# Isolating barium(II), copper(II) and zinc(II) from aqueous solution [closed]

Q15 Each metallic ion was separated from an aqueous solution containing $$\ce{Ba^2+},$$ $$\ce{Cu^2+},$$ and $$\ce{Zn^2+}$$ by the procedure shown in the following figure. From ①–⑥ in the table below choose the correct combination of ions that predominates in the precipitates a, b and the filtrate c.

For the precipitate a, I know the answer is $$\ce{Ba^2+}$$ ion because it has a highly negative standard reduction potential than the other ions, so it tends to displace the sulfuric acid to form insoluble barium sulfate.

But I don't know which one can displace $$\ce{NaOH}.$$ $$\ce{Na+}$$ ion reduction potential is a lot more negative than both $$\ce{Cu^2+}$$ and $$\ce{Zn^2+}$$ ions. I suspect there is something in 'added in excess' part, but I don't know how can it be any different.

• Reduction potential is quite irrelevant here. You might want to review what do you know about the amphoteric properties of Zn. – Ivan Neretin Oct 14 '19 at 10:29
• $\ce{NaOH}$ is added in excess to first neutralize the sulfuric acid from the first step and then to make the pH go up beyond 7. – TAR86 Oct 14 '19 at 10:35
• @AegisNine I suggest you look up the properties of Zn(OH)4 dianion – Waylander Oct 14 '19 at 10:39
• So, this related to properties of transition metal, and not reduction potential, but Ba is not d-block metal, how can I compare it? This is an undergrad admission exam, so is it really necessary to know the specific properties of a compound? – AegisNine Oct 14 '19 at 11:22
• Of course both Cu and Zn will react with NaOH. But they will react in different ways. – Ivan Neretin Oct 14 '19 at 12:29

Once $$\ce{Ba^2+}$$ is eliminated as poorly soluble barium(II) sulfate, there is a solution of soluble copper(II) and zinc(II) sulfates. Considering this has been done by careful/gradual addition of sulfuric acid, and the fact that both remaining sulfates undergo hydrolysis, the solution is going to be slightly acidic — this is one reason for sodium hydroxide "added in excess".

Another reason, as both Ivan and Waylander pointed out in the comments, is the difference between reactivity of copper(II) and zinc(II) in alkali medium. Long story short, you are probably supposed to know the specific properties of the common transition metals cations as the reduction potential isn't really helpful here.

### Copper(II)

Copper(II) with $$\ce{NaOH}$$ first forms colloidal copper(II) hydroxide; the gel is then solidifies to light-blue amorphous substance upon subsequent addition of hydroxide:

$$\ce{CuSO4(aq) + 2 NaOH(aq) -> Cu(OH)2(s) + Na2SO4(aq)}$$

$$\ce{Cu(OH)2(s)}$$ appears to be precipitate b. Sodium tetrahydroxocuprate(II) $$\ce{Na2[Cu(OH)4]}$$ doesn't form here: it's a very moisture-sensitive and in the air quickly turns dark-brown:

$$\ce{Na2[Cu(OH)4](aq) -> CuO(s) + H2O(l) + 2 Na(OH)(aq)}$$

### Zinc(II)

Zinc(II), on the other hand, with $$\ce{NaOH}$$ first forms white precipitate of zinc(II) hydroxide:

$$\ce{\ce{ZnSO4(aq) + 2 NaOH(aq) -> Zn(OH)2(s) + Na2SO4(aq)}}$$

which is then dissolved again forming colorless soluble sodium terahydroxozincate(II) (filtrate c):

$$\ce{Zn(OH)2(s) + 2 NaOH(aq) -> Na2[Zn(OH)4](aq)}$$

Excess of $$\ce{NaOH}$$ is also necessary to keep $$\ce{Na2[Zn(OH)4]}$$ in solution: upon diluting (lowering pH, adding water/acid), terahydroxozincate(II) decomposes to zinc(II) hydroxide.

Note: chemical reactions are adapted from [1].

### References

1. R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)