Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case).
The dissociation of a weak acid, HA is as follows:
HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)
From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.
Let X = initial concentration of HA
The Ka expression for HA is:
Note that each concentration is raised by the stoichiometric coefficient: [HA], [H3O+] and [A–] are raised to 1.
[H3O+] = [A–] = x = 1.8 x 10-3.
We can then plug this into the Ka expression:
In a 0.050 M solution of a weak monoprotic acid, [H+] = 1.8 x 10-3. What is its Ka?
A. 3.6 x 10-2
B. 9.0 x 10-5
C. 6.7 x 10-5
D. 1.6 x 10-7
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.