What is the pH of a 0.136 M monoprotic acid whose Ka is 5.7 10-4?

3.8 Ratings (589 Votes)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â HA --------> H⁺ + A^-

IÂ Â Â Â Â Â Â Â Â Â Â Â Â 0.136Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â 0

CÂ Â Â Â Â Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â +x

EÂ Â Â Â Â Â Â Â Â Â Â Â Â 0.136-xÂ Â Â Â Â Â Â Â +xÂ Â Â Â Â +x

Â Â Â Â Â Â Â Â Ka = [H⁺][A^-]/[HA]

Â Â Â Â Â Â Â 5.7*10^-4Â Â =Â Â x*x/0.136-x

Â Â Â Â Â Â Â 5.7*10^-4 *(0.136-x) = x²

Â Â Â Â Â Â Â Â Â Â xÂ Â = 0.0085

Â Â Â Â Â Â Â Â [H+]Â Â = x = 0.0085M

Â Â Â Â Â Â PHÂ Â Â = -log[H+]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = -log0.0085

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.0705

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