Question
Ca(OH)^2 does not completely dissociate. It only partially dissotiates in water to form OH- and Ca+2. An equilibrium mixture of Ca(OH) and its dissolved ions is found to have a pH= 10.3 what is Kc for this dissociation?
Answers
DrBob222
pH = 10.3.
pH + pOH = pKw = 14; therefore, pOH = 3.7 and pOH = -log(OH^-).
(OH^-) = 2E-4
............Ca(OH)2 ==> Ca^2+ + 2OH^-
................x.........x........2x
Kc = (Ca^2+)(OH^-)^2
(Ca^2+) = 1/2(2E-4)
(OH^-) = 2E-4
Solve for Kc.
pH + pOH = pKw = 14; therefore, pOH = 3.7 and pOH = -log(OH^-).
(OH^-) = 2E-4
............Ca(OH)2 ==> Ca^2+ + 2OH^-
................x.........x........2x
Kc = (Ca^2+)(OH^-)^2
(Ca^2+) = 1/2(2E-4)
(OH^-) = 2E-4
Solve for Kc.