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SS3: CHEMISTRY - 1ST TERM

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  1. Volumetric Analysis (Titration) | Week 1
    3 Topics
    |
    1 Quiz
  2. Heat of Neutralization & Redox Titrations | Week 2
    2 Topics
    |
    1 Quiz
  3. Test for Common Gases | Week 3
    1 Topic
  4. Qualitative Analysis I | Week 4
    2 Topics
  5. Qualitative Analysis II | Week 5
    1 Topic
  6. Qualitative Analysis III | Week 6
    2 Topics
  7. Qualitative Analysis IV - Anions | Week 7
    1 Topic
  8. Test for Fat and Oil; Protein; Starch | Week 8
    3 Topics
  9. Petroleum I | Week 9
    4 Topics
    |
    1 Quiz
  10. Petroleum II | Week 10
    4 Topics
    |
    1 Quiz



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Oxidation-Reduction Titration or Redox titration is a volumetric analysis involving an oxidizing agent and reducing agent. In this titration, an oxidizing agent (also known as the oxidant) is titrated against a reducing agent (otherwise known as the reductant).

The unknown concentration of the analyte is then determined using calculations. At the end of it, the oxidizing agent becomes reduced and the reducing agent becomes oxidized. Examples of redox titration are potassium tetraoxomanganate (VII) – iron(II) and  thiosulphate–iodine titrations.

Redox Titrations of Acidified KMnO4 Solution and a Fe2+ Salt

(1). When 0.011 moldm–3 potassium tetraoxomanganate (VII), KMnO4(aq) solution acidified with teraoxosulphate (VI) acid was titrated against a solution containing 14.0 g of an iron (II) salt in 500cm3 of solution, 21.30 cm3 of acidified KMnO4(aq) oxidized 25.00 cm3 of the iron (II) salt solution. The equation for the reaction is: 

MnO4(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe2+(aq) + 4H2O(l)

(a) From the information provided, calculate the:

(i) moles of MnO4used in the titration.

(ii) moles of Fe2+ titrated

(iii) concentration of Fe2+ in mol dm–3

(iv) concentration of Fe2+ in gdm–3

(v) percentage of iron in the salt

(b) Give reason(s) why

(i) it is necessary to acidify KMnO4(aq) in the titration

(ii) hydrochloric acid cannot be used to acidify KMnO4 solution in the titration

Results and Calculations

Molar concentration of oxidant, Cox = 0.0110 moldm-3

Volume of oxidant, Vox used = 21.30 cm3

(a)(i). Number of moles of the oxidant = molar concentration x volume in dm3

= \(\scriptsize 0.0110 moldm^{-3} \; \times \; \normalsize \frac{21.30cm^3}{1000cm^3} \\ \scriptsize = 0.00234 \; moles \)

(ii). From the balanced equation for the reaction:

1 mole of MnO4(aq) will oxidize 5 moles of Fe2+(aq)

Therefore, the number of moles of B titrated = 0.00234 х 5 = 0.0117 moles

(iii). The concentration of B in moldm-3;

25 cm3 of B contains 0.0117 moles

1000 cm3 will contain = \( \frac{1000}{25} \; \times \; \scriptsize = 0.0117 \scriptsize = 0.468 \; mol dm^{-3} \)

Alternatively:

\( \frac {Cox \; \times \; Vox}{Cred \; \times \; Vred} = \frac{nox}{nred} \\ \frac {0.110 \; \times \; 21.30}{Cred \; \times \; 25} = \frac{1}{5} = \)

Cred = \( \frac{0.110 \; \times \; 21.30 \; \times \; 5}{25 \; \times \; 1} \\ \scriptsize = 0.4686 \; mol dm^{-3}\)

(iv). concentration of the Fe2+ in gdm–3 = concentration of Fe2+ in moldm–3 х molar mass

0.468 moldm–3 х 56 gmol–1 = 26.21 gdm–3

(v). percentage of iron in the salt is: 

500 cm3 contains 14 g of the salt

∴ 1000 cm3 = \( \scriptsize 1000 \; \times \; \normalsize \frac {14}{500} \)

= 28 g of the salt

Therefore, percentage of iron in the salt

= \( \frac {26.21 \; \times \; 100}{28.0} \\ \scriptsize = 93.61\%\)

b(i). It is necessary to acidify the oxidant so as to prevent unwanted reactions which may introduce errors to the titration.

(ii). Dilute HCl(aq) cannot be used because in the presences of the strong oxidizing agent, tetraoxomanganate (VII) ion, the chloride ions in hydrochloric acid are oxidized to chlorine gas.

(2). A is a solution containing 16.20 g dm-3 of impure ethanedioic acid. 

B is 0.100 mol dm-3 sodium hydroxide solution.

On titrating A against B, using phenolphthalein as indicator, 31.50cm3 of A was found to react completely with 25.00 cm3 portions of B.

From information provided above, calculate the: 

i. concentration of A in mol dm-3;
ii. the percentage purity of the ethanedioic acid;

The equation for the reaction involved is:

H2C2O4(aq)+2NaOH(aq) → Na2C2O4(aq) + 2H2O(l)

[H = 1; C = 12; O = 16]

Solution

CA =?
VA = 31.50 cm3
nA = 2
CB = 0.100 moldm-3
VB = 25.00 cm3
nB = 1

(i) \( \frac {C_A V_A}{C_B V_B} = \frac{n_A}{n_B} \\ \scriptsize C_A = \normalsize \frac{C_B V_B n_A}{V_A n_B} \\ \scriptsize C_A = \normalsize \frac{0.1 \; \times \; 25 \; \times \; 2}{31.50 \; \times \; 1} \scriptsize = 0.159 moldm^{-3} \)

The concentration of H2C2O4(aqis 0.159 moldm-3

(ii). The percentage purity of the ethanedioic acid

 Concentration in gdm-3 = Concentration in moldm-3 x molar mass

= 0.159 moldm-3 x 90 gmol-1

= 14.31g

Percentage purity = \( \frac{purity}{impure} = \frac{100}{1} \\ = \frac{14.31}{16.20} = \frac{100}{1} \\ \scriptsize 88.3 \% \)

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