HNO₃ acts as the .......... .
In the reaction H₂SO₄ (aq) + HNO₃ (aq) ⇌ HSO₄^- (aq) + NO₂⁺ (aq) + H₂O (l), H₂SO₄ acts as the acid and HNO₃ acts as the base.
Give the equations linking pH and [H⁺].
pH = −log[H⁺]
[H⁺] = 10^−pH
where [H⁺] is in mol dm⁻³. pH values are normally given to 2 decimal places.
True or False?
The pH scale is linear — a decrease of 1 pH unit corresponds to a decrease of 1 in [H⁺].
False.
The pH scale is logarithmic (base 10). A decrease of 1 pH unit corresponds to a 10-fold increase in [H⁺]. For example, pH 5 is 10 times more acidic than pH 6.
A solution has [H⁺] = 1.60 × 10^-4 mol dm^-3. Its pH = −log(.......... ) = .......... . A solution with pH = 3.10 has [H⁺] = 10.......... = .......... mol dm^-3.
A solution has [H⁺] = 1.60 × 10^-4 mol dm^-3. Its pH = −log(1.60 × 10^-4) = 3.80. A solution with pH = 3.10 has [H⁺] = 10^-3.10 = 7.94 × 10^-4 mol dm^-3.
Why is the [H⁺] from a strong acid assumed to equal the acid concentration when calculating pH?
Strong acids are completely ionised in solution (HA → H⁺ + A^-), so every mole of acid produces one mole of H⁺. The tiny contribution of H⁺ from water is negligible in comparison.
For 0.01 mol dm^-3 HCl: [H⁺] = .......... mol dm^-3, so pH = −log(.......... ) = .......... .
For 0.01 mol dm^-3 HCl: [H⁺] = 0.01 mol dm^-3, so pH = −log(0.01) = 2.00.
True or False?
0.1 mol dm^-3 sulfuric acid gives [H⁺] = 0.2 mol dm^-3 because H₂SO₄ is diprotic and fully ionises in both steps.
False.
H₂SO₄ is a diprotic acid, but the second ionisation (HSO₄^- ⇌ SO₄^2- + H⁺) is only partial. This means [H⁺] is greater than the acid concentration but less than twice it. At A Level, the pH of dilute H₂SO₄ is calculated assuming only the first (complete) ionisation contributes.
10.0 cm³ of a solution at pH 1.0 is diluted to 1000.0 cm³ with distilled water. What is the pH of the final solution?
The volume increased by a factor of 100, so [H⁺] decreased by a factor of 100 (= 10²).
Each factor of 10 decrease in [H⁺] increases pH by 1, so pH increases by 2.
Final pH = 3.
pH
A diprotic acid is an acid that can donate two protons (H⁺ ions) per molecule in aqueous solution, such as sulfuric acid (H₂SO₄) or carbonic acid (H₂CO₃).
What is the ionic product of water, K_w?
Ionic product of water is the equilibrium constant for the dissociation of water: K_w = [H⁺][OH^-], which equals 1 x 10^-14 mol² dm^-6 at 298 K.
Complete the expression for the ionic product of water and give its value at 298 K.
K_w = [.......... ][.......... ] = .......... mol² dm^-6
K_w = [H⁺][OH^-] = 1 x 10^-14 mol² dm^-6
True or False?
As temperature increases, the value of K_w increases and the pH of pure water decreases.
True.
The dissociation of water is endothermic. Increasing temperature shifts the equilibrium right, producing more H⁺ and OH^- ions, raising K_w and lowering the pH of pure water below 7.
Why is pure water neutral even when its pH falls below 7 at higher temperatures?
Neutrality requires [H⁺] = [OH^-], not pH = 7. At higher temperatures K_w increases but [H⁺] still equals [OH^-], so the solution remains neutral even though pH < 7.
For a strong base fully ionised in solution, the steps to find pH are:
1. [OH^-] = .......... (equal to base concentration)
2. [H⁺] = K_w ÷ ..........
3. pH = ..........
1. [OH^-] = [BOH]
2. [H⁺] = K_w ÷ [OH^-]
3. pH = -log[H⁺]
Calculate the pH of 0.15 mol dm^-3 NaOH. (K_w = 1 x 10^-14 mol² dm^-6)
[H⁺] = (1 x 10^-14) ÷ 0.15 = 6.67 x 10^-14 mol dm^-3
pH = -log(6.67 x 10^-14) = 13.18
True or False?
The concentration of OH^- ions in a strong base solution is always much greater than [H⁺], but H⁺ ions are still present due to the ionisation of water.
True.
Even in strongly alkaline solutions, water ionises to produce a small concentration of H⁺ ions. This is why K_w can be used to find [H⁺] from [OH^-].
pOH
A titration curve is a graph of pH against volume of titrant added during an acid–base titration, used to determine the equivalence point and choose a suitable indicator.

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