Please answer this problem about calculating the pH?

Question

A 0.0560 g quantity of acetic acid dissolved in enough H2O to make 50.0 ml of solution. Calculate the pH of the acetic acid solution (Ka or the acid dissociation constant=1.8 x 10^-5)

CH3COOH--->CH3COO^- H^

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Thanks for the help!

Answer 1

First calculate the Molarity of CH3COOH:

Molar Mass of CH3COOH = 60g/mol

#mols = 0.056/60 = 0.0009mols

Volume = 0.05L

Molarity = mols / Volume = 0.0009/0.05 = 0.019M

Then figure out the Molarity of H+:

CH3COOH ----&gt; CH3COO- + H+

0.019M                0               0

-x                       +x            +x

0.019 -x               x              x  

ka = products/ reactants

ka = [CH3COO-][H+]/[CH3COOH]

1.8 X 10^-5 = x^2/(0.019 -x)

we can ignore -x since reactant is favored.

so: 1.8 X 10^-5 = x^2/(0.019)

isolate and solve for x and you get:

[H+] = x= 5.8X10^-4

pH= -log[H+]

pH = 3.23