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ChemistryNumerical+3Question 27 of 54

JEE Advanced 2018 · Paper 2, Chemistry question 9

To measure the quantity of MnCl2\mathrm{MnCl_2} dissolved in an aqueous solution, it was completely converted to KMnO4\mathrm{KMnO_4} using the reaction,

MnCl2+K2S2O8+H2O→KMnO4+H2SO4+HCl(equation not balanced).\mathrm{MnCl_2 + K_2S_2O_8 + H_2O \rightarrow KMnO_4 + H_2SO_4 + HCl} \quad \text{(equation not balanced)}.

Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2\mathrm{MnCl_2} (in mg) present in the initial solution is ____.

(Atomic weights in g mol−1^{-1}: Mn = 55, Cl = 35.5)

Answer and worked solution

Official answer: 120

Oxalic acid. 225 mg is 22590=2.5\dfrac{225}{90}=2.5 mmol.

Titration.
2 MnOX4X−+5 CX2OX4X2−+16 HX+→2 MnX2++10 COX2+8 HX2O\ce{2MnO4- + 5C2O4^{2-} + 16H+ -> 2Mn^{2+} + 10CO2 + 8H2O}
So MnOX4X−=25×2.5=1.0\ce{MnO4-}=\dfrac25\times2.5=1.0 mmol.

Mn balance. Each MnClX2\ce{MnCl2} gave one MnOX4X−\ce{MnO4-}, so there was 1.0 mmol MnClX2\ce{MnCl2} (M=55+71=126M=55+71=126):
1.0×126=126 mg.1.0\times126=126\text{ mg}.

Answer: 126 (key: 120 to 130)

Topics:Redox titrationsImportant compounds (KMnO4, K2Cr2O7)Mole concept
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