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Unformatted text preview: KW = 10-14 = [H3O+] [OH-]
[H3O+] = [H3O+]water + [H3O+]HCl
[OH-] = [H3O+]water Lets set x = [H3O+]water then
[H3O+] = x + 1.0 x 10-8
[OH-] =x 10-14 = (x + 1.0 x 10-8) ( x) Our equation can be rearranged as x x2 + 10-8 x - 10-14 = 0
This quadratic expression can be solved by:
x= -b + b2 - 4ac
2a Only the positive root is meaningful in
equilibrium problems. = -10-8 + [ (10-8)2 + 4x10-14]1/2 / 2
= 1.9 x 10-7 M pH = 6.72 So adding a small amount of HCl to water
DOES make it acidic.
While this approach is more time
consuming, you’ll find it very useful as our
problems get more complex. ...
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- Fall '08