Question

To what final concentration of NH3 must a solution be adjusted to just dissolve 0.060mol of NiC2O4 (Ksp = 4×10−10) in 1.0 L of solution? (Hint: You can neglect the hydrolysis of C2O42− because the solution will be quite basic.)

Answer 1

Nickel ions and oxalate precipitate: \(Ni^{2+} + C_2O_4^{2-} \rightarrow NiC_2O_{4\;(s)} \); \(Ksp = 4×10^{−10}\)

the Ksp is pretty low, so none of really is dissolved

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this is to check how much is undissolved

\(Ksp = 4×10^{−10}=[Ni^{2+}][C_2O_4^{2-}]\rightarrow 4×10^{−10}=x^2\rightarrow x=0.00002 M\)

in 1 liter, you have 0.06 mol - 0.00002mol=0.59998 moles undissolved.

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nickel forms a complex ion with ammonia: \(Ni^{2+}+ 6\;NH_3 \rightarrow [Ni(NH_3)_6]^{2+}\)

at 25 degrees celsius, the formation constant is \(K_f=5.5*10^8\)

The Kf is sufficiently high, we can assume that when these come together in solution it all associates.

So the final concentration of \(NH_3\) should be the same as \(Ni^{2+}\) in the solution.
If you wanna be nit-picky \([NH_3]\)=0.59998 M