8.1 Salts

Salts
■ Salt

	►	An ionic compound consists of positive ions such as metal ions or ammonium ions derived from the base and negative ions derived from the acid.
	►	Positive ions and negative ions are held together by strong ionic bonds when in the solid state.
	►	Some examples of salt and its ion contents are shown in Table. Salts Formula Positive ion Negative ion Sodium chloride NaCl Na+ Cl- Potassium nitrate KNO3 K+ NO-3 Magnesium sulphate MgSO4 Mg2+ SO2-4 Calcium nitrate Ca(NO3)2 Ca2+ NO-3 Zinc chloride ZN(Cl)2 Zn2+ Cl- Copper (II) sulphate CuSO4 Cu2+ SO2-4 Aluminium nitrate Al(NO3)3 Al3+ NO-3 Ammonium sulphate (NH4)SO4 NH+4 SO2-4

■ Definition of salt

	►	An ionic compound resulting from the replacement of hydrogen atoms in an acid by a metal ion or ammonium ion, NH+4
	►	The animation below shows some examples of salts.

■ This video contains information on the definition of salts.

Soluble salt and insoluble salt
■ Salts can be divided into

	►	soluble salts
	►	insoluble salts

■ The solubility of a salt is important

	►	to separate a salt from a mixture of salt.
	►	to prepare a salt sample.
	►	to identify cations and anions through qualitative analysis.

■ General guidelines for solubility of salt:

	►	All alkali metal salts such as sodium and potassium are soluble in water.
	►	All nitrates are soluble in water.
	►	All chlorides are water soluble except silver chloride (AgCl) and lead chloride (PbCl2).
	►	All sulphates are soluble in water except calcium sulphate (CaSO4), barium sulphate (BaSO4) and lead (II) sulphate (PbSO4)
	►	All the carbonates are insoluble in water except sodium carbonate (Na2CO3), potassium carbonate (K2CO3) and ammonium carbonate ((NH4)2CO3).

■ The preparation of salt crystals in laboratory depends on

	►	solubility of salt.
	►	if soluble, whether it is a sodium, potassium or ammonium salt.

■ This video presents a brief summary of several solubility rules.

Preparation of soluble salt
■ Preparation of soluble salt

	►	Soluble salts from Group 1 metal (potassium and sodium salts)
	►	Soluble salts that are not from Group 1 metal (not potassium and sodium salts)

■ This video contains information on the preparation of soluble salt .

■ Preparation of soluble salt potassium and sodium salts

	►	Through neutralization process between acidic solution and alkaline solution.
	►	Titration method can be used to obtain accurate quantities of reactant needed.
	►	Example: Salt Chemical equation Potassium nitrate KOH(aq) + HNO3(aq) → KNO3(aq) + H2O(l) Sodium chloride NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Sodium sulphate NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)

■ Preparation of soluble salt (non-potassium and non-sodium salts)

►

Preparation method Chemical equation and example (a) reaction between acid and metal oxide Acid + Metal oxide → Salt + Water Example : MgO(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) (b) reaction between acid and metal hydroxide Acid + Metal hydroxide → Salt + Water Example : Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) (c) reaction between acid and metal Acid + Metal → Salt + Hydrogen Example : Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) (d) reaction between acid and metal carbonate Acid + Metal carbonate → Salt + Water + Carbon dioxide Example : MgCO3(s) + HCl(aq) → MgCl2(aq) + H2O(l) + CO2(g)

■ Recrystallization method

	►	Soluble salts which contain impurities can be purified by recrystallisation.
	►	Steps in recrystallization method. ○ Step 1: Dissolving the salt in some distilled water. ○ Step 2: Heating it to obtain a saturated solution. ○ Step 3: Hot saturated solution is cooled to get salt crystal.

■ The following flow chart shows the steps involved in the preparation of soluble salts.

Laboratory Activity 8.1.1 : Preparation of soluble salt potassium and sodium salts

Laboratory Activity 8.1.2 : Preparation of non-potassium and non-sodium soluble salts

Physical properties of crystals
■ Salts

	►	Ionic compounds composed of ions which are arranged in a close, orderly manner at fix positions.
	►	Each cell unit is arranged repeatedly many times until a geometric shape is formed called crystal.

■ Crystal

►

A homogeneous solid with fixed shapes.

■ Physical properties of crystal

	►	Similar geometric shapes, for example cuboid, tetragonal, monoclinic and hexagonal.
	►	Even surfaces, straight edges and sharp tips.
	►	Angles between corresponding surfaces are fixed and equal.
	►	Crystals are hard but brittle.
	►	Can cut into certain shapes as their particles are arranged in a close, orderly manner.

■ The size of crystal depends on the rate of crystallization.

	►	Lower rate will produce bigger crystal size.
	►	Even though the size of are different, the shape remains the same.

Preparation of insoluble salt
■ Insoluble salts are prepared through precipitation reaction (or double dissociation reaction).

	►	Involves the exchange of ions. ○ Salt solution (contains cation) + Salt solution (contain anion) → Insoluble salt
	►	Salt Chemical equation Lead chloride PbNO3(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) Ionic equation: Pb2+(aq) + 2Cl-(aq) → PbCl2(s) Barium sulphate Ba(NO3)2(aq) + H2SO4(aq) → BaSO4(s) + 2HNO3(aq) Ionic equation: Ba2+(aq) + SO2-4(aq) → BaSO2(s) Argentum chloride AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) Ionic equation: Ag+(aq) + Cl-(aq) → AgCl(s)

■ The following flow chart shows the steps involved in the preparation of insoluble salts.

■ This video contains information on the preparation of insoluble salt.

Laboratory Activity 8.1.3 : Preparation of insoluble salts

The summary of preparation of salts
■ The animation below summarizes the steps involved in the preparation of soluble salts and insoluble salts.

Construction an ionic equation through the continuous variation method
■ Ionic equation

	►	Shows the actual reaction between two ions.
	►	Shows the actual reaction between ions, atom or molecules.

■ Continuous variation method

	►	Used to construct the ionic equation for the formation of an insoluble salt.
	►	Determine the number of mole ions which are reacting in a chemical reaction.
	►	Involves the reaction between a fixed volume solution and another solution whose volume increased evenly.
	►	Example: a mol of Xb+ ions has combined with b mol of Ya- ions to form a compound with molecular formula XaYb ○ aXb+ + bYa- → XaYb ○ Thus, by finding the ration of a to b, the empirical formula of the salt can be written.

Laboratory Activity 8.1.4 : Constructing an ionic equation through the continuous variation method

Numerical problems involving stoichiometric reaction in the preparation of salts
■ Steps in determining the empirical formula of a salt or a chemical equation for salt formation.

	►	Step 1: Find the number of moles of cations and anions
	►	Step 2: Calculated the simple ratio of the number of moles of cation to anion
	►	Step 3: Write the empirical formula
	►	Step 4: Build the chemical equation for salt formation based on the simple ratio.

✍ Worked-example 8.1(a) Copper(II) carbonate decomposes when heated as follows: CuCO3(s) → CuO(s) + CO2(g). Calculate : the mass of copper oxide, the volume of CO2 formed in this reaction when 12.4 g of CuCO3 is used?, [Relative atomic mass: C, 12; O, 16; Cu, 64; 1 mol of gas occupies 22.4dm3 at s.t.p] Solution: a. Number of moles of CuCO3$\text{=}\frac{12.4}{(64+12+48)}$ $\text{=}\frac{12.4}{124}$ = 0.1mol From the equation, 1 mole of CuCO3 produces 1 mole of CuO. Thus, number of moles of CuO formed = 0.1 mol Mass of CuO = 0.1 X (64 + 16) = 0.1 X 80 = 8.0g b. From the equation, 1 mole of CuCO3 produces 1 mole of CO2. Thus, number of moles of CO2 formed = 0.1 mol Volume of CO2 formed = 0.1 X 22.4 = 2.24dm3

✍ Worked-example 8.1(b) What is the volume of 0.5 M hydrochloric acid required to produce 0.448dm3 hydrogen gas when reacts with excess zinc pallette. [Relative atomic mass: C, 12; O, 16; Cu, 64; 1 mol of gas occupies 22.4dm3 at s.t.p] Solution: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Number of moles of hydrogen gas $\text{=}\frac{0.448}{224}$ = 0.02mol From the equation, 1 mol of hydrogen gas produced by 2 mol of hydrochloric acid. Thus, 0.02 mol of hydrogen gas produced by 0.04 mol of hydrochloric acid. Number of moles $\text{=}\frac{\mathrm{MV}}{1000}$ 0.04 $\text{=}\frac{\mathrm{0.5V}}{1000}$ V = 0.04$\mathrm{\times }\frac{1000}{0.5}$ V = 80cm3 Volume of hydrochloric acid required = 80cm3

✍ Worked-example 8.1(c) Barium nitrate solution reacts with sulphuric acid as follows: Ba(NO3)2(aq) + H2SO4 (ak) → BaSO4(s) + 2HNO3 (aq) Calculate : What volume of sulphuric acid, 1 M is required to react with 100cm3 Ba(NO3)2 , 0.5 M? Calculate the mass of BaSO4 which is formed in this reaction. [Relative atomic mass: O, 16; S, 32; Ba, 137] Solution: a. From the equation, 1 mole of Ba(NO3)2 reacts with 1 mol of H2SO4, thus: $\frac{{\text{M}}_1{\text{V}}_1}{{\text{M}}_2{\text{V}}_2}\mathrm{=}\frac{{\text{n}}_1}{{\text{n}}_2}$ $\frac{1\mathrm{\times }{\text{V}}_2}{0.5\mathrm{\times }100}\mathrm{=}\frac{1}{1}$ V2 = 0.5 X 100 V2 = 50cm3 b. Number of moles of Ba(NO3)2$\frac{0.5\mathrm{\times }100}{1000}$= 0.05mol From the equation, 1 mol of Ba(NO3)2 produces 1 mol of BaSO4. Thus, 0.05 mole of Ba(NO3)2 produces 0.05 mole mol of BaSO4. Mass of BaSO4 = 0.05 x (137 + 64 + 4 x 16) = 11.65 g

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