Salt Preparation

What are the 5 solubility rules?

Back of the flashcard!

 

What would be observed for the water level at X after some time?

 

A)  fall and remain constant

 

B)  fall and rise back to X

 

C)  rise and fall back to X

 

D)  rise and remain constant

 

 

Answer: C

 

- In a porous pot qns, a gas diffuses into the pot while another gas diffuses out i.e. CO₂ diffuses in and Ne diffuses out.

 

- CO₂ has a larger M_r than Ne → Ne diffuses out faster than CO₂ diffuses in.

 

- After a short while, there will be lesser as particles inside the pot → pressure in the pot decrease → water level at X will rise

 

- When the concentration of Ne inside and outside the pot becomes equal, Ne will stop diffusing out while CO₂ will continue to diffuse in. So the pressure inside the pot will slowly increase and the water level will gradually fall back to X.

 

 

 

What would be observed for the water level at x after some time?

 

A)  becomes higher

 

B)  becomes lower

 

C)  fluctuates about x

 

D)  remains unchanged

 

 

Answer: B

 

- In this qns, the difficulty comes in that a mixture of gases is present inside the porous pot.

 

- For a mixture of gases, we need to use the average relative molecular mass which will be between that of the 2 gases i.e. between 44 (CO₂) and 71 (Cl₂).

 

- Comparing the values to that of the gas outside the pot: ammonia (relative mass of 17), it is clear that the gas outside the pot will diffuse in faster than the gas inside the pot diffusing out.

 

- After some time, there will be more gas particles inside the pot → pressure is higher → water level at x will become lower.

 

Common mistake: students will put down the relative mass of chlorine to be 35.5. Bear in mind that chlorine exists as a diatomic molecule i.e. Cl₂.

 

Qualitative Analysis - Thinking Questions

<< Qualitative Analysis

 

Test for Cations

 

[Test for Ca²⁺] On adding aq NaOH, a white ppt is formed. The white ppt is Ca(OH)₂. Isn’t Ca(OH)₂ an alkali/ soluble in water? Isn’t Ca(OH)₂ limewater? So why would there be a precipitate?

 

 

Ca(OH)₂ is sparingly/ slightly soluble in water (~1.7 g in 1 dm³ of water). Compare with NaOH (~1100 g in 1 dm³ of water) which is very soluble.

 

 

Under the conditions of the laboratory test, the amount of Ca(OH)₂ produced is greater than the water can dissolve, hence it precipitates out as solid Ca(OH)₂.

 

[Test for Ca²⁺] On adding aq NH₃, no ppt is formed. Since NH₃ can ionise in water to produce OH^–, shouldn’t Ca(OH)₂ (white ppt) be formed too?

 

For a precipitate to form, there must be sufficient amounts of the constituent ions in solution. Since aq NH₃ is a weak alkali, the amount of OH^– is very little and insufficient for solid Ca(OH)₂ to form.

 

Benzoic acid is widely used in the food industry as a food preservative. The melting and boiling points of benzoic acid are 122 °C and 249 °C respectively.

 

Which of the following will occur to the particles of benzoic acid when it is cooled from 400 °C to 100 °C?

 

	distance between particles	energy of particles
A)	decreases	decreases
B)	decreases	increases
C)	increases	decreases
D)	increases	increases

 

 

Answer: A

 

In this question, the melting and boiling points given are redundant. The key concept here is that the substance is cooled.

 

For any substance that is cooled, the distance between particles will decrease and the energy of particles will decrease.

 

Similarly for any substance that is heated, the distance between particles will increase and the energy of particles will increase.

 

Nitrogen dioxide is a dark brown gas and is denser than air.

 

A gas jar containing nitrogen dioxide is sealed with a glass plate and is then inverted on top of a gas jar containing air.

 

 

The glass plate is removed. Which one of the following correctly describes the colours inside the gas jars after a long period of time?

 

A)  brown in both upper and lower jars

 

B)  dark brown in lower jar and colourless in upper jar

 

C)  dark brown in lower jar and light brown in upper jar

 

D)  light brown in lower jar and dark brown in upper jar

 

 

Answer: A

 

When gases are mixed, given sufficient period of time, a uniform/ homogeneous mixture will be obtained.

 

Common misconception: shouldn't the denser gas sink to the bottom jar?

 

In a gas, the particles are moving so rapidly and randomly that the effect of density is negligible.

 

Just imagine a chaotic crisis situation where everyone is running for their lives, a group of friends will not be able to stay together and become lost in the crowd.

 

HCl is very soluble in water, whereas Cl₂ is only slightly soluble in water. Both gases can be dried using concentrated sulfuric acid.

 

Which diagram represents the correct method of obtaining pure dry Cl₂ from a mixture of damp Cl₂ and HCl?

 

 

Answer: B

 

- For a gas to be passed through a solution/ drying agent, the in-tube must be submerged into the solution → C and D shows an incorrect setup.

 

- In the correct setup (B), the mixture of gases is passed into water → HCl is dissolved and only Cl₂ is passed into the 2nd tube.
In the 2nd tube, damp Cl₂ is passed into conc. sulfuric acid (drying agent) → water is removed and dry Cl₂ is obtained.

 

- Setup in A is erroneous as the gas after drying with conc. sulfuric acid is passed into water again.

 

 

The apparatus can be used to show the diffusion of gases.

 

 

Which pair of gases A and B would cause a fall in the water level at the right side of the U-shaped tube?

 

A)  A = carbon monoxide, B = nitrogen

 

B)  A = fluorine, B = argon

 

C)  A = nitrogen dioxide, B = chlorine

 

D)  A = oxygen, B = neon

 

 

Answer: D

 

- In a porous pot qns, a gas diffuses into the pot while another gas diffuses out i.e. A diffuses in and B diffuses out.

 

- If gas A diffuses in faster than gas B diffuses out → over time there will be more gas particles inside the pot → pressure in the pot increase.
- If gas A diffuses in slower than gas B diffuses out → over time there will be less gas particles inside the pot → pressure in the pot decrease.

 

- For the water level at the right side to fall, the pressure in the pot must decrease i.e. gas A must diffuse in slower than gas B diffuses out.
- We can hence deduce that gas A must have a large relative molecular mass (i.e. be heavier) than gas B.

 

Common mistake: students will put down the relative mass of nitrogen to be 14, oxygen to be 16, chlorine to be 35.5 and fluorine to be 19. Bear in mind that these gases exists as a diatomic molecule i.e. N₂, O₂, Cl₂, F₂.

 

The time taken for 30 cm³ of a gas C_xH_y to diffuse through a porous barrier was the same as that for 30 cm³ of nitrogen gas under the same conditions.

 

Determine the value of x and y.

 

A)  x = 1, y = 2

 

B)  x = 1, y = 4

 

C)  x = 2, y = 4

 

D)  x = 3, y = 6

 

 

Answer: C

 

- Recall: rate of diffusion of a gas is dependent on its mass.
- Since C_xH_y takes the same amount of time to diffuse as nitrogen gas, C_xH_y must have the same relative molecular mass as nitrogen gas i.e. 14x2 = 28.

 

Common mistake: students will put down the relative mass of nitrogen to be 14. Bear in mind that nitrogen exists as diatomic molecules i.e. N₂.

 

The table shows the boiling points of some gases in air.

 

Gas	Boiling Point/ °C
Argon	-186
Helium	-269
Neon	-246
Nitrogen	-196
Oxygen	-183

 

When air is cooled to -200 °C, which gases will liquefy?

 

A)  argon, helium, neon

 

B)  argon, nitrogen, oxygen

 

C)  helium, neon, nitrogen

 

D)  helium, neon

 

 

Answer: B

 

- For such questions, drawing a number line can be very useful i.e.

 

(1) draw a line to represent temperature range (left to right = increasing temperature),
(2) mark out 2 points and indicate melting point and boiling point,
(3) write the states (solid, liquid, gas) in the corresponding regions.

 

 

- Next, mark out the temperature(s) given in the question.

 

 

- As shown in the diagram above, for a gas to be liquid at -200 °C, the boiling point must be higher than -200 °C.
- Do take note that since the temperatures are negative, the larger the numerical value, the lower is the temperature.

 

20 cm³ of an aq 1.0 mol/dm³ solution of the hydroxide of metal X requires 40 cm³ of aq 0.25 mol/dm³ of sulfuric acid for complete neutralisation.

 

What is the formula for the sulfate of X?

 

A)  X₂SO₄

 

B)  XSO₄

 

C)  X₂(SO₄)₃

 

D)  X(SO₄)₂

 

 

Answer: A

 

Approach: calculate the no. of moles of X hydroxide and sulfuric acid and then compare the mole ratio.

 

By comparing mole ratio: 2 mol of X hydroxide reacted with 1 mol of sulfuric acid → X hydroxide must be XOH (monobasic).

 

2XOH + H₂SO₄ → X₂SO₄ + 2H₂O

 

Essentially, neutralisation is represented by H⁺ + OH^– → H₂O
i.e. 1 mol H⁺ reacts with 1 mol OH^–, hence:

 

one mol of monobasic acid (one H⁺) will react with one mol monobasic base (one OH^–)
one mol of dibasic acid (two H⁺)will react with two mol monobasic base (two OH^–)
two mol of monobasic acid (two H⁺) will react with one mol dibasic base (two OH^–)
...

 

Four solutions were added to separate test-tubes of aqueous sodium carbonate and the observations were recorded.

 

Which of the following observations is incorrect?

 

A)  add: barium chloride | observation: white ppt formed

 

B)  add: copper(II) sulfate | observation: blue ppt formed

 

C)  add: dilute hydrochloric acid | observation: effervescence produced

 

D)  add: sodium hydroxide | observation: white ppt formed

 

 

Answer: D

 

- D is incorrect as no ppt can be formed. Recall: all sodium salts are soluble.
- A is correct → white ppt is barium carbonate.
- C is correct → carbon dioxide is formed.

 

Common question: why is B correct? Shouldn't copper(II) carbonate be green?

 

Explanation: In aqueous solutions, carbonate ions can react with water to produce hydroxide ions; they are weak alkalis.
CO₃^2– + H₂O → HCO₃^– + OH^–
So essentially, copper(II) ions will react with the hydroxide ions to form copper(II) hydroxide which is a blue ppt.

 

Further reading: why is copper(II) hydroxide formed instead of copper(II) carbonate since both are insoluble?
Ans: copper(II) hydroxide (K_sp ~ 10^–20) is more insoluble than copper(II) carbonate (K_sp ~ 10^–10) so it is preferentially formed. The smaller the K_sp value, the more insoluble a salt is.

 

All ammonium salts on heating with sodium hydroxide produce ammonia gas.

 

For which ammonium salt can the greatest mass of ammonia be obtained?

 

A)  0.5 mol (NH₄)₃PO₄

 

B)  0.5 mol (NH₄)₂SO₄

 

C)  1.0 mol NH₄Cl

 

D)  1.0 mol NH₄NO₃

 

 

Answer: A

 

Ionic eqn: NH₄⁺ + OH⁻ → NH₃ + H₂O
→ 1 mol NH₄⁺ will produce 1 mol NH₃

 

- the larger the no. of moles of NH₄⁺ present, the larger the no. of moles of NH₃ produced, the greater the mass of ammonia obtained.

 

No. of moles of NH₄⁺ present in:
A = 0.5 x 3 = 1.5 mol
B = 0.5 x 2 = 1.0 mol
C = 1 x 1 = 1.0 mol
D = 1 x 1 = 1.0 mol

 

Zinc nitrate crystals decompose on heating according to the following equation:

 

 2Zn(NO₃)₂.6H₂O (s) → 2ZnO (s) + 4NO₂ (g) + O₂ (g) + 12H₂O (g)

 

If 14.85 g of zinc nitrate crystals are heated strongly and then allowed to cool, what is the total volume of gas obtained at room temperature and pressure?

 

A)  2.4 dm³

 

B)  3.0 dm³

 

C)  5.1 dm³

 

D)  10.2 dm³

 

 

Answer: B

 

- At room temp and pressure, water is no longer a gas
→ 2 moles of Zn(NO₃)₂.6H₂O will produce only 5 moles of gases.

 

M_r of Zn(NO₃)₂.6H₂O = 297
No. of moles of Zn(NO₃)₂.6H₂O = 14.85/297 = 0.05 mol
No. of moles of gases = 5/2 x 0.05 = 0.125 mol
Volume of gases = 0.125 x 24 = 3.0 dm³

 

Which of the following mixtures contain the least number of ions in solution?

 

A)  1 mol Ba(NO₃)₂ + 1 mol NaCl

 

B)  1 mol Ba(NO₃)₂ + 1 mol Na₂SO₄

 

C)  1 mol Ca(NO₃)₂ + 1 mol NaCl

 

D)  1 mol NH₄NO₃ + 1 mol Na₂SO₄

 

 

Answer: B

 

Concept: When a precipitate (insoluble salt) is formed, the ions are removed from solution.

 

To determine the no. of ions in solution, we need to first determine the no. of ions produced for 1 mole of each salt
e.g. Ba(NO₃)₂ → Ba³⁺2+ + 2NO₃^– (i.e 3 moles of ions)

 

	No. of ions
A	5
B	4 since BaSO4 is insoluble
C	5
D	5

 

 

Two acids X and Y of the same concentrations were reacted with 0.1 mole of aqueous sodium hydroxide.

 

Which one of the following statements explains why a larger volume of X would be required to achieve the end-point as compared to Y?

 

A)  X is a strong acid and Y is a weak acid.

 

B)  X is a weak acid and Y is a strong acid.

 

C)  X is a monobasic acid and Y is a dibasic acid.

 

D)  X is a dibasic acid and Y is a monobasic acid.

 

 

Answer: C

 

- Strong acids ionise completely in water. Weak acids ionise partially in water.
- However, when acids react with sodium hydroxide, all the hydrogen ions will be completely reacted (regardless of whether the acid is strong or weak).
→ A and B are not correct.

 

- Monobasic acid contains 1 mole of hydrogen ion per mole of acid; dibasic acid contains 2 moles of hydrogen ions per mole of acid.
- Since a dibasic acid contains more hydrogen ions, we will require less of it to react away 1 mole of sodium hydroxide.
→ Y is a dibasic acid.

 

What happens when molten silver chloride is electrolysed using inert electrodes?

 

A)  chloride ions gain electrons at the positive electrode

 

B)  chloride ions lose electrons at the negative electrode

 

C)  silver ions gain electrons at the negative electrode

 

D)  silver ions lose electrons at the positive electrode

 

 

Answer: C

 

For molten (liquid) silver chloride, there are only 2 ions present: silver (Ag⁺) and chloride (Cl^–) ions.

 

Silver ions (Ag⁺) are attracted to the negative electrode (cathode) and is reduced (gains electrons)

 

Chloride ions (Cl^–) are attracted to the positive electrode (anode) and is oxidised (loses electrons)

 

In the atmosphere, nitrogen can undergo a series of reactions to form nitric acid:

 

 N₂ + O₂ → 2NO
 2NO + O₂ → 2NO₂
 4NO₂ + 2H₂O → 4HNO₃

 

How many moles of nitric acid are formed from 0.25 moles of nitrogen?

 

A)  0.25

 

B)  0.50

 

C)  1.00

 

D)  4.00

 

 

Answer: B

 

Simplest approach:

 

Use reacting ratio for each equation one at a time i.e.

 

N₂:NO = 1:2
→ 0.25 mol N₂ gives 0.50 mol NO

 

NO:NO₂ = 2:2
→ 0.50 mol NO gives 0.50 mol NO₂

 

NO₂:HNO₃ = 4:4
→ 0.50 mol NO₂ gives 0.50 mol HNO₃

 

When the product of a reaction between two gases is added to water, a solution of pH > 7 is formed.

 

Which of the following could be the two gases?

 

A)  carbon monoxide and oxygen

 

B)  hydrogen and chlorine

 

C)  hydrogen and nitrogen

 

D)  hydrogen and oxygen

 

 

Answer: C

 

pH>7 suggests that an alkaline gas is formed

 

The range of temperatures where both compounds can conduct electricity would be where both compounds exist as liquids.

 

A and B give acidic gases (HCl and CO₂)

 

C gives an alkaline gas (NH₃)

 

D gives a neutral gas (H₂O)