![Solve for the pH of a 1.00L solution that is 0.25 M HC3H502 and 0.45 M KC3H502 when 0.100 moles of NaOH are added. [base] pH](http://img.homeworklib.com/questions/a58f7eb0-7594-11ea-8500-bdec8b37eec1.png?x-oss-process=image/resize,w_560)
Homework Answers
after 0.100 moles NaOH added following reaction takes place.
[HC3H5O2] = 0.25 - 0.100 = 0.15 M
[C3H5O2] = 0.45 + 0.100 = 0.55 M
pH = pKa + log [C3H5O2] / [HC3H5O2]
pKa = - log Ka = - log 1.3 x 10-5
pH 4.89
pH = 4.89 + log [0.55] / [0.15]
pH = 5.45
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