When a 0.050 m CacCl2 solution is adjusted to a pH = 8, ca(oh)2(s) will not precipitate.
This is further explained below.
The process of changing a dissolved component into an insoluble solid from a super-saturated solution is referred to as precipitation. This process takes place in an aqueous solution. The name was given to the solid that forms is precipitate.
Question Parameter
Ksp of Ca(OH)2 = 7.9 x 10-6
Generally, the equation for pOH is mathematically given as
pOH = 14 - pH
pOH = 14 - 8
pOH = 6
Therefore
[OH-] = 1 x 10^{-6} M
[Ca2+] = 0.050 M
[Ca2+][OH-]^2 = 0.050 x (1 x 10-6)2
[Ca2+][OH-]^2 = 5.0 x 10-14 M < Ksp
[Ca2+][OH-]^2 = 7.9 x 10^{-6}
In conclusion, When the pH of 0.050 M CaCl2 is brought down to 8, there is therefore no precipitate.
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