Calculate the pH of a 0.09 M solution of ascorbic acid (K_a1= 7.9 × 10^-5; K_a2 is 1.6 × 10^-12). A) 11.4 B) 2.6 C) 5.1 D) 8.9 E) 11.8

Question

Quizplus · Accepted Answer

Ascorbic acid (vitamin C) is a weak acid, so we need to use the equilibrium expression to calculate the pH:
H3C6H5O7 ⇌ H+ + H2C6H6O6
Ka = [H+][H2C6H6O6]/[H3C6H5O7]
Ka = (1.6 × 10^-12)
[H3C6H5O7] = 0.09 M
Let x be the concentration of H+.
Ka = x^2 / (0.09 - x) 
We can assume that x is very small compared to 0.09 M, so we can simplify the expression:
Ka = x^2 / 0.09
x^2 = Ka * 0.09
x = sqrt(Ka * 0.09) = 1.3 × 10^-7 M
pH = -log[H+] = -log(1.3 × 10^-7) = 6.1
Therefore, the answer is B) 2.6.

Calculate the PH of a 0