Answer to Question #162052 in General Chemistry for Gelo

Question #162052

If 30.0 ml of 0.20M Ethanol solution reacts with 1.80 g potassium permanganate, what will be the final pH of the solution? Assume a complete reaction and that only the products of the reaction contribute to the final pH.

Expert's answer

CH3CH2OH + KMnO4 ---> CH3COOH + MnO2

Mole of ethanol= C x V

= 0.20 x 30/100 = 6.0 x 10-3mole

Molar mass of ethanol= 46g/mol

Mass of ethanol= mole x molar mass

= 6x10-3 x 46

= 0.276g

Mass of KMnO4= 1.08g

Molar mass of KMnO4= 158g/mol

Molar mass of ethanoic acid= 60g/mol

Now let's find the limiting reagent

158g of KMnO4 reacts with 46g of ethanol

1.08g of KMnO4 should react with 46/158 x 1.08 = 0.314g of ethanol

Since only 0.276g of Ethanol is present, it is the limiting reagent

46g of ethanol yields 60g of ethaoic acid

0.276g of ethanol will yield

60x0.276/46 = 0.36g

Mole of ethanoic acid= 0.36/60= 6.0 x 10-3mole

Concentration of ethanoic acid= mole/volume = 6.0 x 10-3/0.03 = 0.20M

Ionization of ethanoic acid is given as

CH3COOH ------> CH3COO- + H+

Ka= 1.8 x 10-5

Equilibrium concentrations are

[CH3COOH] = 0.20-x

[CH3COO-] = x

[H+] = x

Ka= [CH3COO-][H+]/[CH3COOH]

1.8 x 10-5 = x²/0.20-x

Since ethanoic acid is a weak acid, 0.20-x is approximately 0.20

x²= 0.20 x 1.8 x 10-5

x= 1.9 x 10-3

[H+] = 1.9 x 10-3M

pH= -log[H+]

pH= -log[1.9 x 10-3]

pH= 2.72

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