AIM – (a) To prepare 250ml of M/20 solution of Mohr’s salt. (b) Using this calculate the molarity and strength of the given KMnO_{4} solution.
APPARATUS AND CHEMICALS REQUIRED Mohr’s salt, weighing bottle, weight box, volumetric flask, funnel, distilled water, chemical balance, dilute H_{2}SO_{4}, beakers, conical flask, funnel, burette, pipette, clamp stand, tile, KMnO_{4} solution.
THEORY (a) Mohr’s salt having the formula FeSO_{4}.(NH_{4})_{2}SO_{4}.6H_{2}O has molar mass 392gmol^{1}. It is a primary standard. Its equivalent mass is 392/1 = 392 as its n factor is 1 as per the following reaction: Fe^{2+}_{ → }Fe^{3+} + e^{}
PROCEDURE:
(b) THEORY
IONIC EQUATIONS INVOLVED: Reduction Half: MnO_{4}^{} + 8H^{+} + 5e^{} → Mn^{2+} + 4H_{2}O Oxidation Half: 5Fe^{2+ } → 5Fe^{3+ }+ 5e^{} Overall Equation: MnO_{4}^{} + 8H^{+} + 5Fe^{2+ } → Mn^{2+} + 5Fe^{3+} + 4H_{2}O
INDICATOR KMnO_{4} acts as a self indicator.
END POINT Colourless to light pink (KMnO_{4} in the burette)
PROCEDURE 1. Fill the burette with KMnO_{4} solution. 2. Pipette out 10ml. of Mohr’s salt solution into the conical flask. 3. Add half a test tube of dil. H_{2}SO_{4}. 4. Keep a glazed tile under the burette and place the conical flask on it. 5. Note down the initial reading of the burette. 6. Run down the KMnO_{4} solution into the conical flask drop wise with shaking. 7. Stop the titration when a permanent pink colour is obtained in the solution. 8. This is the end point. Note down the final burette reading. 9. Repeat the experiment until three concordant values are obtained. 10. OBSERVATION TABLE: (TO BE PUT UP ON THE BLANK SIDE USING A PENCIL) Volume of Mohr’s salt solution taken =
Concordant Value = 8.8mL
CALCULATIONS: (TO BE PUT UP ON THE BLANK SIDE USING A PENCIL) Calculation of amount of Mohr’s Salt to be weighed to prepare 100ml M/20 solution: Molecular Mass of Mohr’s Salt = 392g/mole 1000 cm^{3} of 1M KMnO_{4 }require 392g Mohr’s Salt. 250 cm^{3 }of M/40 KMnO_{4} require =392/40g = 4.9g
Using formula: N_{1}M_{1}V_{1} = N_{2}M_{2}V_{2} Where N_{1}=5 (for KMnO_{4}), V_{1}=8.8mL_{ } , M_{1} =? N_{2 }=1 (for Mohr’s salt), V_{2 }=_{ }10ml, M_{2} = 1/20M M_{1} = [1*(1/20)*10]/[5*8.8] = 1/88M = 0.01M
Strength = M X Molar Mass = 158 *( 1/88) = 1.79g/L
RESULT (ON RULED SIDE) The Molarity of KMnO_{4} = 0.01M And the strength of KMnO_{4} = 1.79g/L
