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The ionization of benzoic acid is represented by thisequation.C6H5COOH(aq) -->H+(aq)...
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`C_6H_5 COOH harr H^+ + C_6H_5 COO^-`
`K_a = ([H^+][C_6H_5 COO^-])/[[C_6H_5 COOH]]`
Using mole ratio `[H^+]=[C_6H_5 COO^-]=1.7xx10^-3`
Final `[C_6H_5 COOH] = 0.045-1.7xx10^(-3)=0.0433`
`K_a = ([1.7xx10^(-3)][1.7xx10^(-3)])/[0.0433]=6.67xx10^(-5)`
So the `K_a` of acid is `6.67xx10^(-5)M.`
Posted by jeew-m on August 27, 2013 at 9:24 AM (Answer #1)
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