Biochemistry

weak acid, HA, is 0.1 % ionized in a 0.2 M solution.
(a) What is the equilibrium constant for the dissociation of the acid (Ka)?
(b) What is the pH of the solution?

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1. ..................HA ==> H^+ + A^-
I...............0.2M.........0.........0
C................-x............x..........x
E..............0.2-x...........x.........x
HA = 0.2 M and 0.1% ionized; therefore,
(H^+) = (A^-) = 0.2 M x 0.001 = 2E-4 and (HA) = 0.2-2E-4
Ka = (H^+)(A^-)/(HA)
Substitute the numbers into the Ka expression and solve for Ka.

b part. pH = -log(H^+). You know (H^+) from above, calculate pH. Post your work if you get stuck.

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DrBob222

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