##### Weak Acid Dissociation Constant

 Chemistry Tutor: None Selected Time limit: 1 Day

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?

Feb 25th, 2015

QUESTION titration of 100mL of 0.100 acetic acid (Ka=1.8 x10^-5) w/ 0.100 M NaOH. What's pH at equivalence?

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 (called hydrolysis).

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  - x)

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)

Feb 25th, 2015

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Feb 25th, 2015
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Feb 25th, 2015
Dec 3rd, 2016
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