The Ka value for ethanoic acid is 1.8x10^-5, what would be the pH of a 2.0x10^-2 M solution of ethanoic acid?
See your other post below.
To calculate the pH of a solution of ethanoic acid, we need to use the Ka value and the concentration of the acid. The Ka value represents the acid dissociation constant, which measures the strength of the acid.
The Ka expression for the dissociation of ethanoic acid is:
Ka = [H3O+][CH3COO-] / [CH3COOH]
Since ethanoic acid is a weak acid, it does not fully dissociate in water. We can assume that the concentration of the hydronium ion, H3O+, formed by the dissociation of ethanoic acid is approximately equal to the concentration of the dissociated acid, [CH3COO-].
Therefore, we can simplify the expression to:
Ka = [H3O+]^2 / [CH3COOH]
Rearranging the equation, we have:
[H3O+]^2 = Ka * [CH3COOH]
Taking the square root of both sides:
[H3O+] = sqrt(Ka * [CH3COOH])
Now we can substitute the given values into the equation. The concentration of ethanoic acid, [CH3COOH], is 2.0x10^-2 M, and the Ka value is 1.8x10^-5.
[H3O+] = sqrt((1.8x10^-5) * (2.0x10^-2))
[H3O+] = sqrt(3.6x10^-7)
[H3O+] ≈ 6.0x10^-4 M
The pH of a solution is defined as the negative logarithm (base 10) of the hydronium ion concentration:
pH = -log[H3O+]
Substituting the value of [H3O+] into the equation:
pH = -log(6.0x10^-4)
Using a calculator, we find:
pH ≈ 3.22
Therefore, the pH of a 2.0x10^-2 M solution of ethanoic acid is approximately 3.22.