Posted by Val on Friday, March 19, 2010 at 8:52pm.
Use the data in this table to calculate the solubility of each sparingly soluble substance in its respective solution.
(a) silver bromide in 0.066 M NaBr(aq)
mol · L-1
I know how to do this one, found it to be 1.167E-11 which is correct.
(b) nickel(II) hydroxide in 0.256 M NiSO4(aq); For the purpose of this calculation, ignore the autoprotolysis of water.
mol · L-1
Ksp for Ni(OH)2 is 6.5E-18
Here's what I started doing:
NiSO4 -->NI^+ + SO4^-
Ksp = [Ni+][SO4-]
Ni(OH)2--> Ni^2+ + 2OH^-
Ksp = [Ni^2+][OH-]^2
so... [OH-]^2 = Ksp of [Ni^2+][OH-}^2 over [Ni+] from the NiSO4
but I'm not sure what to do because the first compound has Ni+ and the second has Ni 2+ ..
- Chemistry - DrBob222, Friday, March 19, 2010 at 10:08pm
I think your only problem is that you think Ni in NiSO4 is +1 but you know SO4 is -2; therefore, Ni must be +2 so Ni in Ni(OH)2 and Ni in NiSO4 are the same animal.
(Ni^+2)(OH^-)^2 = Ksp
I also think you can avoid a lot of confusion if you set up a chart; at least until you get the hang of these problems. This is a common ion problem (the Ni^+2 is the common ion) and all of them are done alike.
Ni(OH)2 ==> Ni^+ + 2OH^-
Let solubility of Ni(OH)2 be S and that is what the problem ask for. At equilibrium,
(Ni^+2) = S from Ni(OH)2 and 0.256 M from NiSO4 to make a total of S+0.256.
(OH^-) = 2S
Then we substitute into Ksp expression as follows:
Ksp = (S+0.256)(2S)^2
Solve for S using the quadratic OR make a simplifying assumption that S+0.256 = 0.256 and solve for S. Making the assumption is the easy way, then check the final answer to see if that assumption is valid.
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