.......Ca(OH)2 ==> Ca^2+ + 2OH^-
I.......solid.......0.......0
C.......solid.......x.......2x
E.......solid......x.......2x
(H^+) for pH 4 is 1E-4
Ksp = (Ca^2+)(OH^-)
From the above, (Ca^2+) = x
(OH^-) = x + 1E-4. Substitute and solve for x = solubility.
Note: for the total OH^-, the 2x is that from Ca(OH)2 and the 10^-4 is contributed by the solution buffered at pH of 4. So total is 2x+1E-4. Sometimes the 2x can be ignored; sometimes it can't be ignored (which avoids a quadratic equation) and the quadratic equation must be solved.
Calculate the molar solubility of calcium hydroxide in a solution buffered at each pH.
a. pH=4
b. pH= 7
c. pH=9
1 answer