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Chemistry
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In a 6.22×10-2 M solution, a monoprotic acid, is 27.5% dissociated. 

Calculate Ka for this acid.

b.In a 6.22×10-2 M solution, a monoprotic acid, is 27.5% dissociated. 

Calculate Ka for this acid.

c. 

Feb 13th, 2015

Please note that the chemical reaction of your problem, is:

HA = H+  +   A-

where HA is the monoprotic acid.

Now if I call with y the concentration of H+, and A- at Equilibrium, I have, by definition:

y/ 0.0622 = 0.275, so I can write:

y = 0.275*0.0622 = 0.0171 mol/liter

Now, please note that the Equilibrium acid constant is given by the subsequent formula:

Ka = [H+]*[A-] / [HA]

then we have:

Ka = y^2/ (0.0622-y)

and substituting y=0.0171 into the above equation, I get:

Ka = 6.5*10^(-3) mol/liter

Feb 13th, 2015

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