0.233 mol/0.1 L = 0.0233 M But I get 2.33 !! that will change the pH. Please give me my question back if the previous answer was wrong. Step 1 Given that, Percentage of acetic acid solution = 14% It means, 14 g of acetic acid dissolved in 100 g or 100 mL of water. Moles of acetic acid in 14 g = 14 g/60.052 g/mol = 0.233 mol Molarity of 0.233 mol of 100 mL acetic acid solution = 0.233 mol/0.1 L = 0.0233 M Ka value of acetic acid = 1.7×10-5 Step 2 Now, acetic acid dissociates in water as follows; CH3COOH <----> CH3COO- + H+ Ka = [CH3COO- ][H+]/[CH3COOH] Let, [CH3COO- ] = [H+] = x Then we have, 1.7×10-5 = x.x/0.0233M x = 0.629×10-3 M Now, pH of the acetic acid solution will be pH = -log[H+] = -log0.629×10-3 M = 2.8
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0.233 mol/0.1 L = 0.0233 M
But I get 2.33 !! that will change the pH.
Please give me my question back if the previous answer was wrong.
Given that,
Percentage of acetic acid solution = 14%
It means, 14 g of acetic acid dissolved in 100 g or 100 mL of water.
Moles of acetic acid in 14 g = 14 g/60.052 g/mol
= 0.233 mol
Molarity of 0.233 mol of 100 mL acetic acid solution = 0.233 mol/0.1 L = 0.0233 M
Ka value of acetic acid = 1.7×10-5
Now, acetic acid dissociates in water as follows;
CH3COOH <----> CH3COO- + H+
Ka = [CH3COO- ][H+]/[CH3COOH]
Let, [CH3COO- ] = [H+] = x
Then we have,
1.7×10-5 = x.x/0.0233M
x = 0.629×10-3 M
Now, pH of the acetic acid solution will be
pH = -log[H+]
= -log0.629×10-3 M
= 2.8
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