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In a quantitative analysis, a methanol (CH3OH) contaminated water sample was titrated with 0.0021 mol L- potassium permanganate (KMnO4). 50.00 mL samples of the water to be tested were acidified by sulfuric acid, then titrated with the permanganate solution. The results are shown below. Burette reading, ml 1st titration 2nd titration 3rd titration 4th titration Final volume 12.40 19.60 26.60 17.25 Initial volume 4.45 12.50 19.60 10.15 Titre 7.95 7.10 7.00 7.10 The complete equation for the redox titration reaction is: 4MnO4- + 12H+ + 5CH3OH → 4Mn2+ + 11H2O + 5HCOOH a. [5] Calculate the concentration of the methanol in mol L-1.​

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In a REDOX titration, one specie is oxidized while the other is reduced. The concentration of methanol is 0.012  mol L-1. Methanol is the oxidizing agent while permanganate is the reducing agent.

The average titre value is; [tex]\frac{7.95 + 7.10 + 7.00 + 7.10}{4}[/tex] = 7.29 mL

Equation of the reaction is:

[tex]4MnO4- + 12H+ + 5CH3OH ----> 4Mn2+ + 11H2O + 5HCOOH[/tex]

Concentration of oxidizing agent = CA = ?

Concentration of reducing agent = CB = 0.0021 mol L-1

Volume of oxidizing agent =  VA= 7.29 mL

Volume of reducing agent = VB = 50.00 mL

Number of moles of oxidizing agent NA = 4

Number of moles of reducing agent NB = 5

Note that NA and NB are obtained from the balanced reaction equation

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA =  CBVBNA/VANB

CA = 0.0021 mol L-1 * 50.00 mL * 4/7.29 mL * 5

CA= 0.012  mol L-1

For a comprehensive definition of redox titration see

https://brainly.com/question/24018439

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