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+ ..
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.
To calculate the solubility of nickel(II) hydroxide (Ni(OH)2) in 0.256 M NiSO4(aq), you need to consider the stoichiometric relationship between the two compounds.
First, let's write the balanced equation for the dissociation of nickel(II) hydroxide:
Ni(OH)2(s) ⇌ Ni^2+(aq) + 2OH^-(aq)
The solubility product constant (Ksp) expression for this reaction is:
Ksp = [Ni^2+][OH-]^2
Now, we can see that in the given NiSO4(aq) solution, NiSO4 dissociates to form Ni^2+ ions:
NiSO4(aq) → Ni^2+(aq) + SO4^2-(aq)
Since the concentration of NiSO4 is provided as 0.256 M, we can assume that the concentration of Ni^2+ is also 0.256 M.
Therefore, you can substitute the concentration of Ni^2+ from the NiSO4 solution into the Ksp expression of Ni(OH)2 to calculate the solubility of Ni(OH)2 in the NiSO4 solution.
Ksp = [Ni^2+][OH-]^2
Using the given Ksp value of 6.5E-18, and the concentration of Ni^2+ as 0.256 M, we can rearrange this equation to solve for [OH-]:
[OH-]^2 = Ksp / [Ni^2+]
[OH-]^2 = (6.5E-18) / (0.256)
Now, take the square root of both sides to isolate [OH-]:
[OH-] = sqrt[(6.5E-18) / (0.256)]
Finally, you can calculate the solubility of Ni(OH)2 in the NiSO4 solution by multiplying the concentration of [OH-] by 2 (since the stoichiometry shows that 2 OH- ions are produced for every Ni(OH)2 molecule that dissolves):
Solubility of Ni(OH)2 = 2 × [OH-]