1[H+] = √( Ka . c) = √(10 -4 ( 0.1) = 3.16 x 10-3
Thus , [OH-] = 10-14 / (3.16x10-3) = 3.16x10-12.
This is due to water , which is equal to the [H+] contribution by water.
so , % contribution by water is 10-7 .
Q.1
Find the % of H+ from water at 250C in 0.1 M CH3COOH take Ka of acid as 10-4 Kw=10-14 and volume = 1L.
(Give complete soln.)
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4 Answers
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1i too got 10^-7 % contribution frm water but i did in a diff method ..... took common ion into consideration...
i didn get srinath's method :(
1357srinath had neglected the [H+] from water initially and you have neglected it in equations.
that's the only difference
