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Calculate the molar solubility of Nickel (II) Iodate in 3.0M NH3.?
Calculate the molar solubility of Nickel (II) Iodate in 3.0M NH3.
Ksp of Ni(IO3)2 = 2.7x10^-3 and Kf of Ni(NH3)6 (3+) = 2.0x10^8

The answer is 0.49M I just have no idea how to solve. For other problems I've combined the equations and multiplied Ksp with Kf to get an overall K value. Still have trouble getting 0.49M tho. Anyone help?
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caominone

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ÍõÌÎ0715: ½ð±Ò+5, ¡ïÓаïÖú 2013-03-05 09:00:02
fds329: ½ð±Ò+2, ¸ÐлӦÖú¡£ÐÁ¿àÁËO(¡É_¡É)O 2013-03-11 11:54:48
your idea is wrong,you should use kf divide ks,write the balance equation,

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2Â¥2013-03-04 23:57:23
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jiagle

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ÍõÌÎ0715: ½ð±Ò+15, ¡ï¡ï¡ï¡ï¡ï×î¼Ñ´ð°¸ 2013-03-05 09:00:10
fds329: ½ð±Ò+3, ICEPI+1, ¸ÐлӦÖú¡£ÐÁ¿àÁËO(¡É_¡É)O 2013-03-11 11:54:59
Ni(IO3)2(s) + 6NH3 = [Ni(NH3)6](2+) + 2IO3(-)
                       3.0           0                    0
                       3.0-6x       x                    2x
       x*(2x)^2 / (3.0 - 6x)^6 = Kf * Ksp
Nowton Iteration  (x< 3.0/6 =0.5)
       x = 0.4838 ~= 0.48mol/L
3Â¥2013-03-05 08:29:02
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