help_wanted
Harmless
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how do i do this problem?
it's a solubility problem i am doing for chemistry class
When 0.075 g of KOH is dissolved in 1.00 L of 1.0 x 10-3 M Cu(NO3)2, a precipitate of Cu(OH)2 is formed. True or False? (Ksp of Cu(OH)2 is 2.2 x
10-20)
how bout this one??
The solubility of silver dichromate at a given temperature is 0.00864 g/100 mL. Calculate the Ksp at this temperature.
[Edited on 11-4-2004 by help_wanted]
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Biochem
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I'm not too sure and kinda rusty but I think you find the moles of KOH and then moles of Cu(NO3)2 by writing the reaction equation and balancing
it. Afterwards use this equation
Ksp = [xA]^a * [xB]^b where A and B are the solubility values of reactants and a and b are the coefficients.
[Edited on 11-4-2004 by Biochem]
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help_wanted
Harmless
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tried that biochem
not getting it...
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Biochem
Harmless
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Beats me then.
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unionised
International Hazard
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0.075g KOH is 0.001337 moles. The solution contains 0.001 moles of copper nitrate and the reaction is
Cu(NO3)2 +2KOH --> Cu(OH)2 +2 KNO3
(That means that there is not enough KOH to react with all the copper salt.)
In order for the copper to precipitate, the product of the concentrations of (hydroxide squared) and copper needs to excede the solubillity product.
The total copper is unchanged at 0.001 moles. The hydroxide concentration is 0.001337 molar
0.001337*0.001337*0.001 = 1.788*10^-9
That's bigger than Ksp so the hydroxide will precipitate. The statement is true.
Do you know what a solubillity product is?
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help_wanted
Harmless
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no never heard that term in class...
and thanx for the reply it cleared it up for me
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