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Chemistry
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An aqueous solution of a monoprotic acid HA (Ka = 1.45×10-1) has a pH of 4.75. 

What is the molarity of the solution? (i.e. [HA]+[A-].)
Feb 9th, 2015

pH = 4.75 

[H+] =10 ^ -4.75 = 1.78e-5


HAH+A-
InitialM00
Change -x +x +x 
EquilibriumM -1.78e-51.78e-5
1.78e-5

Ka = 1.45e-1 = 1.78e-5^2 / (M - 1.78e-5)

1.45e-1 M -2.581e-6 = 3.1684e-10

M = 1.78e-5M

Feb 9th, 2015

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