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Problem: Calculate the percent ionization of 0.0080 M butanoic acid in a solution containing 0.080 M sodium butanoate.

🤓 Based on our data, we think this question is relevant for Professor Wade's class at OSU.

FREE Expert Solution

The dissociation of C3H7COOH is as follows:

C3H7COOH(aq) + H2O(l)  H3O+(aq) + C3H7COO(aq)Ka = 1.5 × 10–5

[C3H7COOH] = 0.0080 M


The dissociation of C3H7COONa is as follows:

C3H7COONa(aq)  Na+(aq) + C3H7COO(aq)

[C3H7COONa] = 0.080 M

[C3H7COONa] = [C3H7COO+] = 0.080 M


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Problem Details

Calculate the percent ionization of 0.0080 M butanoic acid in a solution containing 0.080 M sodium butanoate.

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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.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Wade's class at OSU.