The ionization of benzoic acid is represented by thisequation.C6H5COOH(aq) -->H+(aq) + C6H5COO–(aq)If a 0.045 M solution of benzoic acid has an [H+] =1.7xl0^(-3), what is the Ka of benzoic...

The ionization of benzoic acid is represented by this
equation.
C6H5COOH(aq) -->H+(aq) + C6H5COO–(aq)
If a 0.045 M solution of benzoic acid has an [H+] =1.7xl0^(-3),

what is the Ka of benzoic acid?

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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]]`

`[H^+]=1.7xx10^(-3)`

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.`

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