# What is the pH of a 0.1 M solution of acetic acid?

Tushar Chandra | Certified Educator

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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

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## Related Questions

utkr940 | Student

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+)

Calculation

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+]

= 1.32*10^-3

pH=-log[H30+]

pH= -log(1.32*10^-3)

pH= 2.88

=>pH of 0.1M acetic acid is 2.88

hOPE IT HELPS..!