What is the pH of a solution containing 0.30 M lactic acid and 0.50 M of lactate ion (K_a for lactic acid = 1.38 $\times$ 10^-4)?

Question

Quizplus · Accepted Answer

First, we need to set up the equilibrium equation for lactic acid and its conjugate base, lactate ion:
Lactic acid ⇌ lactate ion + H+
Ka = [lactate ion][H+]/[lactic acid]

We are given the concentrations of lactic acid and lactate ion. Let x be the concentration of H+ ions produced. Then, at equilibrium, we have:
[lactic acid] = 0.30 M - x
[lactate ion] = 0.50 M + x
[H+] = x

Substituting these concentrations into the equilibrium equation, we get:
Ka = (0.50 + x)(x)/(0.30 - x)

Since Ka = 1.38 x 10^-4, we can solve for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac))/2a
where a = 1, b = 0.88, c = -2.28 x 10^-5
x = 0.000408 M

Therefore, the pH of the solution is:
pH = -log[H+] = -log(0.000408) = 3.39

The closest answer choice to this value is D) 4.08. However, since we are given answer choices in whole numbers, we need to round our answer to the nearest whole number, which is 4.

What Is the PH of a Solution Containing 0 ×\times× 10-4)? A)9

What Is the PH of a Solution Containing 0 $\times$ 10^-4)?
A)9