A) Enter the the Ksp expression for the solid AB2 in terms of the molar solubility x.
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 =
Sorry = 4x^3
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
Dr A 4x^3 was correct in MC
I am unsure how they got part D on this, perhaps some can figure it out.
Dr. A set the problem up correctly but he did not give you the correct answer:
A) Ksp = (x)(2x)^2 = 4x^3 NOT 4x^2
B) Ksp = (2x)^2*(3x)^3 = (4x^2)(9x^3) = 36x^5 NOT 108x^5
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