Asked by hello
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
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
Answers
Answered by
DrBob222
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.
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.
Answered by
go to hell ANONYMOUS
said it twice for added effect......
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