Reactions Of Ions In Aqueous Solution

## Read an aqueous cation as a hydrated species - The copper aqua illustration shows water ligands donating through oxygen around a metal ion. Cu²⁺(aq) is shorthand for a hydrated species, not a bare gas-phase ion. - The six-water model is[Cu(H₂O)₆]²⁺. Iron(II), iron(III) and aluminium(III) also form the aqua complexes considered in this lesson. - Aluminium is included because of its aqueous-ion chemistry; it is not a first-row transition metal by the AQA definition. The ideal copper geometry does not assert six equal measured Cu–O distances. ## Separate Lewis bonding from proton transfer - Water acts as a Lewis base when its oxygen atom donates an electron pair to the central metal, which acts as a Lewis acid. The donor is an oxygen atom in H₂O, not an O₂ molecule. - A Brønsted–Lowry base accepts a proton. Ammonia can therefore be both an electron-pair donor and a proton acceptor. - Later aqua-ion reactions remove protons from O–H groups. They need not break every metal–oxygen bond or reduce the metal ion. ## Keep solvent and spectator ions in balanced bookkeeping - An explicit anhydrous model is CuSO₄(s)+6H₂O(l)→[Cu(H₂O)₆]²⁺(aq)+SO₄²⁻(aq). Water supplies the ligands and total charge remains zero. - Shorthand ionic equations often suppress the hydration waters. Use one convention consistently rather than treating aq as an atom-containing species to balance. - Counter-ions such as nitrate can be spectators in the later base reaction. Their number still matters when writing a complete dissolution equation. ## Explain stronger aqua acidity with charge and size - In the comparable iron pair, Fe³⁺ has greater charge density than Fe²⁺. It attracts electron density from coordinated water more strongly. - This polarises the O–H bonds and makes proton loss easier. The acidity belongs to coordinated water in the complex, not to a metal ion emitting an H atom it never contained. - The course pattern compares appropriate common aqua ions. It is not a rule that every 3+ ion is smaller or more acidic than every 2+ ion in all ligand environments. ## Write hydrolysis as an equilibrium - A coordinated water can lose H⁺:[Fe(H₂O)₆]³⁺⇌[Fe(H₂O)₅(OH)]²⁺+H⁺. The residual complex is one charge less positive. - Writing H₃O⁺ instead requires an extra water molecule on the reactant side. Both conventions conserve atoms and charge. - Iron remains +3. Proton loss changes the ligand charge; it does not require electron transfer to or from the metal. ## Compare acid strength without equating pKa and pH - Lower pKa means larger Ka for the specified dissociation. The supplied aqua-acid example at 2.2 has a larger first-step Ka than the comparison acid at 4.8. - The Ka ratio is 10^(4.8−2.2), about 398. This ratio is a comparison of dissociation constants, not an automatic 398-fold ratio of hydrogen concentrations. - Actual pH depends on concentration, other equilibria and conditions. The approximate first-step pKa does not describe every hydrolysis or precipitation step. ## Track successive proton loss and a hydrated solid model - Each formal proton loss converts one neutral water ligand to OH⁻ and lowers the complex charge by one. The metal oxidation state is unchanged. - With suitable base, neutral hydrated hydroxide precipitates. The course may write M(H₂O)₃(OH)₃(s) or use the shorter M(OH)₃(s) bookkeeping. - A hydrated hydroxide solid is not necessarily made of isolated molecules with precisely this microscopic structure. Do not mix the short and hydrated equations without balancing the water. ## Precipitate divalent hydroxides with base - Hydroxide removes protons from coordinated water. For the two course divalent ions, the usual net stoichiometry is two OH⁻ per metal centre. - M²⁺+2OH⁻→M(OH)₂(s) is the shortened equation. The hydrated equation explicitly retains four waters and releases two. - The equation predicts proton equivalents and charge balance. Whether visible solid forms also depends on concentrations and solubility conditions. ## Use three base equivalents for the trivalent hydroxide stage - Iron(III) and aluminium(III) aqua ions form hydroxide-containing precipitates when enough suitable base removes three protons per metal. - For this stage, M³⁺+3OH⁻→M(OH)₃(s) conserves atoms and charge. The metal stays at+3 rather than becoming a neutral metal atom. - Further excess OH⁻ can dissolve aluminium hydroxide. Complete precipitation and subsequent redissolution are separate stages with different stoichiometry. ## Recognise iron(II) hydroxide and oxidation in air - A fresh iron(II) sample commonly gives a green hydroxide precipitate with OH⁻ or ammonia. It usually remains in excess of either reagent under the test conditions. - Air can oxidise iron(II) material to brown iron(III) hydroxide-containing products. One simplified equation is 4Fe(OH)₂+O₂+2H₂O→4Fe(OH)₃. - Record colour promptly and note ageing. A brown surface on an initially green precipitate is not evidence that the initial solution necessarily contained only iron(III). ## Separate iron(III) precipitation from reduction - Iron(III) gives a red-brown hydroxide-containing precipitate with OH⁻ or ammonia. It usually remains in excess of these reagents in the course tests. - Precipitation is acid–base chemistry rather than a reduction to iron metal. In Fe(OH)₃, the three OH⁻ groups still require iron at+3. - Ordinary yellow or orange aqueous iron(III) solutions can include hydrolysed species; pure hexaaquairon(III) is pale violet. Keep solution colour distinct from precipitate colour. ## Recognise copper hydroxide without assuming ammonia behaviour - Copper(II) gives a pale-blue hydroxide-containing precipitate with OH⁻. It normally remains in excess NaOH under the stated school-test conditions. - A solid precipitate is different from the original pale-blue clear solution. Record clarity and physical state as well as colour. - Excess ammonia can dissolve the precipitate through complex formation. Do not transfer the excess-NaOH observation to every alkaline reagent. ## Explain ammonia precipitation by proton acceptance - Ammonia is a weak base in water and can remove protons from coordinated water ligands. The resulting ammonium ions account for the gained protons. - Two NH₃ per divalent aqua ion or three per trivalent ion appear in the stated hydrated precipitation models. These are net acid–base balances, not assertions of elementary collision steps. - Do not invent free NH₄OH molecules as a necessary reagent. Use NH₃, NH₄⁺, water and hydroxide equilibria consistently. ## Explain the copper–ammonia observation sequence - The three-tube illustration shows the conceptual sequence:left clear pale-blue aqua solution, middle pale-blue precipitate, right clear deep-blue excess-ammonia solution. - Initially ammonia acts as a base. In excess, it supplies ligands to form[Cu(NH₃)₄(H₂O)₂]²⁺, tetraamminediaquacopper(II). Copper remains +2 and coordination number 6. - One shortened dissolution model is Cu(OH)₂+4NH₃+2H₂O⇌[Cu(NH₃)₄(H₂O)₂]²⁺+2OH⁻. The water balance differs from the explicitly hydrated-solid version. ## Distinguish aluminium’s excess-hydroxide and ammonia results - Aluminium(III) commonly gives a white hydroxide precipitate with either OH⁻ or ammonia. In excess NaOH, it dissolves to a colourless aluminate-containing solution. - The white precipitate remains in usual excess ammonia. Ammonia’s weak-base equilibrium does not reproduce the high hydroxide availability of excess strong base. - Within the four course ions, the combined tests support aluminium. A white precipitate alone is insufficient when other possible cations are not excluded. ## Show aluminium hydroxide acting as a base toward acid - Aluminium hydroxide dissolves in suitable acid by consuming protons. The short ionic equation is Al(OH)₃+3H⁺→Al³⁺+3H₂O. - When the retained hydration waters are explicit, Al(H₂O)₃(OH)₃+3H⁺→[Al(H₂O)₆]³⁺is an equivalent balanced convention. - This acid reaction demonstrates basic behaviour. Amphoterism requires the additional suitable reaction with base. ## Form tetrahydroxoaluminate in excess hydroxide - Fresh aluminium hydroxide can dissolve in sufficient excess aqueous NaOH to give[Al(OH)₄]⁻. Heating is not a universal requirement for this fresh precipitate. - The short model consumes one additional OH⁻ per Al(OH)₃ after the precipitation stage. Aluminium remains at oxidation state +3. - The tetrahydroxo representation is the course model; do not silently mix it with a different hydrated-aluminate convention and lose waters or charges. ## Demonstrate amphoterism using both reactions - An amphoteric hydroxide reacts with suitable acid and base. For Al(OH)₃, acid gives an aqua-ion solution and excess OH⁻ gives aluminate. - A precipitate dissolving in ammonia is not by itself proof of amphoterism; ligand complex formation can be responsible, as in the copper example. - Use fresh portions and identify the reagent conditions. A missing visible change in one unsuitable reagent does not settle every possible acid–base reaction. ## Qualify the divalent carbonate precipitation model - The course usesM²⁺+CO₃²⁻→MCO₃(s) for divalent carbonate precipitation. Iron(II) and copper(II) give characteristic coloured solids in ordinary tests. - The copper precipitate is commonly a basic carbonate rather than pure anhydrous CuCO₃. Iron solids can also have condition-dependent composition and ageing. - Use a stated simplified formula for formal stoichiometry, but do not infer a precise real precipitate mass or crystal identity without composition evidence. ## Explain carbonate’s base role with trivalent aqua ions - The more acidic Fe(III) and Al(III) aqua systems can protonate carbonate, releasing CO₂ while hydroxide-containing solid forms. They do not simply give the usual stable M₂(CO₃)₃ precipitate in these tests. - The shortened model is 2M³⁺+3CO₃²⁻+3H₂O→2M(OH)₃+3CO₂. Explicit hydrated-solid equations retain a different number of water molecules. - The familiar Fe(III) solid is red-brown and aluminium solid white. The outcome depends on conditions; avoid a universal statement about every 3+ ion and every carbonate environment. ## Combine independent tests for a copper inference - The supplied unknownX pattern combines solution appearance, precipitation and excess-ammonia dissolution. Within the stated candidates, it supports Cu²⁺. - Use the combination instead of a single colour word. A pale-blue solid with NaOH is different evidence from a clear deep-blue ammine solution. - The inference assumes a suitable single-ion sample and comparable reagents. Mixtures or other ligands can complicate the result. ## State the limits of an aluminium identification - Within the four-ion course set, the paired excess-reagent behaviour of sampleB supports Al³⁺. The restricted candidate set is part of the inference. - SampleA’s white solid alone is insufficient to establish aluminium. Other cations, including ions outside this lesson, can produce white precipitates. - A negative observation needs suitable reagent concentration and sufficient time. Record the test conditions before treating no visible change as decisive. ## Calculate a limiting-reagent precipitate amount - The supplied 0.00600 mol OH⁻can precipitate 0.00200 mol Fe(OH)₃ at the 3:1ratio. Hydroxide is limiting against 0.00300 mol iron(III). - The short model leaves 0.00100 mol iron(III) unprecipitated. It does not imply every added base amount must remove all metal from solution. - For a gravimetric claim, hydration, drying and actual solid composition would need control. Use the explicitly stated formula model for the requested amount. ## Use the carbonate ratio and specified gas conditions - For the supplied amounts, 0.00400 mol Fe³⁺requires 0.00600 mol carbonate, so carbonate is in excess. The CO₂amount is 0.00600 mol. - Using the stated molar gas volume gives 0.00600×24.0=0.144dm³. The value 24.0 dm³mol⁻¹belongs to specified collection conditions, not to every temperature and pressure. - An experimental gas result can also depend on dissolution, leaks and reaction completeness. State the ideal model before comparing with observation. ## Record evidence so that a test remains interpretable - Use separate sample portions when comparing OH⁻, ammonia and carbonate. Reagent carry-over can change acidity, ligands and later precipitates. - A reagent blank helps distinguish a metal-related observation from contamination or pre-added acid. Carbonate bubbles alone do not establish a particular metal ion. - The precipitate illustration compares blue, green, brown and white solids qualitatively. Its additional white samples show why colour alone cannot identify every cation; it is not a set of quantitative yields.