# Enter the the Ksp expression for the solid AB2 in terms of the molar solubility x.?

A) Enter the the Ksp expression for the solid AB2 in terms of the molar solubility x.

A)
For the equilibrium
AB2 <----> A2+ + 2 B-
Ksp = [A2+] [ B-]^2
lex x = moles/l of AB2 that dissolve.
This gives us x moles /L of A2+ and 2x moles/L of B-
Ksp = ( x) ( 2x) ^2 =
= 4x^2
Sorry = 4x^3
B)
C2D3 <----> 2 C3+ + 3D 2-
Ksp = [C3+] ^2 [ D2-] ^3
Let x = moles/ L of C2D3 that dissolve.
We get 2x moles/l of C3+ and 3x moles / L of D2-
Ksp = (2x)^2 ( 3x)^3 = 108 x^5
In fact 2^2 = 4
3^3 = 27
27 x 4 = 108

permit the molar solubilit be S Then [Ca2+] = S [IO3]2- = 2S The solubility product = S x (2S)^2 = 4S^3 = 7.a million x 10^-7 = 710 x 10^-9 which on fixing yields S = 8.9 x 10^-3 and the molar solubility For the subsequent occasion, you think approximately x because of the fact the molar solubility of Ca(IO3)2 in water and then the concentration of IO3 could be (2x + 0.06) in view that NaIO3 is thoroughly soluble and then [Ca2+] = x then writing the solubility product equation x (2x + 0.06) = 7.a million x 10^-7 and remedy for x, it is the molar solubility Ca(IO3)2 interior the presence of 0.06 M NaIO3

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Dr A 4x^3 was correct in MC

I am unsure how they got part D on this, perhaps some can figure it out.

4x^3, 108x^5