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What is the pH of a 0.1 M solution of acetic acid?
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Acetic acid ionizes partially, so you need the or k value of acetic acid (in data book), this should be given in the question. SO YOU NEED TO CALCULATE THE CONCENTRATION OF H+ (OR H3O+)
CH3COOH(aq) + H2O(l) -----> CH3COO-(aq) + H3O+(aq)
K IS DISSOCIATION CONSTANT WHICH IS THE PRODUCT OF CONCENTRATION OF PRODUCTS/PRODUCT OF CONCENTRATION OF REACTANTS
K= [CH3COO-] * [H3O+] , CH3COOH AND H20 ARE not included because its change in concentration is negligible. After partial dissociation, [CH3COOH] IS STILL APPROXIMATELY 0.1M.
[CH3COO-] = [H3O+]
K of CH3COOH=1.74 * 10^-5 (from data book)
=>1.74 * 10^-5 = [H30+]^2 / 0.1 [H30]^2 = 1.74*10^-5 * 0.1
= 1.74*10^-6 [H30+]
=>pH of 0.1M acetic acid is 2.88
hOPE IT HELPS..!
Posted by utkr940 on August 30, 2011 at 1:44 PM (Answer #1)
We can calculate the concentration of H+ ions in a 0.1 M solution of acetic acid by using the pKa or ionization constant which for any compound is the negative logarithm of the equilibrium coefficient of the neutral and charged forms.
pKa = -log(10)[Ka]
Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where  indicates the concentration.
pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]
as [H+] = [CH3COO-]
pKa = log(10)[CH3COOH] - 2*log(10)[H+]
For acetic acid pKa = 4.76 and -log(10)[H+] = pH
4.76 = log(10)(0.1) + 2*pH
=> 2*pH = 4.76 + 1
=> pH = 5.76/2
=> pH = 2.88
The required pH of 0.1 M solution of acetic acid is 2.88
Posted by justaguide on August 31, 2011 at 1:26 AM (Answer #2)
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