Questions on metal-aqua ions and the acidity of their solutions

Question 1

Explain, with equations, why a 0.100 mol dm⁻³ solution of [Fe(H₂O)₆]³⁺ has a lower pH than a 0.100 mol dm⁻³ solution of [Fe(H₂O)₆]²⁺. Refer to charge density in your answer.

Accepted Answer

Indicative content

[Fe(H₂O)₆]³⁺(aq) + H₂O(l) ⇌ [Fe(H₂O)₅(OH)]²⁺(aq) + H₃O⁺(aq), the charge on the complex falling by one as a proton is lost
[Fe(H₂O)₆]²⁺(aq) + H₂O(l) ⇌ [Fe(H₂O)₅(OH)]⁺(aq) + H₃O⁺(aq), the same hydrolysis for the 2+ ion, to a lesser extent
Fe³⁺ has a higher charge and a smaller radius than Fe²⁺, so it has the higher charge density; accept charge/size ratio
Fe³⁺ therefore attracts the lone pairs of its water ligands more strongly, so the O–H bonds in those ligands are polarised more
the O–H bonds are weakened more, so H⁺ is released more readily and the hydrolysis equilibrium lies further to the right
[H₃O⁺] is higher in the iron(III) solution, so its pH is lower: about 1.7, against about 5 for the iron(II) solution

Levels

Level 3 (5–6 marks) both hydrolysis equations are correct with state symbols, the higher charge density of Fe³⁺ is explained through its charge and radius, the stronger polarisation and weakening of the O–H bonds is linked to the position of the equilibrium, and the lower pH is stated with approximate values
Level 2 (3–4 marks) one hydrolysis equation is right and the higher charge density of Fe³⁺ explains the stronger polarisation of the water ligands; the second equation, the shift in the equilibrium or the pH comparison is not there
Level 1 (1–2 marks) Fe³⁺ is said to be more acidic, often because it 'has more H⁺' to release; no correct equation and no mention of the O–H bonds in the ligands

Question 2

Explain, with equations, how sodium carbonate solution can be used to decide whether an unknown solution contains [M(H₂O)₆]²⁺ or [M(H₂O)₆]³⁺ ions, and why the two charges behave differently.

Accepted Answer

Indicative content

a 3+ ion gives effervescence of carbon dioxide and a hydroxide precipitate; a 2+ ion gives a carbonate precipitate and no gas
for the 3+ ion, 2[M(H₂O)₆]³⁺(aq) + 3CO₃²⁻(aq) → 2[M(H₂O)₃(OH)₃](s) + 3CO₂(g) + 3H₂O(l), balanced with state symbols
for the 2+ ion, [M(H₂O)₆]²⁺(aq) + CO₃²⁻(aq) → MCO₃(s) + 6H₂O(l), the metal carbonate precipitating with no gas
the 3+ ion has a higher charge/size ratio, so it polarises the O–H bonds of its water ligands more strongly
H⁺ is released more readily, so the 3+ solution is acidic enough to react with carbonate and release carbon dioxide
the 2+ ion is too weak an acid to release carbon dioxide, so its metal carbonate precipitates; the gas can be confirmed by limewater turning milky

Levels

Level 3 (5–6 marks) both observations are given and matched to the right charge, both equations are correct and balanced, and the difference is explained through the charge/size ratio polarising the water ligands so that only the 3+ solution is acidic enough to release carbon dioxide from carbonate
Level 2 (3–4 marks) effervescence with the 3+ ion and a carbonate precipitate with the 2+ ion are described and one equation is right; the other equation, or the explanation through polarisation and acidity, is thin
Level 1 (1–2 marks) one observation is given, typically carbon dioxide from both ions or a carbonate precipitate from the 3+ ion; no correct equation and no reason for the difference

Question 3

Compare what is seen, and the products formed, when sodium carbonate solution is added to separate solutions of iron(II) sulfate and iron(III) chloride. Give the equation, with state symbols, for the iron(III) reaction.

Accepted Answer

B1 both solutions give a precipitate, green with iron(II) and brown with iron(III) B1 only the iron(III) solution gives effervescence, of carbon dioxide B1 the iron(II) precipitate is the carbonate, FeCO₃ B1 the iron(III) precipitate is the hydroxide, [Fe(H₂O)₃(OH)₃], not a carbonate; reject Fe₂(CO₃)₃ B1 2[Fe(H₂O)₆]³⁺(aq) + 3CO₃²⁻(aq) → 2[Fe(H₂O)₃(OH)₃](s) + 3CO₂(g) + 3H₂O(l)

Question 4

Give the formula of the metal-aqua ion and the colour of the solution for iron(III) nitrate, copper(II) sulfate and aluminium sulfate, and state why the aluminium sulfate solution is colourless.

Accepted Answer

B1 iron(III) nitrate contains [Fe(H₂O)₆]³⁺ and its solution is yellow-brown; allow pale violet for the unhydrolysed ion B1 copper(II) sulfate contains [Cu(H₂O)₆]²⁺, giving a blue solution B1 aluminium sulfate contains [Al(H₂O)₆]³⁺, giving a colourless solution B1 Al³⁺ has no d electrons, so no electron can be promoted and no visible light is absorbed

Question 5

Complete the equation, with state symbols, for the reaction that makes a solution of [Al(H₂O)₆]³⁺ acidic, [Al(H₂O)₆]³⁺(aq) + H₂O(l) ⇌, and identify the Brønsted–Lowry acid and the Brønsted–Lowry base on its left-hand side.

Accepted Answer

B1 the products are [Al(H₂O)₅(OH)]²⁺ and H₃O⁺, with the complex's charge falling from 3+ to 2+ B1 both products are shown as (aq) and the equilibrium arrow is kept; allow H⁺(aq) only if the H₂O(l) on the left is removed B1 the acid is [Al(H₂O)₆]³⁺, which donates a proton from one of its water ligands B1 the base is the solvent water molecule, which accepts the proton and becomes H₃O⁺

Question 6

Describe the bonding and the shape of the [Fe(H₂O)₆]²⁺ ion.

Accepted Answer

B1 six water molecules act as ligands around the central Fe²⁺ ion B1 each water molecule forms a coordinate bond by donating a lone pair from its oxygen atom to the iron ion B1 the six oxygen atoms are arranged octahedrally around the iron, with bond angles of 90°

Question 7

Explain why a solution of iron(III) chloride is acidic, although iron(III) chloride is a salt.

Accepted Answer

B1 in water the Fe³⁺ ion is surrounded by six water ligands as [Fe(H₂O)₆]³⁺ B1 the metal ion attracts electron density from the oxygen atoms, polarising and weakening the O–H bonds of the water ligands B1 a water ligand loses H⁺ to a solvent water molecule, giving [Fe(H₂O)₅(OH)]²⁺ and H₃O⁺, so the solution is acidic

Question 8

Calculate the pH of a solution in which the concentration of [Al(H₂O)₆]³⁺ is 0.0500 mol dm⁻³. Ka for [Al(H₂O)₆]³⁺ is 1.0 × 10⁻⁵ mol dm⁻³.

Accepted Answer

M1 Ka = [H⁺]² ÷ [Al(H₂O)₆³⁺], taking [H⁺] equal to the concentration of [Al(H₂O)₅(OH)]²⁺ and the aqua ion concentration as unchanged M1 [H⁺]² = 1.0 × 10⁻⁵ × 0.0500 = 5.0 × 10⁻⁷, so [H⁺] = 7.07 × 10⁻⁴ mol dm⁻³ A1 pH = −log₁₀(7.07 × 10⁻⁴) = 3.15, to two decimal places

Question 9

Calculate Ka for [Fe(H₂O)₆]²⁺, given that a 0.100 mol dm⁻³ solution of iron(II) sulfate has a pH of 5.25.

Accepted Answer

M1 [H⁺] = 10 to the power −5.25 = 5.62 × 10⁻⁶ mol dm⁻³ M1 Ka = [H⁺]² ÷ [Fe(H₂O)₆²⁺] = (5.62 × 10⁻⁶)² ÷ 0.100 A1 Ka = 3.2 × 10⁻¹⁰ mol dm⁻³, to two significant figures

Question 10

Calculate the pH of a 0.250 mol dm⁻³ solution of copper(II) sulfate. Ka for [Cu(H₂O)₆]²⁺ is 3.2 × 10⁻⁸ mol dm⁻³.

Accepted Answer

M1 [H⁺]² = Ka × [Cu(H₂O)₆²⁺] = 3.2 × 10⁻⁸ × 0.250 = 8.0 × 10⁻⁹ M1 [H⁺] = square root of 8.0 × 10⁻⁹ = 8.94 × 10⁻⁵ mol dm⁻³ A1 pH = −log₁₀(8.94 × 10⁻⁵) = 4.05, to two decimal places

Question 11

Explain why sodium carbonate solution releases carbon dioxide from a solution of [Al(H₂O)₆]³⁺ but not from a solution of [Cu(H₂O)₆]²⁺.

Accepted Answer

B1 [Al(H₂O)₆]³⁺ is the stronger acid because its higher charge/size ratio polarises the O–H bonds more, so its solution holds more H₃O⁺ B1 that H₃O⁺ reacts with carbonate ions to give carbon dioxide, and the aluminium hydroxide precipitates B1 [Cu(H₂O)₆]²⁺ is too weak an acid to release carbon dioxide, so the carbonate ions precipitate copper(II) carbonate instead

Question 12

Suggest why a yellow-brown solution of iron(III) nitrate turns almost colourless, with a faint violet tint, when concentrated nitric acid is added.

Accepted Answer

B1 the yellow-brown colour is largely due to the hydrolysed complex [Fe(H₂O)₅(OH)]²⁺ B1 adding acid raises the concentration of H₃O⁺, so the hydrolysis equilibrium shifts to the left, by Le Chatelier's principle B1 the solution then contains mainly [Fe(H₂O)₆]³⁺, which is pale violet

Question 13

Evaluate the use of the weak-acid approximation to find the pH of a 0.0500 mol dm⁻³ solution of [Fe(H₂O)₆]³⁺, for which Ka is 6.3 × 10⁻³ mol dm⁻³.

Accepted Answer

B1 the approximation gives [H⁺] = square root of (6.3 × 10⁻³ × 0.0500) = 1.8 × 10⁻² mol dm⁻³ and a pH of 1.75 B1 that [H⁺] is about 35% of 0.0500, so the aqua ion concentration has not stayed close to its starting value B1 the approximation is therefore poor here; it overestimates [H⁺], and the true pH is higher than 1.75

Question 14

Calculate the pH of the solution made when 10.0 cm³ of 0.200 mol dm⁻³ copper(II) sulfate solution is diluted to 100.0 cm³ with water. Ka for [Cu(H₂O)₆]²⁺ is 3.2 × 10⁻⁸ mol dm⁻³.

Accepted Answer

M1 the concentration of [Cu(H₂O)₆]²⁺ after dilution is 0.200 × 10.0 ÷ 100.0 = 0.0200 mol dm⁻³ M1 [H⁺]² = 3.2 × 10⁻⁸ × 0.0200 = 6.4 × 10⁻¹⁰, so [H⁺] = 2.53 × 10⁻⁵ mol dm⁻³ A1 pH = −log₁₀(2.53 × 10⁻⁵) = 4.60, to two decimal places

Question 15

Explain why diluting a solution of [Al(H₂O)₆]³⁺ tenfold raises its pH by about 0.5, whereas diluting hydrochloric acid tenfold raises its pH by 1.

Accepted Answer

B1 hydrochloric acid is fully ionised, so [H⁺] falls tenfold and the pH rises by log₁₀10 = 1 B1 for the aqua ion [H⁺] = √(Ka × c), so a tenfold fall in c lowers [H⁺] only by √10, about 3.16, a pH rise of 0.5 B1 on dilution the hydrolysis equilibrium shifts to the right, so a larger fraction of the aqua ions is hydrolysed and [H⁺] falls by less than the concentration does

Question 16

Suggest why a solution of magnesium chloride has a pH of about 6.5, between the pH of sodium chloride solution and that of aluminium chloride solution of the same concentration.

Accepted Answer

B1 Mg²⁺ has a higher charge and a smaller radius than Na⁺, so its charge/size ratio is high enough to polarise the O–H bonds of its water ligands a little B1 [Mg(H₂O)₆]²⁺ therefore hydrolyses slightly, releasing a little H₃O⁺, so the solution is very slightly acidic B1 Mg²⁺ has a lower charge and a larger radius than Al³⁺, so it polarises the O–H bonds far less and the hydrolysis lies much further to the left than for [Al(H₂O)₆]³⁺

Question 17

Show that a 0.100 mol dm⁻³ solution of copper(II) sulfate has a pH of 4.25 at 298 K. Ka for [Cu(H₂O)₆]²⁺ is 3.2 × 10⁻⁸ mol dm⁻³.

Accepted Answer

M1 [Cu(H₂O)₆]²⁺(aq) + H₂O(l) ⇌ [Cu(H₂O)₅(OH)]⁺(aq) + H₃O⁺(aq), so Ka = [H⁺]² ÷ 0.100 with the two weak-acid assumptions M1 [H⁺] = √(3.2 × 10⁻⁸ × 0.100) = √(3.2 × 10⁻⁹) = 5.66 × 10⁻⁵ mol dm⁻³ A1 pH = −log₁₀(5.66 × 10⁻⁵) = 4.25; the solution is only weakly acidic because a 2+ ion polarises its water ligands slightly

Question 18

Suggest why a solution of sodium chloride is neutral while a solution of aluminium chloride of the same concentration is acidic.

Accepted Answer

B1 Na⁺ has only a single charge and a larger radius than Al³⁺, so its charge/size ratio is much lower B1 Na⁺ does not polarise the O–H bonds of the surrounding water enough to release H⁺, whereas Al³⁺ does