Asked by Hannah

The solubility product expression for mercury(I) iodide, Hg2I2 is Ksp = ?

My answer is [Hg2^2+^][2I^-^]^2^

Would you agree?
Answered by DrBob222
close.
Hg<sub>2</sub>I<sub>2</sub> ==> Hg<sub>2</sub><sup>++</sup> + 2I<sup>-</sup>

Ksp = [Hg<sub>2</sub><sup>++<sup>]<sup>2</sup>[I<sup>-</sup>]<sup>2</sup>
I hope that turns out ok. It's always so easy to miss turning that on or off. The ++ is the same as 2+
Answered by DrBob222
See what I mean?
Ksp = [Hg<sub>2</sub><sup>++</sup>]<sup>2</sup>[I<sup>-</sup>]<sup>2</sup>
Let's see if this looks ok.
Answered by DrBob222
Still not right.
[Hg<sub>2</sub><sup>++</sup>][I<sup>-</sup>]<sup>2</sup>
Answered by DrBob222
That last one is ok.
Answered by Rohan
Explain betterly
Answered by Shweta
Hg2I2==>hg2^2+2I-.......coz Hg2^2 always exist in dimeric form
Answered by Lola
[hg2 ^2+] [I-]^2
There are no AI answers yet. The ability to request AI answers is coming soon!