February 20, 2017

Homework Help: Chemistry

Posted by Jamie on Sunday, March 18, 2012 at 10:58pm.

In a 1.760 M aqueous solution of a monoprotic acid, 3.21% of the acid is ionized. What is the value of it's Ka?

x=1.760M * 3.21/100 - .0565M

Ka expression is Ka= {[H+][A-]} / [HA]

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