No it would not because it’s still hard
Answer:
Explanation:
Hello!
In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:
Now, we compute the concentration of hydroxyl ions in solution:
![[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%3D10%5E%7B-3.95%7D%3D1.41x10%5E%7B-4%7DM)
Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:
Whose equilibrium expression is:
![Ksp=[M^+][OH^-]](https://tex.z-dn.net/?f=Ksp%3D%5BM%5E%2B%5D%5BOH%5E-%5D)
Therefore, the Ksp for the saturated solution turns out:
Best regards!
<span>XY4Z2-->Square planar (Electron domain geometry: Octahedral) sp3d2
XY4Z-->Seesaw (Electron domain geometry: Trigonal bipyramidal) sp3d
XY5Z-->Square pyramidal (Electron domain geometry: Octahedral) sp3d2
XY2Z3-->Linear (Electron domain geometry: Trigonal bipyramidal) sp3d
XY2Z-->Bent (Electron domain geometry: Trigonal planar) sp2
XY3Z-->Trigonal pyramidal (Electron domain geometry: Tetrahedral) sp3
XY2Z2-->Linear (Electron domain geometry: Tetrahedral) sp3
XY3Z2-->T shaped (Electron domain geometry: Trigonal bipryamidal) sp3d
XY2-->Linear (Electron domain geometry: Linear) sp
XY3 Trigonal planar (Electron geometry: Trigonal planar) sp2
XY4-->Tetrahedral (Electron domain geometry: tetrahedral) sp3
XY5-->Trigonal bipyramidal (Electron domain geometry: Trigonal bipyramidal) sp3d
XY6-->Octahedral (Electron domain geometry: Octahedral) sp3d2</span>
Is soluble in water but not soluble in acetonitrile.
So i think its false
I think it will be B. Hope it helps :)
