Question

0.250 moles of a weak acid, HA, is dissolved in pure water to make up 500 ml of a solution. As the HA dissolves, the following equilibrium is established:

HA(aq) + H₂O (l) = H₃O⁺(aq) + A^- (aq) K=3.5 * 10^-5 (25 C)

A). Calculate the concentrations of HA, H3O⁺, and A^- present at equilibrium

B). Calculate the pH of the solution at equilibrium.

Answer 1

[HA] = number of moles of HA / volume in L

= 0.250 moles/0.500 L

= 0.500 M

HA dissociates as:

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

0.500 0 0

0.500-x x x

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

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((3.5*10^-5)*0.500) = 4.183*10^-3

since c is much greater than x, our assumption is correct

so, x = 4.183*10^-3 M

so,[H+] = x = 4.183*10^-3 M

[A-] = x = 4.183*10^-3 M

[HA] = 0.500 - x = 0.500 - 4.183*10^-3 M = 0.496 M

B)

use:

pH = -log [H+]

= -log (4.183*10^-3)

= 2.38