Question

1. The pH of a .02 M solution of monoprotic acid is 6.20. Find the K_a of the acid

2. If .08 moles of a monoprotic acid are dissolved in 2 L of aqueous solution, it is then found that 14% of the acid has dissociated at equilibrium. Calculate the K_a for this monoprotic acid.

3. Consider the reaction: N₂O_{4 (g)}⇌ 2 NO_2(g) taking place in a sealed, fixed volume container. Suppose initially [N₂O₄ (g)] ₀ = 0 and [NO₂(g)] = .6 M. Further, suppose that at some temperature the reaction proceeds to equilibrium and [NO₂(g)]_e= .15 M, i.e. the equilibrium concentration of [NO₂(g)]_e= .15 M. Sketch a reasonable plot of molar concentrations of each substance as a function of time showing the progress of reaction to reach equilibrium. (Make sure the axes are labeled and show linear increments along each access)

Thanks.

Answer 1

1)

use:

pH = -log [H+]

6.2 = -log [H+]

[H+] = 6.31*10^-7 M

HA dissociates as:

HA -----> H+ + A-

2*10^-2 0 0

2*10^-2-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Ka = 6.31*10^-7*6.31*10^-7/(0.02-6.31*10^-7)

Ka = 1.991*10^-11

Answer: 1.99*10^-11

Only 1 question at a time please