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Concentration (in g dm–3)Concentration (in mol dm–3) = ——————————————— … (1.1) Relative molecular mass
Concentration (in dm–3) = Concentration (in mol dm–3) � Relative molecular mass … (1.2)If the reaction between X and Y to form P and Q is represented by the chemical equation:
aX + bY → cP + dQThen,
(M1V
1)
X a
————– = — … (1.3) (M
2V
2)
Y b
Example
Calculate the concentration in mol dm–3 of the following solutions:(a) F1 is hydrochloric acid of concentration 0.913 g dm–3.(b) F2 is a solution containing 3.4 g of OH– per dm3.(c) F3 is a solution containing 2.38 g MnO
4– per dm3.
(d) F4 is a solution containing 0.775 g KMnO4 per 250 cm3.
F3 is 0.10 mol dm–3 sodium hydroxide solution.F4 is an acid with molecular formula, HXO
4.
25.0 cm3 of F4 required 21.00 cm3 of F3 for complete neutralisation. (a) Write the equation for the reaction between HXO
4 and NaOH.
(b) Calculate the concentration of HXO4 in solution F4.
(c) Hence calculate (i) the relative molecular mass of X(OH)2 and (ii) the relative atomic mass of X.
(The concentration of HXO4 is 8.44 g dm–3)
(d) Suggest an identity for X.
Solution
(a) HXO4 + NaOH → NaXO
4 + H
2O
(M1V
1)
HXO4 1(b) ——————————— = —
(M2V
2)
NaOH 1
0.1 � 21.00 Concentration of HXO
4 = ——————————— = 0.084 mol dm–3
25
(c) (i) Concentration (in g dm–3) = Concentration (in mol dm–3) � Relative molecular mass 8.44
Relative molecular mass = —————— = 100.5 0.084
(ii) 100.5 = 1 + X + (4 � 16)X = 35.5
(d) X is chlorine.
Example 3 You are asked to determine the accurate concentration of a monoprotic acid, HX in F5 solution from the following experiment. The rough concentration of HX is about 1.0 mol dm–3.By means of a pipette, 50.0 cm3 of F5 is transferred into a 250 cm3 volumetric flask. Distilled water is then added and make up to the mark on the volumetric flask. The solution is labelled as F6 solution.F7 is prepared by dissolving 2.65 g of anhydrous sodium carbonate in250 cm3 solution. 25.0 cm3 of F7 required 20.95 cm3 of F6 for complete neutralisation using methyl orange as indicator.(a) Calculate the concentration of sodium carbonate in F7 solution.(b) Write the chemical equation for the reaction between sodium carbonate and HX.(c) Calculate the concentration of F6.(d) Hence, calculate the accurate concentration of HX in F5.
Solution
(a) Relative molecular mass of Na2CO
3 = 106
2.65 � 4Concentration of Na
2CO
3 = ———————— = 0.1 mol dm–3
106
(b) 2HX + Na2CO
3 → 2NaX + CO
2 + H
2O
(M1V
1)
HX 2
(c) ——————————— = — (M
2V
2)
Na2CO3 1
2 � 0.1 � 25.0Concentration of HX in F6 = ———————————— = 0.239 mol dm–3
20.95 250(d) Concentration of HX in F5 = 0.239 � ——— = 1.20 mol dm–3
50
NaOH solution is not used to standardise acid because it is a deliquescent solid. Thus, it is difficult to find the accurate mass of NaOH as it absorbs the moisture from the air (deliquescent) during weighing.
1. A half-equation is a chemical equation containing electrons.
Fe2+ → Fe3+ + e– … oxidation reactionCl
2 + 2e– → 2Cl– … reduction reaction
2. The following steps are used to balance half-equations for redox reactions.Step 1 : Determine the oxidation number of the atom that undergoes oxidation or reduction.Step 2 : Balance the half-equation in terms of the total charge and the number of atoms on
both sides of the equation.Step 3 : Calculate the number of oxygen atoms on both sides of the equation and add water (if
necessary) on the side of equation that has insufficient number of oxygen atoms.
Example 1Balance the equation : MnO
4– + H+ → Mn2+
Step 1 The oxidation number of Mn decreases from +7 to +2. Hence, 5e– must be added to the left-hand side of the equation.
MnO4– + H+ + 5e– → Mn2+ … not balanced
Step 2 Balance the total charge and number of atoms on both sides of the equation
MnO4– + 8H+ + 5e– → Mn2+ + 4H
2O … balanced
Total charge on LHS of equation = –1 + 8 � (+1) + 5 � (–1) = +2Total charge on RHS of equation = +2Total number of atoms on LHS of equation = 1Mn + 4O + 8HTotal number of atoms on RHS of equation = 1Mn + 4O + 8HHence, the half-equation is a balanced equation.
Example 2Balance the equation : XO
2– → XO
3–
Step 1 The oxidation number of X increases from +3 to +5. Hence, 2e– must be added to the right-hand side of the equation.
XO2– → XO
3– + 2e– … not balanced
Steps 2 and 3
XO2– + H
2O → XO
3– + 2e– + 2H+ … balanced
Total charge on LHS of equation = –1 Total charge on RHS of equation = (–1) + 2 � (–1) + 2 � (+1) = –1Total number of atoms on LHS of equation = 1X + 2H + 3OTotal number of atoms on RHS of equation = 1X + 3O + 2HHence, the half-equation is a balanced equation.
Balancing ionic equations for redox reactions
Step 1 : Write the half-equations for the oxidation and reduction reactions.Step 2 : Combine the half-equations to get an ionic equation that does not contain electrons.
Example 3Balance the equation for the reaction between HCOO– and MnO
4–.
Step 1 Write the half-equations for HCOO– and MnO4–
25.0 cm3 of F1 required 18.50 cm3 of F2 for complete neutralisation.25.0 cm3 of F1 required 33.60 cm3 of F3 for complete redox reaction.(a) Calculate the concentration (in g dm–3) of H
2C
2O
4 in F1 solution
(b) Calculate the concentration (in mol dm–3) of C2O
42– ions in F1 solution.
(c) Hence, calculate the concentration (in g dm–3) of Na2C
3OH+, reduces iron(III) ion, Fe3+, to iron(II) ion, Fe2+. In the
following experiment, you are asked to determine the chemical equation for the reaction between hydroxyammonium ion and iron(III) ion. F4 is a solution prepared by boiling 1.56 g of hydroxyammonium sulphate, NH
3OH+HSO
4– ,with
excess iron(III) ammonium sulphate and dilute sulphuric acid. The reaction mixture is then made up to 250 cm3 with distilled water. F5 is potassium manganate(VII) solution containing 1.58 g of KMnO
4 per 500 cm3.
In a titration experiment, 25.0 cm3 of F4 required 24.4 cm3 of F5 for complete reaction.(a) Calculate the concentration of (i) NH
3OH+HSO
4–, (ii) KMnO
4 in mol dm–3.
(b) Calculate the concentration (in mol dm-3) of Fe2+ ions produced in F4.(c) Hence, calculate the number of moles of Fe3+ that react with 1 mol of NH
3OH+.
(d) The half-equation for the oxidation of hydroxylamine to nitrogen is:
2NH2OH → N
2O + H
2O + 4H+ + 4e–
Hence, write a balanced redox equation between NH3OH+ ions and Fe3+ ions.
1. The half-equations for the redox reaction between iodine and thiosulphate are
2S2O
32– → S
4O
62– + 2e– … (1)
I2 + 2e– → 2I– … (2)
2. The ionic equation for the reaction between sodium thiosulphate and iodine is
2S2O
32– + I
2 → S
4O
62– + 2I– … (3)
This reaction is used to determine the concentration of iodine by titration against sodium thiosulphate. In some cases, the reaction is used to determine the concentration of an oxidising agent. For example, the oxidising agent is added to potassium iodide solution. The iodine liberated is titrated against sodium thiosulphate. From this, the concentration of the oxidising agent can be deduced indirectly.
3. Iodine has a low solubility in water but dissolves readily in potassium iodide solution.
I2(s) + KI(aq) I
3–(aq) + K+(aq)
In most direct titrations with iodine, a solution of iodine in potassium iodide is used. For simplicity, equation (3) is commonly used for calculation.
Important equations for iodine-sodium thiosulphate titration
Note: a = number of moles of the first reactant; b = number of moles for thiosulphate
Ionic equation aRatio of — b
2S2O
32– + I
2 → S
4O
62– + 2I–
In the following reactions, iodine produced by the reaction is titrated against sodium thiosulphate solution.
H2O
2 + 2H+ + 2I–
→ 2H
2O + I
2
(H2O
2 ≡ I
2 ≡ 2S
2O
32–)
IO3– + 6H+ + 5I– → 3H
2O + 3I
2
(IO3– ≡ 3I
2 ≡ 6S
2O
32–)
2Cu2+ + 4I– → Cu
2I
2 + I
2
(2Cu2+ ≡ I
2 ≡ 2S
2O
32–)
1—2
1—2
1—6
1—1
Precautions in iodine-sodium thiosulphate titrations
1. Iodine is a volatile substance. Hence the iodine solution prepared should be titrated immediately and quickly to avoid loss of iodine due to evaporation.
2. In I2 – S
2O
32– titration, starch is used as an indicator. The starch solution should not be added
at the beginning of the titration where there is a high concentration of iodine. This is because iodine is adsorbed onto the starch molecule and may remain adsorbed even at the end point.
3. The starch solution should be added towards the end of the titration when the reaction mixture turns pale yellow. The reaction mixture in the conical flask should be shaken vigorously but carefully as the titration proceeds.
4. The starch solution will produce a dark blue colour with iodine and when the end point is reached, the solution turns colourless abruptly.
5. When the solution in the conical flask is left aside after titration, the solution becomes blue again. Ignore this, because iodine is formed due to the atmospheric oxidation of excess potassium iodide in the reaction mixture.
6. Starch solution is unstable and should be prepared fresh for each titration.
Example 1To determine the concentration of iodine solutionF1 contains 20.0 g of Na
2S
2O
3.5H
2O per dm3.
F2 is an iodine solution.25.0 cm3 of F2 required 22.5 cm3 of F1 for complete reaction.Calculate (a) the concentration (in mol dm–3) of iodine in solution F2, (b) the mass of iodine required to prepare 200 cm3 of F2.
Solution
(a) Relative molecular mass of Na2S
2O
3.5H
2O = 248
20.0Concentration of Na
2S
2O
3.5H
2O = ———— = 0.0806 mol dm–3
248
2S2O
32– + I
2 → S
4O
62– + 2I–
(M1V
1)
I2 1
———————————— = — (M
2V
2)
S2O32– 2
1 0.0806 � 22.5Concentration of I
2 = — � ————————— = 0.0363 mol dm–3
2 25.0
(b) Relative molecular mass of I2 = 254
200Mass of iodine = (0.0363 � 254) � ———— = 1.84 g
1000
Example 2 To determine the percentage purity of hydrated sodium sulphite, Na
2SO
3.7H
2O
F3 is a solution containing 11.1 g of hydrated sodium sulphite, Na2SO
3.7H
2O per dm3.
F4 is 0.050 mol dm–3 iodine.F5 is 0.025 mol dm–3 sodium thiosulphate solution.25.0 cm3 of F4 is mixed with 25.0 cm3 of F3. The resulting solution required 18.90 cm3 of F5 for complete reaction. Sulphite ion reacts with iodine as represented by the equation
∆H = mc∆Twhere m = mass of solution, c = specific heat capacity (4.2 J g–1 K–1) ∆T = maximum rise or fall in temperature
Example 1 To determine the heat of reaction between a metal hydrogen carbonate and dilute hydrochloric acid.F1 is 1.0 mol dm–3 hydrochloric acid.F2 is a metal hydrogen carbonate (MHCO
3).
50.0 cm3 of F1 is placed in a polystyrene cup. 4.7 g of F2 is added to F1. The mixture is constantly stirred and the temperature recorded.
1. The density of solution is usually assumed to be 1.0 g cm–3.
Hence, the mass of solution = volume of solution used.
2. The mass of solution does not include the mass of solid added to water. For example, when 1.0 g of NaOH is added to 100 cm3 of water, the mass of solution is 100 g and not 101 g.
Take NoteTake Note
Temperature of F1 before mixing
Lowest temperature reached after mixing
Temperature change
30.0 °C
25.0 °C
5.0 °C
(a) Calculate the amount of heat absorbed in the experiment.(b) Calculate the number of moles of HCl used.(c) Calculate the number of moles of MHCO
3 used.
(d) Calculate the number of moles of MHCO3 reacted.
(The relative molecular mass of the metal hydrogen carbonate = 100)(e) Calculate the enthalpy change for the reaction.
1.0 � 50.0(b) Number of moles of HCl used = —————————— 1000
= 0.05
4.7(c) Number of moles of MHCO
3 used = ———
100
= 0.047
(d) MHCO3(aq) + HCl(aq) → MCl(aq) + H
2O(l) + CO
2(g); ∆H (+)ve
HCl is in excess. Number of moles of MHCO
3 reacted = 0.047
1.05 (e) ∆H = ————— = 22.3 kJ mol–1
0.047
Example 2To determine the partition of an organic acid, HOOC(CH
2)
nCOOH between water and ether.
F3 is a solution of an organic acid, HOOC(CH2)
nCOOH.
F4 is 0.1 mol dm–3 NaOH solution.F5 is 0.01 mol dm–3 NaOH solution.40.0 cm3 of F3 is added to a bottle, followed by 60.0 cm3 of ether. The bottle is tightly closed and shaken vigorously. After 30 minutes, 10.0 cm3 of the ether layer is pipetted out and titrated with F5. Then, 10.0 cm3 of aqueous layer is pipetted out and titrated with F4.
Table 3 Reactions with aqueous ammoniaNote: The asterisk (*) shows the cation that will form a precipitate with NH
3(aq) but the precipitate
will dissolve when NH4Cl is added.
1. Ammonia is a weak base. In aqueous solution, it undergoes partial dissociation to form NH4
+ and OH– ions.NH3(aq) + H2O(l) NH4
+(aq) + OH–(aq)2. Ca2+(aq) forms a
precipitate with NaOH(aq) but it does not form a precipitate with NH3(aq). This is because the concentration of OH– ions in aqueous ammonia is very low. The ionic product of Ca(OH)2 for ammonia is lower than the solubility product (Ksp) of Ca(OH)2. Hence, precipitation of Ca(OH)2 cannot occur.
3. Some metal hydroxides, such as Mg(OH)2, dissolve in NH4Cl, because of the common ion effect. In the presence of NH4Cl, the concentration of OH– is decreased due to the common ion, NH4
+. The metal hydroxide dissolves because the ionic product is less than the solubility product.
Take NoteTake NoteObservation
White precipitate, insoluble in excess NH
3(aq)
White precipitate, soluble in excess NH
3(aq)
White precipitate, turns rapidly to brown colour; insoluble in excess NH
3(aq)
Dirty green precipitate, insoluble in excess NH
3(aq)
Brown precipitate, insoluble in excess NH
3(aq)
Bluish-green precipitate, insoluble in excess NH
3(aq)
Green precipitate, soluble in excess NH
3(aq) to form
light blue solution
Blue precipitate, soluble in excess NH
3(aq) to form
dark blue solution
Inference
Pb2+, *Mg2+, Al3+ present
*Zn2+ present
*Mn2+ present
*Fe2+ present
Fe3+ present
Cr3+ present
*Ni2+ present
*Cu2+ present
Explanation
The metal hydroxides are formed
Zn(OH)2 is soluble in
NH3 due to the formation
of [Zn(NH3)
4]2+
Mn(OH)2 is oxidised
easily to Mn(OH)3
Fe(OH)2 formed and
oxidised by air to brown Fe(OH)
3
Fe(OH)3 formed
Cr(OH)3 formed
Ni(OH)2 formed which
dissolves in NH3(aq) to
form the complex ion [Ni(NH
3)
6]2+
Cu(OH)2 formed which
dissolves in NH3(aq) to
form the complex ion, [Cu(NH
3)
4]2+
Table 4 Reactions with aqueous iron(III) chloride, FeCl3(aq)
(I) Formation of precipitate
Observation
Buff-coloured (yellowish-brown) precipitate
Brown precipitate; CO2 liberated
Inference
C6H
5COO– (benzoate) present
CO32– present
Explanation
Formation of iron(III) benzoate, (C
6H
5COO)
3Fe
Formation of iron(III) carbonate; CO
2 liberated because FeCl
3(aq)
undergoes hydrolysis to produce H+ ions
(II) Colour change without precipitation
Observation
Solution turns red; Brown precipitate formed on heating