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.
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
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