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

Calculate K_{a} for this acid.

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

c.

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

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up