Unit 8 Checkpoint Question 1 - NEW CED Compatible

Question

Vitamin C, Ascorbic acid H2C6H6O6, is a weak acid with a Ka of 7.9 x 10-5

A) Write an equation showing the dissociation of one H+ from the acid.

B) Identify the acid and conjugate base.

C) Build a RICE table and find the pH of a 0.100 M solution of Ascorbic acid. Be sure to show all

your work.

Answer 1

A) H2C6H6O6 ==> H⁺ + HC6H6O6^-

B) acid = H2C6H6O6; conjugate base = HC6H6O6^-

C) H2C6H6O6 ==> H⁺ + HC6H6O6^- Reaction

.......0.100 ................0............0.............Initial

.......-x.......................+x..........+x...........Change

......0.100-x................x.............x............Equilibrium

Ka = [H⁺][HC6H6O6^-] / [H2C6H6O6]

7.9x10^-5 = (x)(x) / 0.100 - x (assume x is small and ignore it in the denominator)

x² = 7.9x10^-6

x = 2.8x10^-3 M H⁺ (note: this is small, i.e. ~ 3% compared to 0.100 M so our assumption above was valid)

pH = -log [H⁺]

pH = -log 2.8x10^-3

pH = 2.55