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In today’s class, we will be talking about the hydrolysis of salt. Enjoy the class!
HYDROLYSIS OF SALT
Some salts undergo hydrolysis in water to give an acidic or alkaline medium (solution) e.g. Na2CO3, NaHCO3, AlCl3, Na2S, NH4Cl, CH3COONa etc.
Na2CO3 + H2O → NaOH + H2CO3.
AlCl3 + H2O → Al (OH)3 + HCl.
Na2S + H2O → NaOH + H2S
Hydrolysis of salt occurs when a salt reacts with water e.g salt of strong acid and weak base gives an acidic solution. The change in the PH of the solution is due to hydrolysis.
USES OF SALTS
- NH4Cl is used as an electrolyte in a dry cell (Leclanche cell)
- CaCO3 is used as medicine to neutralise acidity in the stomach
- CaCl2 is used as antifreeze while fused CaCl2 is used as a drying agent and also in the desiccator.
- CaSO4 is used for making plaster of Paris.
- CuSO4 is used in dyeing and calico printing.
- MgSO4 is used as a laxative.
- KNO3 is used for making gunpowder, matches and soil fertilizer.
- NaCl is used for preserving food and in glazing pottery.
- ZnCl2 is used in petroleum refining
- Define salt?
- List the five main types of salts giving two examples each
- Name four salts and state the use of each of them
|S/NO||SOLUBLE SALTS||INSOLUBLE SALTS|
|1.||All Na+, K+ and NH4+ salt|
|2.||All trioxonitrate (v)|
|3.||All chloride except||PbCl2, HgCl2 and AgCl are soluble in hot water.|
|4.||Trioxocarbonate (iv) of Na+, K+ and NH4+||All other trioxocarbonate (iv).|
|5.||Trioxosulphate (vi) of Na+, K+, NH4+ & Cu2+||All other trioxosulphate (IV)|
|6.||Sulphide of Na+, K+ and NH4+||All other sulphides.|
|7.||All tetraoxosulphate (vi) except||PbSO4, BaSO4 and CaSO4 are slightly soluble in H2O|
|8.||All hydrogen trioxocarbonate (iv)|
|S/NO||SOLUBLE BASE/ALKALIS||INSOLUBLE BASE/ALKALIS|
|1.||K2O, Na2O are very soluble
MgO, CaO are slightly soluble
|Other oxides are insoluble|
|2.||NaOH, KOH, Ca(OH)2 are very soluble
Mg (OH)2 is slightly soluble.
|Other hydroxides are insoluble.|
METHOD OF PREPARATION OF SALTS
The method of preparing a particular salt depends on its:
- Solubility in water
- Stability to heat.
It is necessary, therefore, for us to become familiar with the simple rules of solubility indicated above. Knowing the solubility of the salt enables us to determine which method to be used.
Soluble salts can be prepared by the following method:
- The action of dilute acid on a metal.
- Neutralization of an acid by an alkali
- The action of dilute acid on an insoluble base.
- The action of dilute acid on trioxocarbonate (IV).
RECOVERING SOLUBLE SALTS FROM SOLUTION
This can be done by:
- Heating to dryness: This is used to prepare soluble salts which are not destroyed or decomposed by heat e.g. most chlorides such as NaCl, ZnCl2, FeCl2 and FeCl3 are recovered by heating.
- Crystallization: This is used to prepare salt which is easily decomposed or destroyed by dry heating. All trioxonitrate (V) and tetraoxosulphate (VI) are recovered by crystallization.
Insoluble salts can be prepared by the following method:
- Double decomposition or precipitation.
Pb (NO3)2(aq) + 2NaCl (aq) → 2NaNO3 (aq) + PbCl2(s)
AgNO3 (aq) + NH4Cl (aq) → NH4NO3 (aq) + AgCl(s)
- Direct combination of 2 elements.
Fe(s) + S(s) → FeS(s)
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
ANHYDROUS AND HYDRATED SALT
- Anhydrous salts: are salts which do not contain water and cannot be crystallized out from the aqueous solution.
- Hydrated salts/salts with water of crystallization: are salts which combine chemically with water. The water molecule is loosely held to the salt molecule and when heated, such salt loses their water of crystallization. The water attached is known as the water of crystallization e.g
Cu(NO3)2.3H2O: Copper (ii) trioxonitrate (v) trihydrate.
MgSO4.7H2O: Magnesium tetraoxosulphate (vi) heptahydrate.
FeSO4.7H2O: Iron (ii) tetraoxosulphate (vi) heptahydrate.
|Salts without water of crystallization||Salts with water of crystallization|
Calculation of water of crystallization
14g of hydrated H2C2O4.xH2O was heated to give an anhydrous salt weighing 9.99g.
(a). Calculate the value of x.
(b). Give the formula of the hydrated salt.
(c). Calculate the % of the water of crystallization.
(a). Mass of hydrated salt = Molar mass of hydrated salt
. Mass of water molecule Molar mass of water molecule
14 = (90+18x)
14 = (90 +18x)
14(18x) = 4.01 (90 + 18x)
252x = 360.9 + 72.18x
252x – 72.18x = 360.9
179.82x = 360.9
x = 360.9/179.82
x = 2.007
x = 2 to the nearest whole number.
(b) Formula of hydrated salt = H2C204.2H20.
(c) To calculate the % of water of crystallization:
% of water of crystallization = Mass of water x 100%
. Total mass
= 36 x 100
. (90 + 36)
= 36 x 100
EFFLORESCENCE, DELIQUESCENCE AND HYGROSCOPIC
When a certain compound is exposed to the air, they either lose their water of crystallization or they absorb moisture from their surroundings. The term efflorescent, deliquescent and hygroscopic are used to describe such compound.
EFFLORESCENCE: are substances which on exposure to air, lose some or all of their water of crystallization. The phenomenon or process is efflorescence. There is a loss of weight or mass of the substances.
e.g. Na2CO3.10H2O → Na2CO3.H2O + 9H2O
Other examples are Na2SO4.10H2O, MgSO4.7H2O and CuSO4.5H2O etc.
DELIQUESCENTS: are substances that absorb so much water from the air and form a solution e.g. NaOH, CaCl2, FeCl3, MgCl2, KOH and P4O10. There is a gain in weight.
HYGROSCOPIC: are substances which absorb moisture on exposure to the atmosphere without forming a solution. If they are solids, no solution will be formed but if a liquid absorbs water, it gets diluted. There is little or no difference in mass e.g. Conc. H2SO4, NaNO3, CuO, CaO and anhydrous Na2CO3.
These are substances which have a high affinity for water or moisture. They are either deliquescent or hygroscopic. They remove water molecules to affect physical change. Drying agents are different from dehydrating agents which removes elements of water i.e hydrogen and oxygen atoms or intra-molecular water.
Drying agents which react with gases are not used to dry the gas e.g conc. H2SO4 is not used to dry NH3 and H2S gas.
NH3(g) + H2SO4(aq) → (NH4)2SO4(aq)
H2S(g) + H2SO4(aq) → 2H2O(l) + SO2(g) + S(s)
|Conc. H2SO4||All gases except NH3 & H2S|
|Fused CaCl2||All gases except NH3|
|CaO or quicklime||For ammonia|
|P2O5||All gases except ammonia|
|Silica gel||All gases|
Salts are usually dried in a desiccator.
- Using balanced equations, state two methods of preparing: (a) Soluble salt (b) insoluble salt
- How can soluble salts be recovered from their solution?
- Calculate the percentage of water in sodium trioxocarbonate (iv) heptahydrate
- What is the number of molecules in 6.4g of sulphur (iv) oxide (NA=6.0X1023/mol)
- Write an equation to show the acid formed when phosphorus (v) oxide is dissolved in cold water and name the acid formed
- Differentiate between a base and an alkali
- Define: Efflorescence, Deliquescence and Hygroscopy
New School Chemistry for Senior Secondary Schools by O.Y Ababio pages 100-101 and 108-115.
- The two types of bonds that exist in H3O+ are a. covalent and ionic b. co-ordinate covalent and covalent c. metallic and ionic d. polar covalent and metallic
- How many moles of hydrogen ions are there in 50cm3 of 0.20moldm-3 H2SO4? A. 0.01 B. 0.02 C. 0.10 D. 0.20
- Which of these is not recovered through dry heating (evaporation)? A. NaClO3 B. NH4NO3 C. CuHSO4 D. NaHCO3.
- Which pH value indicates a basic solution? A. -1 B. 3 C. 9 D. 7
- All common gases are dried using P2O5 except A. NO2 B. NH3 C. SO2 D. H2S
- Give the reason for each of the following:
- Sodium salts cannot be prepared by double decomposition
- Na2CO3(aq) which is a salt solution, turns red litmus blue.
- 1.34g of hydrated Na2SO4 was heated to give an anhydrous salt weighing 0.71g.
- Calculate the number of molecules of water of crystallization
- Give the formula of the hydrated salt [Na=23, S=32, O=16, H=1].
In our next class, we will be talking about Carbon and its Properties. We hope you enjoyed the class.
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