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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
Calculate the percent ionization of 0.0080 M butanoic acid in a solution containing 0.080 M sodium butanoate.
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Based on our data, we think this problem is relevant for Professor Wade's class at OSU.