The **final pH **of the **solution **is **2. 72**

<h3>How to determine the final pH</h3>

CH3CH2OH + KMnO4 ---> CH3COOH + MnO2

Mole of ethanol= Concentration * Volume = 0.20 * 30/100 = 6.0 *10-3mole

Molar mass of ethanol= 46g/mol

Mass of ethanol= mole * molar mass = 6 * 10-3 * 46 = 0.276g

Mass of KMnO4= 1.08g

Molar mass of KMnO4= 158g/mol

Molar mass of ethanoic acid formed = 60g/mol

Let's determine the limiting reagent

158g of KMnO4 reacts with 46g of ethanol

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

With 0.276g of Ethanol is present, it is said to be the limiting reagent

46g of ethanol yields 60g/mol of ethanoic acid

Then, 0.276g of ethanol will yield = 60 * 0.276 /46 = 0.36g

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

To find the concentration of ethanoic acid= mole/volume

= 6.0 * 10-3/0.03 = 0.20M

Ionization of ethanoic acid will give;

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

Ka= 1.8 * 10-5

Then, equilibrium concentrations will be

[CH3COOH] = 0.20-x

[CH3COO-] = x

[H+] = x

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

Substitute values into the equilibrium equation

1.8 * 10-5 = x²/0.20-x

Ethanoic acid is a weak acid, so 0.20-x is approximately 0.20

x²= 0.20 * 1.8 * 10-5

x= 1.9 * 10-3

[H+] = 1.9 * 10-3M

To find the **pH**, use the formula;

** pH= -log[H+]**

Substitute the value of H+

pH= -log[1.9 x 10-3]

**pH= 2.72**

Therefore, the **final pH **of the **solution **is **2. 72**

**Learn **more about **pH **here:

brainly.com/question/13557815

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