A sample of Cd (OH)2 (s) is added to pure water and allowed to come to equilibrium at 25 degrees C. The concentration of Cd^2+ is 1.7 x 10 ^ -5 M at equilibrium. What is the value of Ksp for Cd(OH)2?
The answer is 2.0 x 10^ -14
Can someone show me how to get that answer please? Thank you so much!
2007-04-12 10:50:59 · 2 answers · asked by hitomiki 3 in Science & Mathematics ➔ Chemistry
Cd(OH)(2) ----> Cd(2+) + 2(OH)(-)
Complete dissociation, due to the strength of OH(-)
[(OH)(-)] = 2 * [Cd(2+)], from the reaction equation
=2 * 1.7 x 10^ -5 M = 3.4 x 10^-5 M
K(sp) = [Cd(2+)][(OH)(-)]^2 = 2.0 x 10^-14 [M^3]
2007-04-12 11:04:42 · answer #1 · answered by Anonymous · 0⤊ 0⤋
[OH-] is twice that of [Cd2+], so when you square it, you get
Ksp = [Cd2+] x (2 x [Cd2+] squared) = 4 [Cd2+] cubed.
2007-04-12 18:03:27 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋