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+ ..
Addendum: What i've tried doing is...
[OH-]^2= 6.5E-18 / .256 = 2.54E-17 and then finding the square root of that, which is 5.04E-9. this is what makes most sense to me...but is apparently incorrect...
Answered below.
To calculate the solubility of nickel(II) hydroxide (Ni(OH)2) in the given solution of 0.256 M NiSO4(aq), you need to determine the concentration of the hydroxide ions, OH-.
Here's a step-by-step approach to solving this:
1. Write the balanced chemical equation for the dissociation of Ni(OH)2:
Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH-(aq)
2. Write the expression for the solubility product constant, Ksp, using the concentrations of the dissociated ions:
Ksp = [Ni2+][OH-]^2
3. Since you want to find the concentration of the hydroxide ions, rearrange the equation:
[OH-]^2 = Ksp / [Ni2+]
4. Substitute the given Ksp value (6.5E-18) and the concentration of Ni2+ from NiSO4 (0.256 M) into the equation:
[OH-]^2 = 6.5E-18 / 0.256
5. Calculate the square root of the result to find the concentration of OH-:
[OH-] = √(6.5E-18 / 0.256)
6. Evaluating this expression will give you the concentration of OH-. In this case, the value you should get is 1.09E-8 mol/L.
So, the solubility of nickel(II) hydroxide (Ni(OH)2) in 0.256 M NiSO4(aq) solution is 1.09E-8 mol/L.