Preparation of Salts Worked ExamplesIGCSE Chemistry 0620

(a)(i) All ammonium salts are soluble → soluble. (ii) Carbonates are insoluble except those of sodium, potassium and ammonium → calcium carbonate is insoluble.

(iii) All nitrates are soluble → soluble. (iv) Hydroxides are insoluble except those of sodium, potassium, ammonium (and calcium, partially) → magnesium hydroxide is insoluble.

(v) Potassium salts are all soluble → soluble. (vi) Chlorides are soluble except those of lead and silver → barium chloride is soluble.

(b) Swap the partners: the possible new salts are lead(II) sulfate and ammonium nitrate. Ammonium nitrate is soluble, but sulfates of barium, calcium and lead are insoluble, so the precipitate is lead(II) sulfate.

(a) While acid remains: fizzing (effervescence / bubbles) and the white solid disappears (dissolves). The solution stays colourless.

(b) When the carbonate is in excess, the fizzing stops when more solid is added and some white solid remains undissolved at the bottom.

(c) All of the sulfuric acid has been used up (has reacted), so there is nothing left for the extra magnesium carbonate to react with.

(d) The residue is the solid left in the filter paper: the excess magnesium carbonate. The filtrate is the solution that passes through: aqueous magnesium sulfate.

(e) Acid + carbonate → salt + water + carbon dioxide.

Barium carbonate is insoluble (carbonates are insoluble except those of sodium, potassium and ammonium), so it is made by precipitation from two soluble salts. Barium nitrate and sodium carbonate are both soluble.

The barium carbonate is the precipitate, so it collects as the residue. It is washed with distilled water to remove soluble sodium nitrate, then dried. Answer: C.

B is the method for a soluble salt: the barium carbonate is thrown away on the filter paper and the filtrate gives sodium nitrate crystals. D leaves sodium nitrate mixed with the product. A would react the barium carbonate away, making barium nitrate.

Precipitation needs two soluble salts: one supplying the cation ($\text{Ca}^{2+}$) and one supplying the anion ($\text{SO}_4^{\,2-}$). When the solutions are mixed, the insoluble salt comes out of solution.

Calcium nitrate (all nitrates soluble) and potassium sulfate (all potassium salts soluble) are both soluble. The other product, potassium nitrate, stays in solution. Answer: B.

A: calcium carbonate is insoluble, so there are no free calcium ions. C: barium sulfate is insoluble, so there are no free sulfate ions. D: calcium hydroxide is only partially soluble and lead(II) sulfate is insoluble.

(a) A hydrated substance is chemically combined with water. An anhydrous substance contains no water.

(b) Water of crystallisation is the water molecules present in hydrated crystals — for example the $6\text{H}_2\text{O}$ in $\text{CoCl}_2 \cdot 6\text{H}_2\text{O}$ or the $5\text{H}_2\text{O}$ in $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$.

(c) Constant mass shows that all the water of crystallisation has been driven off. If some remained, the mass of water would be too small and $x$ too low.

(d) Mass of water lost, and moles of each part. $M_r(\text{CoCl}_2) = 59 + 2 \times 35.5 = 130$; $M_r(\text{H}_2\text{O}) = 18$.

Divide by the smaller number of moles to find the ratio.

(a) Titration. Both reactants are soluble, so an excess of one cannot be removed by filtering; the exact volumes must be found so that neither is left over.

(b) Read the bottom of the meniscus: initial $1.40\text{ cm}^3$, final $21.40\text{ cm}^3$.

(c) $\text{KOH} + \text{HNO}_3 \rightarrow \text{KNO}_3 + \text{H}_2\text{O}$: the ratio is $1:1$.

(d) Repeat the titration without the indicator, mixing $25.0\text{ cm}^3$ of alkali with exactly $20.00\text{ cm}^3$ of acid. Heat the solution to evaporate some water until it reaches the crystallising point (crystals form on a glass rod dipped in and removed). Leave to cool so crystals form, filter them off, and dry them between filter papers or in a warm place.

(e) $0.0200$ mol of $\text{KNO}_3$ forms. $M_r = 39 + 14 + 3 \times 16 = 101$.

(a) Precipitation. X is the residue: barium carbonate. Y is the filtrate: aqueous sodium nitrate (with the excess sodium carbonate).

(b) Only the ions that form the precipitate are shown; sodium and nitrate ions are spectators.

(c) Barium nitrate is the limiting reactant (the carbonate is in excess). The ratio $\text{Ba}^{2+} : \text{BaCO}_3$ is $1:1$; $M_r(\text{BaCO}_3) = 137 + 12 + 48 = 197$.

(d) The unwashed residue is still wet with solution containing soluble sodium nitrate and sodium carbonate; when it dries these solids remain in the sample and add to its mass.

The residue should be washed with distilled water while still in the filter funnel, then left to dry in a warm place (or in a low-temperature oven, or pressed between filter papers).

(a) A saturated solution contains the maximum mass of solute that can dissolve in a given mass of solvent at a stated temperature.

(b) Read up from $60\,^{\circ}\text{C}$ to the curve and across: $68$ g per $100$ g of water.

(c) The graph gives mass per $100$ g of water, so divide each value by $4$ for $25$ g of water. At $80\,^{\circ}\text{C}$: $92 \div 4 = 23$ g dissolved. At $20\,^{\circ}\text{C}$: $32 \div 4 = 8$ g can stay dissolved.

(d) Crystals appear when the solution becomes saturated: find the temperature where the solubility equals $48$ g per $100$ g of water — $40\,^{\circ}\text{C}$.

(a) So that all of the acid reacts. The excess solid is insoluble and is easily removed by filtering, so the zinc chloride solution is not contaminated with acid.

(b) All of the hydrochloric acid has been used up. With no acid left, the remaining zinc carbonate has nothing to react with.

(c) Any insoluble base or the metal itself: zinc oxide, zinc hydroxide or zinc metal. Each reacts with the acid to form zinc chloride and (for the metal) hydrogen, or water.

(d) Two HCl are needed for the two chloride ions in $\text{ZnCl}_2$.

Put a measured volume of dilute sulfuric acid (e.g. $50\text{ cm}^3$ from a measuring cylinder) into a beaker.

Add nickel(II) oxide a little at a time, stirring, until it is in excess (some green solid remains undissolved). The oxide, not the acid, must be in excess so that all the acid is used up.

Warm / heat the mixture gently while stirring, because the reaction is very slow at room temperature; heating increases the rate.

Filter to remove the excess nickel(II) oxide; keep the filtrate.

Heat the filtrate in an evaporating basin to evaporate some of the water until it reaches the crystallising point — do not evaporate to dryness, or the water of crystallisation is lost and anhydrous salt forms.

Leave to cool so crystals form; filter off the crystals, and dry them between filter papers or in a warm place.

(a) Moles of acid, then use the $1:1$ ratio. $M_r(\text{ZnO}) = 65 + 16 = 81$.

(b) The acid is the limiting reactant. One mole of acid gives one mole of crystals; the $M_r$ must include the water of crystallisation: $65 + 32 + 4 \times 16 + 7 \times 18 = 287$.

(c)

(d) Some zinc sulfate stays dissolved in the solution left after crystallisation; some solution is lost on the filter paper or in transfers between containers.

(a) Apply the solubility rules to each possible pair. Barium sulfate and barium carbonate are insoluble; every other combination (chlorides, nitrates, sodium and potassium salts) is soluble. So only the barium solution gives precipitates, and it gives two.

A gives two precipitates, so A is barium nitrate. C gives none with A, so C is potassium chloride (barium chloride is soluble).

B and D are the carbonate and the sulfate. Carbonates fizz with acid (carbon dioxide is given off), so B is sodium carbonate and D is sodium sulfate.

(b)

(c) The barium carbonate precipitate fizzes and dissolves; the barium sulfate precipitate does not change. In the sulfate test, acidifying first removes any carbonate ions, so a white precipitate with barium ions can only be barium sulfate — a carbonate cannot give a false positive.