The Ksp of Ca(OH)2 is 6.5 x 10-6. Will a precipitate form if the pH of a 0.085 M CaCl2 solution is adjusted to 8.8?
Concentration of CaCl2 = 0.085 M
So, [Ca2+] = 0.085 M
Then
pH = 8.8
So, pOH = 14 - 8.8 = 5.2
[OH-] = 10-5.2 M = 6.31× 10-6 M.
Now, Q of Ca(OH)2
= [Ca2+] × [OH-]2
= 0.085 × ( 6.31× 10-6)2
= 3.38 × 10-12 .
Now, Ksp of Ca(OH)2 = 6.5×10-6
As, Ksp > Q , no precipitate will form.
The Ksp of Ca(OH)2 is 6.5 x 10-6. Will a precipitate form if the pH of...
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