We are being asked to calculate the Ka of lactic acid.
Recall: Weak acids possess a Ka value less than 1, while weak bases possess a Kb value less than 1.
The equilibrium expressions of Ka and Kb are the same as other equilibrium constants we’ve seen.
For weak acids, their equilibrium constant is
For weak bases, their equilibrium constant is
The greater the Ka value the stronger the acid, while the greater the Kb the stronger the base.
Ka and Kb are connected by the following equation:
where Kw = 1 x 10-14
It can be found in textbooks or on the internet.
We are given a saturated solution of calcium lactate salt with [Ca2+]=0.26 M and a pH=8.40.
The dissociation of calcium lactate salt in water is:
Ca(Lact)2(s) + H2O(l)→ Ca2+(aq) + 2 Lact-(aq)
Lactic acid is a weak acid found in milk. Its calcium salt is a source of calcium for growing animals. A saturated solution of this salt, which we can represent as Ca(Lact)2 has a [Ca2+]=0.26 M and a pH=8.40.
Assuming the salt is completely dissociated, calculate the Ka of lactic acid.
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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.
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Based on our data, we think this problem is relevant for Professor Morkowchuk's class at UVA.