Trends in Chemical Properties of Group 14 Elements and Their Compounds

Lessons

Senior 4 · Chemistry

Trends in Chemical Properties of Group 14 Elements and Their Compounds

5 lessons

Lesson 4 of 5

Tests for Tin(II), Lead(II), Carbonate and Hydrogencarbonate Ions

25 min

Two white powders

Baking soda (sodium hydrogencarbonate) and washing soda (sodium carbonate) are both white powders that fizz with vinegar, and old paints, pipes and solders may contain lead or tin compounds that look like any other white solid. A chemist cannot tell them apart by eye, but a few test-tube reactions can. This lesson gives the tests for two Group 14 cations, tin(II) and lead(II), and for the two anions that carbon forms with oxygen in water, carbonate and hydrogencarbonate, and explains the chemistry behind each observation.

Definition

Qualitative test: a reaction that shows whether a particular ion is present by a clear observation (a precipitate of a stated colour, a gas, a colour change). Example: a yellow precipitate with potassium iodide solution shows lead(II) ions.

Confirmatory test: a second, different test used to rule out ions that give the same result in the first test. Example: zinc, aluminium, tin(II) and lead(II) hydroxides all dissolve in excess sodium hydroxide, so a further test is needed.

Lead(II) ions, $\mathrm{Pb^{2+}}$

Reagent	Observation	Equation
NaOH(aq), a few drops then excess	white precipitate, dissolves in excess to a colourless solution	$\mathrm{Pb^{2+} + 2OH^- \rightarrow Pb(OH)_2(s)}$ ; $\mathrm{Pb(OH)_2 + 2OH^- \rightarrow [Pb(OH)_4]^{2-}}$
NH₃(aq), a few drops then excess	white precipitate, does not dissolve in excess	$\mathrm{Pb^{2+} + 2OH^- \rightarrow Pb(OH)_2(s)}$
KI(aq)	bright yellow precipitate; it dissolves when the mixture is boiled and reappears as glittering golden plates on cooling ("golden rain")	$\mathrm{Pb^{2+}(aq) + 2I^-(aq) \rightarrow PbI_2(s)}$
dilute HCl or NaCl(aq)	white precipitate, dissolves on heating, reappears on cooling	$\mathrm{Pb^{2+} + 2Cl^- \rightarrow PbCl_2(s)}$
dilute $\mathrm{H_2SO_4}$	white precipitate	$\mathrm{Pb^{2+} + SO_4^{2-} \rightarrow PbSO_4(s)}$

Tin(II) ions, $\mathrm{Sn^{2+}}$

Reagent	Observation	Equation
NaOH(aq), a few drops then excess	white precipitate, dissolves in excess	$\mathrm{Sn^{2+} + 2OH^- \rightarrow Sn(OH)_2(s)}$ ; $\mathrm{Sn(OH)_2 + 2OH^- \rightarrow [Sn(OH)_4]^{2-}}$
NH₃(aq), a few drops then excess	white precipitate, does not dissolve in excess	$\mathrm{Sn^{2+} + 2OH^- \rightarrow Sn(OH)_2(s)}$
dilute $\mathrm{H_2SO_4}$	no precipitate (tin(II) sulphate is soluble)	none
iodine solution, a few drops	brown colour disappears	$\mathrm{Sn^{2+} + I_2 \rightarrow Sn^{4+} + 2I^-}$
acidified potassium manganate(VII), a few drops	purple colour disappears	$\mathrm{5Sn^{2+} + 2MnO_4^- + 16H^+ \rightarrow 5Sn^{4+} + 2Mn^{2+} + 8H_2O}$

The hydroxide tests alone cannot separate tin(II) from lead(II): both hydroxides are amphoteric (Lesson 3). The decisive differences are the yellow lead(II) iodide and the reducing power of tin(II), which follows from the inert pair effect: tin(II) is readily oxidised to the more stable tin(IV). Figure 9.7 puts the tests in order.

Figure 9.7: Telling $\mathrm{Pb^{2+}}$ from $\mathrm{Sn^{2+}}$ . Aluminium ions also pass the first step but give neither a colour change nor a precipitate with iodide.

Carbonate and hydrogencarbonate ions

Both react with dilute acid, giving effervescence of carbon dioxide, which turns limewater (calcium hydroxide solution) milky:

\mathrm{CO_3^{2-}(aq) + 2H^+(aq) \rightarrow CO_2(g) + H_2O(l)} \qquad \mathrm{HCO_3^-(aq) + H^+(aq) \rightarrow CO_2(g) + H_2O(l)}

\mathrm{CO_2(g) + Ca(OH)_2(aq) \rightarrow CaCO_3(s) + H_2O(l)}

Two tests tell them apart:

Magnesium sulphate (or calcium chloride) solution, cold. A carbonate gives a white precipitate at once, because magnesium carbonate is insoluble: $\mathrm{Mg^{2+} + CO_3^{2-} \rightarrow MgCO_3(s)}$ . A hydrogencarbonate gives no precipitate in the cold, because magnesium hydrogencarbonate is soluble; when the mixture is boiled, the hydrogencarbonate decomposes and the white precipitate appears: $\mathrm{Mg^{2+}(aq) + 2HCO_3^-(aq) \xrightarrow{\text{heat}} MgCO_3(s) + CO_2(g) + H_2O(l)}$
Heat the solid (Figure 9.8). Sodium and potassium hydrogencarbonates decompose at gentle heat, giving water droplets and carbon dioxide; the carbonates of sodium and potassium do not decompose at Bunsen temperatures: $\mathrm{2NaHCO_3(s) \rightarrow Na_2CO_3(s) + H_2O(g) + CO_2(g)}$ (Carbonates of most other metals do decompose on strong heating, so this test separates the sodium and potassium salts only.)

Figure 9.8: Heating a hydrogencarbonate (clamp stand not shown). The mouth of the tube points slightly down so that condensed water cannot run back onto the hot glass and crack it.

Worked example 1: identifying four solutions

Four colourless solutions, A to D, are lead(II) nitrate, tin(II) chloride (in dilute HCl), sodium carbonate and sodium hydrogencarbonate. Results: A gives a yellow precipitate with KI(aq). B decolourises iodine solution. C and D both fizz with dilute nitric acid; C gives a white precipitate with cold $\mathrm{MgSO_4(aq)}$ , D gives one only on boiling. Identify A to D and write an ionic equation for each positive result.

Solution.

A: lead(II) nitrate, $\mathrm{Pb^{2+} + 2I^- \rightarrow PbI_2}$ .
B: tin(II) chloride, $\mathrm{Sn^{2+} + I_2 \rightarrow Sn^{4+} + 2I^-}$ (tin(II) is a reducing agent).
C: sodium carbonate, $\mathrm{Mg^{2+} + CO_3^{2-} \rightarrow MgCO_3}$ .
D: sodium hydrogencarbonate, $\mathrm{Mg^{2+} + 2HCO_3^- \rightarrow MgCO_3 + CO_2 + H_2O}$ on boiling.

Note that nitric acid, not hydrochloric or sulphuric acid, was used for the fizz test: either of those could precipitate lead(II) if a lead solution were tested by mistake, hiding the result.

Worked example 2: analysing a mixture by heating

A 5.00 g mixture of sodium carbonate and sodium hydrogencarbonate is heated to constant mass and loses 0.620 g. Calculate the percentage of sodium hydrogencarbonate in the mixture. $M$ : $\mathrm{NaHCO_3}$ 84.01, $\mathrm{Na_2CO_3}$ 105.99 g mol⁻¹.

Solution. Only the hydrogencarbonate decomposes: $\mathrm{2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2}$ . For every $2 \times 84.01 = 168.02$ g of $\mathrm{NaHCO_3}$ , the mass lost is $168.02 - 105.99 = 62.03$ g (the water and carbon dioxide). $m(\mathrm{NaHCO_3}) = 0.620 \times \frac{168.02}{62.03} = 1.68\ \mathrm{g} \qquad \% = \frac{1.68}{5.00} \times 100 = 33.6\%$

Warning

Common mistakes. (1) Saying "the precipitate dissolves in excess ammonia" for lead or tin: neither does; only zinc's does. (2) Using dilute HCl to test a possible lead salt for carbonate: insoluble $\mathrm{PbCl_2}$ can coat the solid and stop the fizzing; use dilute nitric acid. (3) Concluding "carbonate" from fizzing alone: hydrogencarbonates fizz too. (4) Letting limewater suck back into the hot tube: remove the delivery tube before you stop heating.

Safety. Lead compounds are toxic, especially to young people: use small amounts of dilute solutions, wear gloves, and wash your hands; dispose of lead waste as your teacher directs, never down the sink. Potassium manganate(VII) stains and is an oxidiser. Wear eye protection throughout; if a solution splashes into your eyes, rinse with plenty of running water and get medical help.

Tip

Lead is yellow, tin is a reducer. Pb²⁺: yellow $\mathrm{PbI_2}$ (golden rain). Sn²⁺: takes the colour out of iodine and manganate(VII). For the carbon anions: both fizz; only carbonate clouds cold $\mathrm{MgSO_4}$ .

Thoisoi2: Golden Rain, growing crystals of lead iodide (3 min). Watch the bright yellow precipitate form when lead(II) nitrate and potassium iodide solutions are mixed, dissolve in hot water, and come back as glittering crystals on cooling.

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Practice

Describe what you would see when sodium hydroxide solution is added, drop by drop until in excess, to lead(II) nitrate solution. Write two ionic equations.
How would you show that a solution contains tin(II) and not lead(II)? Give two tests and their results.
Write the ionic equation for the reaction of tin(II) ions with acidified manganate(VII) ions, and state which species is oxidised.
Describe two tests that distinguish sodium carbonate from sodium hydrogencarbonate.
Why must the delivery tube be removed from the limewater before heating stops?
Calculate the mass of lead(II) iodide precipitated when excess potassium iodide is added to 25.0 cm³ of 0.100 mol dm⁻³ lead(II) nitrate. $A_{r}$ : Pb 207.2, I 126.9.
Heating 2.52 g of a pure Group 1 hydrogencarbonate, $\mathrm{MHCO_3}$ , to constant mass gave 1.59 g of the carbonate, $\mathrm{M_2CO_3}$ . Identify M.

Show answersHide answers

A white precipitate forms, then dissolves in excess to give a colourless solution: $\mathrm{Pb^{2+} + 2OH^- \rightarrow Pb(OH)_2}$ ; $\mathrm{Pb(OH)_2 + 2OH^- \rightarrow [Pb(OH)_4]^{2-}}$ .
Iodine solution or acidified manganate(VII) is decolourised, because tin(II) is a reducing agent (lead(II) is not). Dilute sulphuric acid gives no precipitate, whereas lead(II) would give white $\mathrm{PbSO_4}$ (and potassium iodide would give yellow $\mathrm{PbI_2}$ ).
$\mathrm{5Sn^{2+} + 2MnO_4^- + 16H^+ \rightarrow 5Sn^{4+} + 2Mn^{2+} + 8H_2O}$ ; tin(II) is oxidised (to tin(IV)).
Cold $\mathrm{MgSO_4(aq)}$ : white precipitate at once with carbonate, only on boiling with hydrogencarbonate. Heating the solids: the hydrogencarbonate gives water droplets and $\mathrm{CO_2}$ (limewater milky); sodium carbonate gives no gas.
As the tube cools, the gas inside contracts and the limewater would be sucked back into the hot tube, which could crack it.
$n(\mathrm{Pb^{2+}}) = 0.0250 \times 0.100 = 2.50 \times 10^{-3}$ mol; $M(\mathrm{PbI_2}) = 461.0$ ; mass $= 2.50 \times 10^{-3} \times 461.0 = 1.15\ \mathrm{g}$ .
$\mathrm{2MHCO_3 \rightarrow M_2CO_3 + H_2O + CO_2}$ . Let the molar mass of M be m: $\frac{2.52}{m + 61.02} \times \frac{1}{2} = \frac{1.59}{2m + 60.01}$ . Solving: $2.52 (2 m + 60.01) = 3.18 (m + 61.02)$ , so $5.04 m + 151.2 = 3.18 m + 194.0$ , $m = 23.0$ : M is sodium.

Quick check

A colourless solution gives a white precipitate with NaOH(aq) that dissolves in excess, and a bright yellow precipitate with KI(aq). Which ion is present?

Key points

$\mathrm{Pb^{2+}}$ and $\mathrm{Sn^{2+}}$ both give white hydroxide precipitates that dissolve in excess NaOH but not in excess NH₃.
$\mathrm{Pb^{2+}}$ : bright yellow $\mathrm{PbI_2}$ with KI (soluble in hot water, "golden rain" on cooling); white $\mathrm{PbCl_2}$ and $\mathrm{PbSO_4}$ .
$\mathrm{Sn^{2+}}$ : a reducing agent; it decolourises iodine solution and acidified manganate(VII).
$\mathrm{CO_3^{2-}}$ and $\mathrm{HCO_3^-}$ both fizz with acid ( $\mathrm{CO_2}$ turns limewater milky). Only carbonate gives a precipitate with cold $\mathrm{MgSO_4}$ ; sodium hydrogencarbonate, but not sodium carbonate, decomposes on gentle heating.

Sources and credits

Image: Lluvia de oro.JPG by Der Kreole, via Wikimedia Commons, licensed under CC BY 3.0.
Image: Sodium bicarbonate.jpg by Thavox, via Wikimedia Commons, public domain.
Lead and health: Lead poisoning and health, World Health Organization; first aid for chemical splashes: NIOSH Pocket Guide, first aid, CDC.
Video: Golden Rain - Growing crystals of lead iodide, Thoisoi2 - Chemical Experiments!, embedded from YouTube.
Written by Ganzaa: the text, apparatus drawing, flowchart, tables, worked examples and Practice items.

1Two white powders

2Lead(II) ions, Pb2+ \mathrm{Pb^{2+}} Pb2+

3Tin(II) ions, Sn2+ \mathrm{Sn^{2+}} Sn2+

4Carbonate and hydrogencarbonate ions

Two white powders

Lead(II) ions, $\mathrm{Pb^{2+}}$

Tin(II) ions, $\mathrm{Sn^{2+}}$

Carbonate and hydrogencarbonate ions