Questions on Group 2 compounds: solubility, uses and the sulfate test

Question 1

Describe how sodium hydroxide solution and dilute sulfuric acid could be used to identify which of four solutions contains Mg²⁺, Ca²⁺, Sr²⁺ and Ba²⁺ ions. Give the expected observations and explain them.

Accepted Answer

Indicative content

add sodium hydroxide solution dropwise to a sample of each solution, and dilute sulfuric acid to a fresh sample of each
with sodium hydroxide, Mg²⁺ gives a white precipitate, Ca²⁺ a slight white precipitate, and Sr²⁺ and Ba²⁺ little or none
with dilute sulfuric acid, Ba²⁺ gives a dense white precipitate, Sr²⁺ a white precipitate, Ca²⁺ a slight one and Mg²⁺ none
the hydroxide results follow because hydroxide solubility increases down the group, so magnesium hydroxide precipitates most
the sulfate results follow because sulfate solubility decreases down the group, so barium sulfate precipitates most
a correct ionic equation, such as Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s) or Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Levels

Level 3 (5–6 marks) both reagents are used on separate samples, the observations for all four ions with each reagent are given, both trends in solubility are used to explain them, and a correct ionic equation is written; the answer would let a student tell the four solutions apart
Level 2 (3–4 marks) the observations with both reagents are mostly right and one solubility trend explains them; the other trend or the ionic equation is not there
Level 1 (1–2 marks) a white precipitate is mentioned for one ion; the two solubility trends are often run the same way, with barium hydroxide called the least soluble

Question 2

Describe the trends in solubility of the Group 2 hydroxides and sulfates shown in the bar chart, and explain how the hydroxide trend affects the pH of their saturated solutions.

Accepted Answer

B1 the hydroxides become more soluble down the group, Mg(OH)₂ being sparingly soluble B1 the sulfates become less soluble down the group, BaSO₄ being insoluble B1 a more soluble hydroxide puts a higher concentration of hydroxide ions into its saturated solution B1 so the pH of the saturated solution rises down the group, from about 10 for Mg(OH)₂ to about 13 for Ba(OH)₂

Question 3

Give one use each of magnesium hydroxide, calcium hydroxide, calcium oxide and barium sulfate, and the property of each compound that makes it suitable.

Accepted Answer

B1 magnesium hydroxide treats indigestion: it neutralises excess stomach acid, and is sparingly soluble so the hydroxide concentration stays low B1 calcium hydroxide treats acidic soil: it is a base that neutralises the acid and raises the pH B1 calcium oxide removes sulfur dioxide from flue gases: it is a basic oxide that reacts with the acidic gas B1 barium sulfate is the barium meal: it is insoluble, so toxic Ba²⁺ is not absorbed, and barium absorbs X-rays

Question 4

Explain why barium sulfate is given as a suspension for a barium meal, although barium ions are toxic, while magnesium sulfate (Epsom salts) can be taken dissolved in water.

Accepted Answer

B1 barium sulfate is insoluble, so it cannot form a solution and is swallowed as a suspension of solid B1 because it does not dissolve, barium ions are not released and are not absorbed into the blood; reject 'barium sulfate is not toxic' B1 barium absorbs X-rays strongly, so the gut lined with barium sulfate shows up on the X-ray B1 magnesium sulfate is soluble and dissolves, and magnesium ions are not toxic at that dose

Question 5

Describe how to test a solution for sulfate ions, giving the reagents, the positive result and the ionic equation, with state symbols, for the reaction.

Accepted Answer

B1 add dilute hydrochloric acid (or dilute nitric acid) to the solution first; reject sulfuric acid B1 then add a few drops of barium chloride solution; allow barium nitrate B1 a white precipitate shows that sulfate ions are present; ignore 'cloudy' unless a precipitate is named B1 Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Question 6

Explain why the solution is acidified before barium chloride is added in the sulfate test, why sulfuric acid cannot be the acid, and why fizzing when the acid is added does not spoil the test.

Accepted Answer

B1 the acid removes carbonate ions, reacting with them to form carbon dioxide and water B1 otherwise carbonate ions would form a white precipitate of barium carbonate, giving a false positive; reject 'removes impurities' B1 sulfuric acid contains sulfate ions, so a white precipitate would form whether or not the sample contained sulfate B1 fizzing shows that carbonate was present; once it has been removed, a white precipitate still shows sulfate

Question 7

Evaluate the use of calcium carbonate rather than calcium oxide to remove sulfur dioxide from power-station flue gases.

Accepted Answer

B1 both react with sulfur dioxide to form calcium sulfite, so both remove it B1 calcium carbonate releases carbon dioxide at the scrubber: CaCO₃(s) + SO₂(g) → CaSO₃(s) + CO₂(g) B1 calcium oxide is made by heating calcium carbonate, which releases the same carbon dioxide at the kiln and uses energy B1 so calcium carbonate is cheaper and releases no more carbon dioxide overall, which makes it the better choice

Question 8

Describe how to compare the thermal stabilities of magnesium carbonate and calcium carbonate in the laboratory, including how to make the test fair and how to avoid suck-back.

Accepted Answer

B1 heat the same amount in moles of each carbonate in identical boiling tubes, with a delivery tube leading into limewater B1 use the same Bunsen flame at the same height and the same volume of limewater for each B1 time how long it takes for the limewater to turn milky; the shorter the time, the less stable the carbonate B1 lift the delivery tube out of the limewater before removing the flame, so cold limewater is not drawn back into the hot tube

Question 9

Describe a test, with its positive result, for each of hydrogen, oxygen, chlorine and sulfur dioxide.

Accepted Answer

B1 hydrogen: a lighted splint at the mouth of the tube gives a squeaky pop B1 oxygen: a glowing splint placed in the gas relights B1 chlorine: damp blue litmus paper turns red and is then bleached white; allow damp litmus is bleached B1 sulfur dioxide: acidified aqueous potassium manganate(VII) turns from purple to colourless; allow acidified potassium dichromate(VI) turning from orange to green

Question 10

Explain how a student could show that a solution of an unknown salt contains nitrate ions, using sodium hydroxide solution and aluminium foil, and why sodium hydroxide is first warmed with the solution on its own.

Accepted Answer

B1 warm the solution with sodium hydroxide alone and test for ammonia with damp red litmus, so that ammonia from any ammonium ions is detected and driven off first B1 then add aluminium foil, or Devarda's alloy, and warm again B1 a gas that turns damp red litmus paper blue is ammonia, and shows that nitrate ions were present B1 aluminium reduces the nitrogen from +5 in NO₃⁻ to −3 in NH₃, so the ammonia can only have come from nitrate

Question 11

Show that the concentration of sulfate ions in a solution is 0.100 mol dm⁻³, given that excess barium chloride solution added to 25.0 cm³ of it gives 0.583 g of dry barium sulfate. Ar values: Ba 137.3, S 32.1, O 16.0.

Accepted Answer

M1 Mr of BaSO₄ = 137.3 + 32.1 + 4 × 16.0 = 233.4 M1 n(BaSO₄) = 0.583 ÷ 233.4 = 2.50 × 10⁻³ mol M1 Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s), and barium chloride is in excess, so n(SO₄²⁻) in the 25.0 cm³ sample = 2.50 × 10⁻³ mol A1 concentration = 2.50 × 10⁻³ ÷ 0.0250 = 0.100 mol dm⁻³, with the working shown in full because the answer is given

Question 12

Calculate the volume, in cm³, of 0.0500 mol dm⁻³ hydrochloric acid neutralised by a tablet containing 0.350 g of magnesium hydroxide. Ar values: H 1.0, O 16.0, Mg 24.3.

Accepted Answer

M1 Mr(Mg(OH)₂) = 58.3, so amount of Mg(OH)₂ = 0.350 ÷ 58.3 = 6.003 × 10⁻³ mol M1 Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l), so amount of HCl = 2 × 6.003 × 10⁻³ = 1.201 × 10⁻² mol A1 volume = 1.201 × 10⁻² ÷ 0.0500 = 0.240 dm³ = 240 cm³, to three significant figures

Question 13

Calculate the mass of calcium oxide, in tonnes, needed to remove 2.50 tonnes of sulfur dioxide from flue gases. Ar values: O 16.0, S 32.1, Ca 40.1.

Accepted Answer

M1 Mr(SO₂) = 64.1, so amount of SO₂ = 2.50 × 10⁶ g ÷ 64.1 = 3.900 × 10⁴ mol M1 CaO(s) + SO₂(g) → CaSO₃(s) is 1 : 1, and Mr(CaO) = 56.1, so mass = 3.900 × 10⁴ × 56.1 g A1 2.19 × 10⁶ g, which is 2.19 tonnes to three significant figures

Question 14

Identify the Group 2 ion, Mg²⁺, Ca²⁺ or Ba²⁺, in each of solutions P, Q and R from the results in the table.

Accepted Answer

B1 P contains Mg²⁺, since magnesium hydroxide is only sparingly soluble and magnesium sulfate is soluble B1 Q contains Ba²⁺, since barium hydroxide is soluble and barium sulfate insoluble B1 R contains Ca²⁺, since calcium hydroxide and calcium sulfate are both sparingly soluble

Question 15

Suggest why a student who added barium chloride solution to tap water from a limestone area saw a white precipitate, and what she should do to find out whether the water contains sulfate ions.

Accepted Answer

B1 water from a limestone area contains carbonate (or hydrogencarbonate) ions B1 these form a white precipitate of barium carbonate, so the result does not show sulfate B1 repeat the test after adding dilute hydrochloric acid until fizzing stops; a white precipitate then shows sulfate

Question 16

Compare the results obtained when sodium hydroxide solution and dilute sulfuric acid are added separately to solutions of magnesium chloride and barium chloride.

Accepted Answer

B1 with sodium hydroxide, magnesium chloride gives a white precipitate but barium chloride gives no change B1 with dilute sulfuric acid, barium chloride gives a white precipitate but magnesium chloride gives no change B1 the results are opposite because hydroxide solubility increases down the group while sulfate solubility decreases

Question 17

Explain why barium carbonate needs a higher temperature to decompose than magnesium carbonate.

Accepted Answer

B1 the Ba²⁺ ion is larger than the Mg²⁺ ion and has the same charge, so its charge density is lower B1 Ba²⁺ polarises (distorts) the electron cloud of the carbonate ion less, so the C–O bonds in the anion are weakened less B1 more energy is needed to break the carbonate ion into an oxide ion and carbon dioxide, so a higher temperature is needed

Question 18

Calculate the volume of oxygen, in cm³ at room temperature and pressure, given off when 1.48 g of magnesium nitrate is fully decomposed by heating. Ar values: N 14.0, O 16.0, Mg 24.3; molar gas volume = 24.0 dm³ mol⁻¹.

Accepted Answer

M1 Mr(Mg(NO₃)₂) = 24.3 + 2 × (14.0 + 3 × 16.0) = 148.3, so amount = 1.48 ÷ 148.3 = 9.980 × 10⁻³ mol M1 2Mg(NO₃)₂(s) → 2MgO(s) + 4NO₂(g) + O₂(g), so amount of O₂ = 9.980 × 10⁻³ ÷ 2 = 4.990 × 10⁻³ mol A1 volume = 4.990 × 10⁻³ × 24.0 = 0.120 dm³ = 120 cm³, to three significant figures

Question 19

Identify the anion in each of solutions X, Y and Z from the results in the table.

Accepted Answer

B1 X contains carbonate ions, CO₃²⁻, because the gas given off with acid is carbon dioxide B1 Y contains sulfite ions, SO₃²⁻, because barium sulfite dissolves in acid and gives sulfur dioxide, which reduces manganate(VII) B1 Z contains sulfate ions, SO₄²⁻, because barium sulfate is insoluble even in excess acid

Question 20

Suggest why damp blue litmus paper held in chlorine turns red and then white, while damp blue litmus paper held in sulfur dioxide turns red and stays red.

Accepted Answer

B1 chlorine reacts with the water on the paper to give hydrochloric acid and chloric(I) acid, which turn the litmus red B1 chloric(I) acid is an oxidising agent and bleaches the dye, so the paper turns white B1 sulfur dioxide dissolves to give sulfuric(IV) acid, which is acidic and turns the litmus red but does not oxidise the dye, so it stays red

Question 21

Suggest why barium hydroxide would not be suitable as an indigestion remedy in place of magnesium hydroxide.

Accepted Answer

B1 barium hydroxide is soluble, so it would give a high concentration of hydroxide ions, a strongly alkaline solution that would damage the gut B1 it would release toxic Ba²⁺ ions into solution, to be absorbed into the blood

Question 22

Suggest why heating lithium nitrate gives a brown gas but heating sodium nitrate does not.

Accepted Answer

B1 Li⁺ is a very small ion with a high charge density, so it polarises the nitrate ion strongly, as a Group 2 ion does, and the nitrate decomposes to the oxide, brown nitrogen dioxide and oxygen B1 Na⁺ is larger, so it polarises the nitrate ion less and sodium nitrate only loses oxygen, forming sodium nitrite with no nitrogen dioxide

Solution	With sodium hydroxide solution	With dilute sulfuric acid
P	White precipitate	No change
Q	No change	Dense white precipitate
R	Slight white precipitate	Slight white precipitate

Solution	Test	Result
X	Dilute hydrochloric acid added	Effervescence; the gas turns limewater milky
Y	Ba²⁺(aq), then excess dilute hydrochloric acid	White precipitate dissolves; the gas decolourises acidified manganate(VII)
Z	Ba²⁺(aq), then excess dilute hydrochloric acid	White precipitate that does not dissolve