Thursday, 19 February 2015

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.
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.

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.

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 :

1. the mass of copper oxide,
2. 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$\text{=}\frac{12.4}{\left(64+12+48\right)}$ $\text{=}\frac{12.4}{124}$ = 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 $\text{=}\frac{MV}{1000}$ 0.04 $\text{=}\frac{0.5V}{1000}$ V = 0.04$\mathrm{×}\frac{1000}{0.5}$ 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 :

1. What volume of sulphuric acid, 1 M is required to react with 100cm3 Ba(NO3)2 , 0.5 M?
2. 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: $\frac{{\text{M}}_{1}{\text{V}}_{1}}{{\text{M}}_{2}{\text{V}}_{2}}\mathrm{=}\frac{{\text{n}}_{1}}{{\text{n}}_{2}}$ $\frac{1\mathrm{×}{\text{V}}_{2}}{0.5\mathrm{×}100}\mathrm{=}\frac{1}{1}$ V2 = 0.5 X 100 V2 = 50cm3 b. Number of moles of Ba(NO3)2$\frac{0.5\mathrm{×}100}{1000}$= 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