Could someone please explain what is special about Zn vs. Fe (and what other elements would behave like Zn or Fe)?
The question:
Ten mL of a 1 M NaOH solution is added to equal volumes of two saturated solutions. One solution contains an excess of Zn(OH)2(s) and the other an excess of Fe(OH)2(s). Why does more Zn(OH)2 dissolve in the first flask, but more Fe(OH)2 precipitate in the second flask?
Ksp of Zn(OH)2: 1.8 x 10^-14
Ksp of Fe(OH)2: 1.1 x 10^-14
A) The Ksp of Zn(OH)2 is greater, allowing more solid to dissolve.
B) The Ksp of Fe(OH)2 is smaller, so its molar solubility is lower.
C) Zn(OH)2 acts as an acid and neutralizes the NaOH, while Fe(OH)2 cannot neutralize the NaOH.
D) Due to the common ion effect, the Zn(OH)2 equilibrium shifts to the right while the Fe(OH)2 equilibrium shifts to the left.
Answer explanation (below, I've underlined the part I'm wondering about):
When excess hydroxide reacts with solid Zn(OH)2 it forms a new soluble coordination compound according to the following reaction: . The Fe(OH)2 does not have the ability to form such a soluble compound, and therefore behaves as predicted by the common ion effect.
Why is this so?
The question:
Ten mL of a 1 M NaOH solution is added to equal volumes of two saturated solutions. One solution contains an excess of Zn(OH)2(s) and the other an excess of Fe(OH)2(s). Why does more Zn(OH)2 dissolve in the first flask, but more Fe(OH)2 precipitate in the second flask?
Ksp of Zn(OH)2: 1.8 x 10^-14
Ksp of Fe(OH)2: 1.1 x 10^-14
A) The Ksp of Zn(OH)2 is greater, allowing more solid to dissolve.
B) The Ksp of Fe(OH)2 is smaller, so its molar solubility is lower.
C) Zn(OH)2 acts as an acid and neutralizes the NaOH, while Fe(OH)2 cannot neutralize the NaOH.
D) Due to the common ion effect, the Zn(OH)2 equilibrium shifts to the right while the Fe(OH)2 equilibrium shifts to the left.
Answer explanation (below, I've underlined the part I'm wondering about):
When excess hydroxide reacts with solid Zn(OH)2 it forms a new soluble coordination compound according to the following reaction: . The Fe(OH)2 does not have the ability to form such a soluble compound, and therefore behaves as predicted by the common ion effect.
Why is this so?
