Proton Transfer Reactions (DP IB Chemistry)

False.

A proton in aqueous solution can be represented as both H⁺ (aq) and H₃O⁺ (aq).

A conjugate acid-base pair is two species that are different from each other by a H⁺ ion.

In the following reaction, the acid and conjugate base pair is H₂CO₃ (aq) and HCO₃^- .

H₂CO₃ (aq) + OH^- (aq) ⇌ HCO₃^- (aq) + H₂O (l)

In the following reaction, the base and conjugate acid pair is CN^- (aq) and HCN (aq) .

H₂PO₄⁻ (aq) + CN^- (aq) ⇌  HCN (aq)  + HPO₄²⁻ (aq)

In the following reaction, the acid and conjugate base pair is HSO₄^- (aq) and SO₄^- (aq).

HSO₄^- (aq) + OH^- (aq) ⇌ SO₄^- (aq) + H₂O (l)

In the following reaction, the acid and conjugate base pair is HCO₃^- (aq) and CO₃^2- (aq).

HCO₃^- (aq) + H₂O (l)  ⇌ CO₃^2- (aq) + H₃O⁺ (aq)

False.

In the following reaction the acid on the left hand side has a conjugate base on the right hand side.

H₂SO₄ + HNO₃ ⇋  H₂NO₃⁺ + HSO₄^-

The conjugate acid in the reaction is CH₃COOH₂⁺.

CH₃COOH + HCl  ⇌ CH₃COOH₂⁺ + Cl^-

The conjugate base of NH₄⁺ is NH₃.

True.

The conjugate base of a strong acid is a weak base.

The relationship between the strength of an acid and its conjugate base is:

The stronger the acid, the weaker its conjugate base.

Amphiprotic is a term describing species that can act both as proton donors and acceptors.

True.

Water is an example of an amphiprotic species.

An amphoteric compound has both basic and acidic character.

True.

All amphiprotic substances are amphoteric.

False.

Not all amphoteric substances are amphiprotic.

For example, aluminium oxide is amphoteric but not amphiprotic.

False.

Amphiprotic substances can both accept or donate protons.

In the following equation is Al₂O₃ is acting as an acid.

Al₂O₃ (s) + 2NaOH (aq) + 3H₂O (l) → 2NaAl(OH)₄ (aq)     

True.

HCO₃⁻ is amphiprotic.

pH is the negative logarithm of the hydrogen ion concentration in mol dm^-3

pH = -log[H⁺]

The pH scale measures how acidic or basic a substance is.

False.

The pH scale is logarithmic.

The pH range for acidic solutions is 0-6 .

The pH range for alkaline solutions is 8-14.

The most accurate way of measuring pH is using a pH meter.

The pH of a neutral solution at 298K is 7.

As [H⁺] increases, pH decreases.

True.

A solution with pH 4 is 10 times more acidic than a solution with pH 5.

The pH of a solution that has an H⁺ concentration of 0.001 mol dm^-3 at 298 Kis pH 3.

In alkaline solutions OH^- ions are in a greater concentration than H⁺ ions.

False.

Using universal indicator is a less accurate method is to measure the pH of a substance than a using a pH meter.

The pH of HCl with a concentration of 0.43 mol dm^-3 is 0.37.

pH = -log[0.43]

pH = 0.37

An equation for the equilibrium established in water is:

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq).

The equation for the ion product of water is:

K_w = [H⁺][OH^-]

False.

The value of K_w changes with temperature.

As temperature increases, K_w increases.

In pure water at 298K, [H⁺] = [OH^-].

True.

In acidic solutions, [H⁺] > [OH^-].

True.

In alkaline solutions, [H⁺] < [OH^-].

As temperature increases, the pH of water decreases.

The ionisation of water is endothermic.

The pH of a solution of sodium hydroxide, NaOH (aq) of concentration 1.0 × 10⁻⁴ mol dm⁻³ is 10.

• K_w = [H⁺] [OH^-]

• Rearranging gives [H⁺]  = K_w ÷ [OH^-]

• [H⁺] = (1.0 × 10⁻¹⁴) ÷ (1.0 × 10⁻⁴) = 1.0 × 10⁻¹⁰ mol dm⁻³

• pH = – log₁₀[1.0 × 10⁻¹⁰ (aq)]

• So the pH = 10

[H₂O (l)] not included in the expression for K_w a s the concentration of water molecules remains constant.

A strong acid is an acid that dissociates almost completely in aqueous solutions.

A weak acid is an acid that partially (or incompletely) dissociates in aqueous solutions.

True.

Weak acids are proton donors and their solutions are poor conductors.

A strong base is a base that dissociates almost completely in aqueous solutions.

True.

HCl, HBr, HI, HNO₃ and H₂SO₄ are all strong acids.

Examples of strong bases include Group 1 hydroxides (e.g. NaOH, KOH)

HI is a stronger acid than HBr.

This is because the HI bond is a longer bond than HBr and is weaker. Therefore H⁺ ions are more easily released.

Strong acids conduct electricity better than weak acids.

A stronger acid has a higher concentration of H⁺ so it conducts electricity better.

CH₃CH₂COOH has a higher pH than HNO₃.

Strong acids react more vigorously because the concentration of H⁺ is greater in strong acids compared to weak acids.

The equation for the ionisation of ethanoic acid is:

CH₃COOH (aq)CH₃COO^- (aq) + H⁺ (aq)

A strong acid has a higher concentration of H⁺ ions so it conducts electricity better than a weak acid.

Weak acids are less acidic than strong acids as they contain fewer H⁺ in solution. This is because they only partially ionise.

Weak bases include:

• NH₃

• Amines

• Some transition metal hydroxides

False.

In general, strong acids produce weak conjugate bases and weak acids produce strong conjugate bases.

The ionic equation to show how a base accepts hydrogen ions is:

OH^– (aq) +  H⁺ (aq)  ⇌  H₂O (l)

HNO₃ dissociates to form H⁺ and NO₃^- ions.

The NO₃^- ion is a weak conjugate base.

True.

A solution of 10 mol dm^-3 HCN is a concentrated weak acid.

A neutralisation reaction is one in which an acid and a base/alkali react together to form water and a salt

A sulfate salt is produced when sulfuric acid is reacted with an alkali.

A nitrate salt is produced when nitric acid is reacted with an alkali.

A chloride salt is produced when hydrochloric acid is reacted with an alkali.

Ethanoic acid is used to make ethanoate salts.

The general equation for the reaction between an acid and a metal hydroxide is:

acid + metal hydroxide → salt + water

The equation for the reaction between hydrochloric acid  + potassium hydrogencarbonate is:

HCl (aq) + KHCO₃ (s) → KCl (aq) + H₂O (l) + CO₂ (g)

The equation for the reaction between nitric acid + copper carbonate is:

2HNO₃ (aq) + CuCO₃ (s) → Cu(NO₃)₂ (aq) + H₂O (l) + CO₂ (g)

The equation for the reaction between sulfuric acid + barium hydroxide is:

H₂SO₄ (aq) + Ba(OH)₂ (aq) → BaSO₄ (s) + 2H₂O (l)

Ammonium sulfate is formed when sulfuric acid reacts with ammonia.

Bases that could be used to form copper(II) chloride are copper(II) oxide, copper(II) hydroxide, copper(II) carbonate.

During a titration, a pH meter can be used and a pH curve plotted.

X is the equivalence or stoichiometric point.

The pH at the equivalence point of a titration between HCl and NaOH is 7.

From a pH curve, you can determine:

• The initial pH of the acid

• The pH at the equivalence point

• The volume of base at the equivalence point

• The range of pH at the vertical section of the curve

The pH at the start of the titration is pH 3.

The pH at equivalence point is pH 8.

The volume of NaOH (aq) that was added to reach the equivalence point is 30 cm³.

The equivalence point strong acid and strong base titration is pH 7.

False.

The equivalence point for a weak acid and strong base titration is above 7.

True.

The equivalence point for a strong acid and weak base titration is below 7.

The half equivalence point is the stage of the titration at which exactly half the amount of weak acid has been neutralised.

X is the half equivalence point.

The region that is labelled X is the buffer region.

pK_a = pH at half equivalence 

AND

pK_b = pOH at half equivalence 

True.

[NH₃ (aq)] = [NH₄⁺ (aq)] at the half equivalence point during a titration involving NH₃ (aq) and HCl (aq)

A weak acid and weak base pH curve is shown.

A weak base and strong acid pH curve is shown below.

The pH at the y-intercept is pH 11.

The pH when [H⁺] is 2.80 x 10^-3 mol dm^-3 is:

pH = -log[H⁺]

pH = -log [2.80 x 10^-3 ]

pH = 2.55

The conversion for pH to H⁺ concentration is:

[H⁺]= 10^-pH .

[H⁺] when the pH is 1.89 is 0.013 mol dm^-3 at 298 K.

[H⁺]= 10^-pH

The pH of 0.75 mol dm^-3 sodium hydroxide, NaOH is:

• [H⁺] = K_w  ÷ [OH^–]

• [H⁺] = (1 x 10^-14) ÷ 0.75 = 1.33 x 10^-14

• pH = -log 1.33 x 10^-14  = 13.88

[OH^-] can be calculated from pOH by:

[OH^-] = 10^-pOH

The equation to calculate [H⁺] from K_w and [OH^-] is:

[H⁺] =

True.

pH can be calculated by pH = 14 - pOH.

To calculate the pH of NaOH with a concentration of 0.49 mol dm³ using the equation pH = 14 - pOH:

• pOH = -log [0.49] = 0.31

• pH = 14 - 0.31 = 13.69

False.

pH + pOH = pK_w

The acid dissociation constant, K_a, for the acid HA is:

K_a =

The base dissociation constant, K_b, for C₆H₅CH₂NH₂ is:

K_b =

The K_a of benzoic acid is 6.61 x 10^–5.

• K_a= 10^-pK_a

• K_a= 10^-4.18

• K_a = 6.61 x 10^–5

The relationship between K_a, K_b and K_w is:

K_a x K_b = K_w

The relationship between pK_a, pK_b and pK_w is:

pK_a + pK_b = pK_w

True .

The larger the pK_a value, the weaker the acid.

False.

The larger the K_b value, the stronger the base.

The acid dissociation constant, K_a, for propanoic acid is:

The pK_a of hydrocyanic acid is 9.21.

• pK_a = -log_pKa

• pK_a = -log 6.17 x 10^-10 = 9.21

The pK_a value of CH(Cl₂)COOH is 1.35.

• pK_a + pK_b = 14

• 14 - pK_b

• 14 - 12.65 = 1.35

The [H⁺] in a 0.15 mol dm^-3 solution whose K_a = 1.00 x 10^-8 at 298 K is 3.87 x 10^-5.

• K_a =  = 1.00 x 10^-8

• [H⁺]² = 1.5 x 10^-9

• [H⁺] = 3.87 x 10^-5 (mol dm^-3)

The [OH^-] of a solution with a [H⁺] of 4.50 x 10^-6 mol dm³ at 298 K is:

• [OH^-] = = 2.22 x 10^-9 mol dm^-3

The acid dissociation constant, K_a, at 298 K for a 0.50 mol dm^-3 solution of propanoic acid with a [H⁺] of 1.75 x 10^-5 is:

• K_a = 

• K_a =

• K_a = 6.13 × 10^-10

The relationship of K_a and K_b for a conjugate acid-base pair is K_a x K_b = K_w.

The pH of a 0.04 mol dm^-3  solution of NaOH at 298 K is:

• [H⁺] =  =  = 2.5 x 10^-13

• pH = -log(2.5 x 10^-13)

• pH = 12.60

The [OH^-] of 0.05 mol dm^-3 ammonia solution, NH₃ (aq) is:

• K_b =

• [OH^-]² = 1.78 x 10^-5 x 0.05

• [OH^-] = 9.43 x 10^-4 mol dm^-3

The Na⁺ and Cl^- ions do not act as Brønsted-Lowry acids or bases as they can not release or accept H⁺ ions.

NH₄Cl is an acidic salt as it is formed from NH₃ (aq) (weak base) and HCl (aq) (strong acid)

An equation to show how NH₄⁺ reacts with water.

NH₄⁺ (aq) + H₂O (l) → H₃O⁺ (aq) + NH₃ (aq) 

CH₃COONa (aq) is an alkaline salt as it is formed from NaOH (aq) (strong base) and CH₃COOH (aq) (weak acid).

An equation to show how CH₃COO^– (aq)⁺ reacts with water is:

CH₃COO^– (aq) + H₂O (l) → CH₃COOH (aq) + OH^- (aq)

True.

If the salt is formed from a weak acid and weak base then its hydrolysis is determined by the relative K_a and K_b values.

False.

If a salt is formed from a strong acid and strong base titration then it is neutral.

The K₂CO₃ salt is alkaline as it is formed from KOH (aq) (strong bae) and H₂CO₃ (aq) (weak acid).

The KNO₃ salt is neutral as it is formed from KOH (aq) (strong base) and HNO₃ (aq) (strong acid).

True.

The NaHCO₃ salt is alkaline by hydrolysis as it is formed from NaOH (aq) (strong base) and H₂CO₃ (aq) (weak acid).

Acid-base indicators are weak acids, where the components of the conjugate acid–base pair have different colours.

An equation for the dissociation of the weak acid HIn.

HIn (aq)   ⇌ H⁺ (aq) + In^– (aq)

The acid dissociation constant, K_a, for the weak acid HIn is:

K_a =

The equilibrium will shift to the left in acidic conditions.

H-Phen   ⇌ H⁺ + Phen^-

True.

The pK_a of an indicator is the same as the pH of its endpoint.

The indicator is a mixture of many indicators with a wide range of colour change is universal indicator.

The equilibrium will shift to the right at a high pH as H⁺ ions will be mopped up by OH^- ions introduced. Therefore the H⁺ ions need to be replenished by the equilibrium shifting to the right hand side.

True.

An appropriate indicator for a titration has an end point range that coincides with the pH at the equivalence point.

The end point is the point where the indicator changes colour during a titration.

The equivalence point in a titration refers to a point at which the added titrant is chemically equivalent to the sample analyte.

OR

At the equivalence point in an acid-base titration, moles of base = moles of acid and the solution only contains salt and water.

Methyl red is suitable for a strong acid and weak base titration.

True.

Phenolphthalein has a pK_a of 9.6 so is suitable for a strong base and weak acid titration.

In weak acid and weak base titrations, there is no sudden pH change at the end-point and thus there are no suitable indicators for these titrations.

True.

The indicator changes colour when the pH = pK_a of the acid.

A buffer solution is a solution which resists changes in pH when small amounts of acid or base are added.

A buffer can consist of weak acid – conjugate base or weak base – conjugate acid.

The equation for the ionisation of sodium ethanoate is:

CH₃COONa + aq → Na⁺ (aq) + CH₃COO^- (aq) 

The equation for the ionisation of ethanoic acid is:

CH₃COOH (aq) ⇌ H⁺  (aq) + CH₃COO^- (aq) 

The composition of a basic buffer is a weak base and its salt.

Water is formed when OH^- are added to an acidic buffer.

OH^– (aq) + H⁺  (aq) → H₂O (l)

True.

When OH^- ions are added to an acidic buffer, CH₃COOH molecules ionise to form more H⁺ and CH₃COO^– until equilibrium is re-established.

When an acid is added to an acidic buffer H⁺ ions react with A^– ions to form more HA until equilibrium is re-established.

The equation to show how NH₃ (aq) reacts with water by:

NH₃ (aq) + H₂O (l) NH₄⁺ (aq) + OH^– (aq)

The equation to show what happens when acid is added to a basic buffer which contains ammonia is:

NH₃ (aq) + H⁺ (aq) NH₄⁺ (aq) 

A basic buffer is made by mixing a solution of a weak base with its salt.

True.

A basic buffer contains NH₃ (aq) and NH₄Cl (aq). If a base is added OH^– ions will combine with the acid NH₄⁺ and form NH₃ and H₂O.

NH₄⁺ (aq) + OH^– (aq) NH₃ (aq) + H₂O (l)

The pH of a buffer solution depends on:

• The pK_a or pK_b of its acid or base.

• The ratio of the concentration of acid or base to the concentration of the conjugate base or acid.

The equation to calculate the [H⁺] of an acidic buffer solution is:

The equation to calculate [OH^-] of a basic buffer is:

[OH^–] = K_b 

False.

A constant temperature must be maintained when using buffers as temperature will influence the pH of the solution

True.

Dilution does not change the ratio of acid or base to the salt concentration as both components will be decreased by the same amount

To calculate [H⁺] of a buffer solution containing 0.45 mol dm^-3 of ethanoic acid and 0.36 mol dm^-3 sodium ethanoate:

• 

• = 2.18 x 10^-5 mol dm^-3