Respuesta :
The final pH of the solution is 2. 72
How to determine the final pH
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:
https://brainly.com/question/13557815
#SPJ1
