A precipitate of copper(II) hydroxide is formed in a reaction. What change will help to dissolve the copper(II) hydroxide?
Cu^2+(aq) + 2OH^-(aq) Cu(OH)"subtext"2(s)
A. Adding NaOH
B. Adding HCl
C. Removing CuCl2
D. Removing H2O
is the answer for this d? removing H2O??
I don't understand how the answer can be D. There is no water in the reaction. What you want to do is to shift the equilibrium to the left (remove the OH^-) and you can do that by adding an acid so I would choose B.
B. Adding HCl
Oh wow, talk about a chemistry trick! Well, to dissolve the copper(II) hydroxide, we need to think about what can break it down. Hmmm...
Adding NaOH (choice A) is actually going to make things worse because it will just add more hydroxide ions, which will only form more copper(II) hydroxide. So, that's a no-go.
Adding HCl (choice B) is the winner here! HCl is an acid, and it will react with the copper(II) hydroxide to form water and a soluble compound, which will dissolve the precipitate. It's like a chemical superhero saving the day!
Removing CuCl2 (choice C) won't do anything because it's not in the equation. So that one is out.
Removing H2O (choice D) might seem tempting, but unfortunately, water is necessary for the reaction to occur. So we can't remove it.
So, the answer is B. Adding HCl will help dissolve the copper(II) hydroxide.
The correct answer is A. Adding NaOH.
When copper(II) hydroxide (Cu(OH)2) precipitate forms, it can be dissolved by adding more hydroxide ions (OH-) to the solution. Sodium hydroxide (NaOH) is a strong base that provides hydroxide ions when dissolved in water. By adding NaOH to the solution containing copper(II) hydroxide, the additional OH- ions will react with the Cu(OH)2 precipitate and form soluble complex ions, thereby dissolving the precipitate.
Therefore, the change that will help to dissolve the copper(II) hydroxide is adding NaOH (choice A).
The correct answer to this question is A. Adding NaOH. Adding sodium hydroxide (NaOH) will help dissolve the copper(II) hydroxide precipitate.
To understand why, let's first consider the reaction:
Cu^2+(aq) + 2OH^-(aq) -> Cu(OH)2(s)
In this reaction, a precipitate of copper(II) hydroxide (Cu(OH)2) is formed when copper ions (Cu^2+) react with hydroxide ions (OH^-). The solid Cu(OH)2 is not soluble in water and forms a precipitate.
To dissolve this precipitate, we need to introduce a species that can react with the Cu(OH)2 to form a soluble complex or compound.
By adding sodium hydroxide (NaOH), hydroxide ions (OH^-) are introduced into the system. These hydroxide ions react with copper(II) hydroxide, forming a soluble complex called sodium cuprate(II):
Cu(OH)2(s) + 2OH^-(aq) -> Na2[Cu(OH)4](aq)
Sodium cuprate(II) is soluble in water, which means it dissolves, forming a solution. Therefore, adding NaOH helps to dissolve the copper(II) hydroxide precipitate.
The other options (B, C, D) do not directly contribute to dissolving the precipitate. Adding hydrochloric acid (HCl) (option B) would actually result in an increase in the concentration of H+(aq) ions, which can potentially cause the copper hydroxide precipitate to convert into copper chloride. Removing CuCl2 (option C) or removing water (option D) would not have any direct effect on the solubility of the copper(II) hydroxide precipitate.