What is the approximate pH of a 0.06 M solution of CH3COOH (given that Ka = 1.78 x 10-5) ? A. 1 B. 3 C. 6 D. 7 E. 9

the answer is 3 which corresponds to B. 3

Explanation:

CH3COOH + H2O ⇄ CH3COO⁻ + H3O⁺

Since CH3COOH is a weak acid, it does not dissociate completely. Therefore an "x" value is being ionized to make the product.

At equilibrium:

[CH3COOH]=0.06 - x (because the "x" value have been ionized)

[CH3COO-] = x

[H3O+] = x

We know that Ka = [CH3COO-][H3O+]/[CH3COOH] from the formula

at equilibrium, Ka = (x*x) / (0.06-x)

because of the small Ka value, we can make a mathematic assumption 0.06 is really higher than "x"

Ka = (x^2) / 0.06

x^2 = (1.78x10^5) (0.06)

x = 0.0010334 = [H3O+]

pH = - log of [H3O+]

pH = - log (0.0010334) = 3.0