f 4.2 X 10-8 6 water established for Cr (OH 14. Fill in the blanks in the follow

Question

f 4.2 X 10-8 6 water established for Cr (OH 14. Fill in the blanks in the following table. Compound cation [anion] Ksp 15-3 (a) BaC2O4 1.6 x 10 (b) Cr(OH) 2.7 x 10-8 6.3 x 10 31 Ksp ar (c) Pb3 (PO4)2 1 x 10 54 27. C 8 x 10 that is sp and Solubility 15. Calculate the molar solubility of the following compounds (a) calcium iodate: K 6.5 x 10-6 x 10 (b) cadmium phosphate: Ksp 2.5 (c) silver oxalate: K sp 5.4 x 10-12 16. Calculate the molar solubility of the following 2 compounds. (a) MgF2 (b) Fe (OH)3 (c) Mg (PO42 17. Calculate the K sp of the following compounds, given

Explanation / Answer

a) Ca(IO3)2 (s) <----> Ca2+ (aq) + 2IO3-(aq)

at equilibrium [Ca2+] = S , [IO3-] = 2S ,

Ksp = [Ca2+] [IO32-]^2 = ( S) ( 2S)^2 = 4S^3

6.5x10^-6 = 4S^3

S = 0.01176 moles/L is the solubilit

b) Cd(PO4) (s) <---> Cd3+ (aq) + PO43-(aq)

at equilibrium if S moles of Cd(PO)4 dissociated then [Cd3+] =[PO43-] = S

Ksp = [Cd3+][PO43-] = (S) (S) = S^2 = 2.5x10^-33

S = 5 x 10^-17 Moles/Liter is solublity

c) Ag2C2O4 (s) <----> 2Ag+ (aq) + C2O42-(aq)

[Ag+] = 2S , [C2O42-] = S

Ksp = [Ag+]^2 [C2O42-] = (2S)^2 ( S) = 4S^3

5.4 x10^-12= 4S^3

S = 1.1 x 10^ - 4 moles/Liter is solubility of salt

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