a. pH = - log(H^+)
b. pH + pOH = pKw = 14. You know pKw and pH, solve for pOH. Then pOH = -log(OH^-)
c.
...................Ca(OH)2 ==> Ca^2+ + 2OH^-
I.....................solid...............0.............0
C...................solid-x.............x.............2x
E...................solid................x.............2x
You know OH^- from b. (Ca^2+) is 1/2 (OH^-) and (Ca^2+) is [Ca(OH)2]
A Ca(OH) 2solution has a pH of 8.0. Determine the following for the solution: a. [H 3 O + ] b. [OH -] c. [Ca(OH) 2 ]
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