Enough of a monoprotic weak acid is dissolved in water to produce a 0.0165 M solution. The pH of the resulting solution is 2.34 . Calculate the Ka for the acid.

![since PH= 2.34 -log (t] = 2.34 Nog (Ft+] = -2.34 Hi] = 102.34 M = 4.57X103M so (H+] = cx = 4.57X10 M. 4.57X10²M - 0.0165 M. -](http://img.homeworklib.com/questions/e682a4e0-8f6e-11eb-b362-edad374b0f86.png?x-oss-process=image/resize,w_560)
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0165 M solution....
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a. Enough of a monoprotic acid is dissolved in water to produce
a 0.0163 M solution. The pH of the resulting solution is 2.51.
Calculate the Ka for the acid. Ka= ________
It's not 5.86 10-4. I tried it and it
wasn't correct.
b. The Ka of a monoprotic weak
acid is 6.61 × 10-3. What is the percent ionization of a 0.108 M
solution of this acid? Percent ionization= ___________
it's not 24.72 %. I tried it's also incorrect.