Question

What would be the pH of a solution of hy-
poiodous acid (HOI) prepared by dissolving
144 grams of the acid in 200 mL of pure water
(H2O)? The Ka of hypoiodous acid is 2¡¿10−11

1. 7
2. 10
3. 13
4. 5
5. 1

Answer 1

HOI ==> H^+ + OI^-

Ka = (H^+)(OI^-)/(HOI)
At equilibrium,
(H^+) = x
(OI^-) = x
(HOI) = moles/L. moles = 144 g/molar mass HOI and L = 0.2. So at equilibrium (HOI) = M-x
Solve for x and convert to pH.

Answer 2

said it twice for added effect......