Carbonates Worked ExamplesIGCSE Chemistry 0620

(a) An observation is something you can see (or hear, feel, smell). Here: fizzing / effervescence (bubbles), the green solid dissolves / disappears, and a blue solution forms (copper(II) nitrate solution is blue). Any two.

(b) Acid + carbonate → salt + water + carbon dioxide. The salt of nitric acid is a nitrate:

copper(II) carbonate + nitric acid → copper(II) nitrate + water + carbon dioxide

(c) Bubble the gas through limewater: it turns milky — so the gas is carbon dioxide.

When two solutions mix, the ions can swap partners. A precipitate forms only if one of the new combinations is insoluble. Rules: sodium, potassium and ammonium salts and all nitrates are soluble; chlorides are soluble except lead and silver; carbonates are insoluble except sodium, potassium and ammonium.

B: the new pair is calcium carbonate (insoluble) and potassium nitrate (soluble). Calcium carbonate precipitates, so B is correct.

A gives ammonium chloride and sodium carbonate, C gives sodium nitrate and potassium carbonate, and D gives sodium chloride and calcium nitrate. All of these are soluble, so no precipitate forms. D is tempting because it contains calcium, but there is no carbonate in it.

(a) Sodium carbonate is $\text{Na}_2\text{CO}_3$. Carbonate needs two $\text{H}^+$, so two HCl, giving two NaCl:

(b) Cancel the spectator ions ($\text{Na}^+$ and $\text{Cl}^-$), which are unchanged in solution. What reacts is carbonate ions with hydrogen ions:

Check the charges: left $-2 + 2(+1) = 0$; right 0. Balanced.

(c)

(d) Ethanoic acid is a weak acid: it is only partially dissociated, so its solution has a much lower concentration of $\text{H}^+$ ions than hydrochloric acid (a strong acid, completely dissociated) of the same concentration.

(a) Magnesium is $\text{Mg}^{2+}$ and nitrate $\text{NO}_3^{\,-}$, so magnesium nitrate is $\text{Mg(NO}_3\text{)}_2$ and two $\text{HNO}_3$ are needed:

(b) When the acid has all reacted: no more fizzing (effervescence) when more magnesium carbonate is added, and white solid remains undissolved at the bottom of the beaker.

(c) Excess solid makes sure all the acid is used up, so no acid is left in the salt. Filtering in step 2 removes the unreacted magnesium carbonate.

(d) Heating to dryness can decompose the salt and gives a powder instead of crystals (it also removes any water of crystallisation). Heating only to the crystallisation point, then cooling, gives crystals.

(a) $M_r$ of $\text{K}_2\text{CO}_3 = 2(39) + 12 + 3(16) = 138$.

Mole ratio $\text{K}_2\text{CO}_3 : \text{CO}_2$ = 1 : 1, so 0.00250 mol of $\text{CO}_2$. One mole of gas is $24 \text{ dm}^3 = 24\,000 \text{ cm}^3$ at r.t.p.

(b) The ratio $\text{K}_2\text{CO}_3 : \text{HCl}$ is 1 : 2.

(c) So that the syringe starts at zero: any air already inside would be added to the reading, and the gas volume measured would be too large.

(a) Read up from 1.5 min to the curve and across: $62 \text{ cm}^3$.

(b) As the acid is used up its concentration decreases, so there are fewer acid particles per unit volume and the frequency of collisions (collisions per second) between acid particles and the marble surface decreases. The rate — the gradient — falls.

(c) The marble chips are in excess, so the hydrochloric acid is the limiting reactant: once it is all used up, no more gas forms and the line is horizontal.

(d) Final volume $= 90 \text{ cm}^3$.

HCl : $\text{CO}_2$ = 2 : 1, and $50.0 \text{ cm}^3 = 0.0500 \text{ dm}^3$.

(a) Heat breaks calcium carbonate down into calcium oxide (quicklime) and carbon dioxide:

(b) Calcium oxide is a basic oxide; silicon dioxide is an acidic oxide. They react together — a neutralisation — to form calcium silicate, the slag, which is molten and runs off.

(c) Mass of pure calcium carbonate (1 tonne = $10^6$ g):

$M_r$ of $\text{CaCO}_3 = 40 + 12 + 48 = 100$; ratio $\text{CaCO}_3 : \text{CaO}$ = 1 : 1; $M_r$ of CaO $= 56$.

(a) To heat to constant mass: the 2nd and 3rd masses are the same, so all the calcium carbonate has decomposed and no more carbon dioxide is lost.

(b) The mass lost is carbon dioxide:

(c) $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$, ratio 1 : 1. $M_r$ of $\text{CO}_2 = 44$ and of $\text{CaCO}_3 = 100$.

Mass of limestone $= 23.62 - 18.62 = 5.00$ g.

(d) The impurities do not lose mass on heating (they do not decompose or give off gas), so all the mass lost is carbon dioxide from calcium carbonate.

(e) Powder has a larger surface area, so it decomposes faster and more completely.

Fizzing with acid and a gas that turns limewater milky shows carbonate ions — this rules out D (sulfates do not fizz with acid).

X dissolves in water. Carbonates are insoluble except sodium, potassium and ammonium, so B (calcium carbonate, insoluble) is ruled out.

Warming with sodium hydroxide gives no ammonia, so X is not an ammonium salt — this rules out A. The answer is C, potassium carbonate.

(a) In the first tube: fizzing / effervescence (and the solid dissolves). In the second tube: the limewater turns milky.

(b) The gas is carbon dioxide. The tube must dip below the surface so the gas bubbles through the limewater and can react with it; otherwise it escapes above the liquid.

(c) A green precipitate insoluble in excess with both reagents, turning brown near the surface on standing, shows iron(II) ions, $\text{Fe}^{2+}$. Combined with the carbonate result, J is iron(II) carbonate, $\text{FeCO}_3$.

(a) Acids react with carbonates to give carbon dioxide. Sodium chloride has no carbonate ions, so it does not react with hydrochloric acid.

(b) $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$, so $\text{Na}_2\text{CO}_3 : \text{CO}_2$ = 1 : 1.

$M_r$ of $\text{Na}_2\text{CO}_3 = 2(23) + 12 + 3(16) = 106$.

(c) Ratio $\text{Na}_2\text{CO}_3 : \text{HCl}$ = 1 : 2.

(d) Less gas was measured than was really made. Carbon dioxide is slightly soluble in water, so some dissolves as it passes through the trough. (Or: some gas escapes before the bung is put in.) A gas syringe would avoid the first problem.

(a) The mass lost is carbon dioxide ($M_r = 44$):

(b) $\text{MCO}_3 \rightarrow \text{MO} + \text{CO}_2$, ratio 1 : 1, so 0.0250 mol of $\text{MCO}_3$ has a mass of 3.10 g.

The carbonate part is $12 + 3(16) = 60$, so $A_r(\text{M}) = 124 - 60 = 64$. The Periodic Table gives copper. Check with the oxide: $\dfrac{2.00}{0.0250} = 80 = 64 + 16$. ✓

(c)

(d) Add aqueous ammonia: a light blue precipitate forms, which dissolves in excess to give a dark blue solution — this confirms $\text{Cu}^{2+}$. (A blue-green flame test also confirms copper(II).)