Titration of 100mL of 0.100 acetic acid (Ka=1.8 x10^-5) w/ 0.100 M NaOH. What's pH at equivalence, and half equivalence?
QUESTION titration of
100mL of 0.100 acetic acid (Ka=1.8 x10^-5) w/ 0.100 M NaOH. What's pH at
ANSWER: OK, so the equivalence point is the point on the titration curve
where the mols of OH- added equals the mols of HA (I'm using HA to refer toanygeneral weak acid).
So first we need to know the number of mols of HA before mixing
mols HA = 0.1 L * 0.100 M = 1E-2 mols
At the equivalence point, all HA and OH- is used up in
neutralization to make H2O
All that is left is A-. This
A- will react with water to create more HA until equilibrium is reached
To find the amounts after equilibrium is reached in hydrolysis,
we need to use kb instead of ka. Kb
= kw/ka = 1E-14/1.8E-5 = 5.6E-10
Using an ICE table, we can find that the A- remaining after equilibrium will be given by
5.6E-10 = x^2 / (1E-2 -
using the method of successive approximations, x=2.366E-6
mols of OH- = x = 2.366-6
volume of KOH added = 1E-2 / 0.100 M = 0.1 L
the total volume = 0.1 L HA + 0.1 L NaOH = 0.2 L
[OH-] = 12.366E-6 / 0.2 L = 1.183E-5 M
pOH = -log [OH-] = -log (1.183E-5) = 4.927
pH = 14 – pOH = 14 – 4.927
pH = 9.1
(with sigfigs in mind)
QUESTION What's pH
at half equivalence?
ANSWER: At the half
equivalence point, the pH = pka, where pka denotes the -log (ka)
-log (1.8E-5) =
4.7 (with sigfigs in mind)
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?