Answer:
the answer is 3 which corresponds to B.3
Step-by-step 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