A weak acid, HA, has a p𝐾a of 4.500 . If a solution of this acid has a pH of 4.154 , what percentage of the acid is not ionized? Assume all H+ in the solution came from the ionization of HA.

Answer 1

Just wanted to try this by an alternate method from that proposed by Anthony T, and this what I've come up with. The answer is different from the previous, so I'll leave it to you to figure out the difference since I'm unable to do so.

HA <==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA] and we can find Ka from the pKa: Ka = 1x10^-4.500 = 3.16x10^-5

3.16x10^-5 = [H+][A-] / [HA]

From the pH, we can find the [H⁺] and [A^-]: [H⁺] = [A^-] = 1x10^-4.154 = 7.01x10^-5 M

3.16x10^-5 = (7.91x10-5)² / [HA]

Solving for the [HA] we have...

[HA] = 6.26x10^-9 / 3.16x10^-5

[HA] = 1.98x10^-4 M

fractional ionization =7.01x10^-5 / 1.98x10^-4 = 0.354 or 35.4% is ionized.

The % not ionized would thus be 100 - 35.4 = 64.6%

Answer 2

Start with the equilibrium expression Ka = [H+]² / [HA] where the approximation that the undissociated acid is approximately equal to the total acid concentration.

Take the log of both sides

log Ka = 2 log [H+] - log [HA]

Multiply by -1

-log Ka = -2 log [H+] + log [HA]

pKa = pH - log [H+] + log [HA]

pka = pH + log [HA] / [H+]

4.500 = 4.154 + log [HA] / [H+]

0.346 = + log [HA] / [H+] which can be rearranged to

-0.346 = log [H+] / [HA] therefore

[H+] / [HA] = 10^-.346 = 0.458 this is the degree of ionization, The fraction of unionized acid is

1 - 0.458 = 0.5420 or 54.20 %