Question

How many of the following hydroxides are soluble in excess \[NaOH\]solution?
\[Fe{{(OH)}_{3}}\],\[Al{{(OH)}_{3}}\],\[Zn{{(OH)}_{3}}\],\[Zn{{(OH)}_{2}}\],\[Ni{{(OH)}_{2}}\],\[Mn{{(OH)}_{2}}\],\[Sn{{(OH)}_{2}}\],\[Sn{{(OH)}_{3}}\],\[Cu{{(OH)}_{2}}\]

Answer 1

Hint: Certain inorganic substances don't react with NaOH. Some can emit combustible gas when they react with NaOH. In excess of NaOH, more polar metal salts can dissolve. More polar salts are d-block metal salts.

Complete Step by Step Solution:
In this question we need to identify those metal hydroxides which are soluble in excess NaOH solution. Since, we know that NaOH is a strong base and in order to be soluble, compounds should be acidic. So, we can say that we can find the lewis acids among the given compound and that will be soluble in excess of NaOH.

When Aluminium hydroxide (\[Al{{(OH)}_{3}}\] ) reacts with NaOH, they both react to give $NaAl{{O}_{2}}$which is soluble along with water. The reaction of Aluminium hydroxide with Sodium Hydroxide is given as:
\[Al{{(OH)}_{3}}+NaOH\to NaAl{{O}_{2}}+2{{H}_{2}}O\], here Sodium Aluminate formed is soluble.

When Zinc hydroxide (\[Zn{{(OH)}_{2}}\]) reacts with NaOH, they both react to give $N{{a}_{2}}Zn{{O}_{2}}$that is Sodium Zincate and it is soluble. It’s formed along with two molecules of water. The reaction of the same is given as:
$Zn{{(OH)}_{2}}+NaOH\to N{{a}_{2}}Zn{{O}_{2}}$, here sodium zincate is formed which is soluble.

When Tin hydroxide or Stannous hydroxide (\[Sn{{(OH)}_{2}}\]) reacts with NaOH, it will react in a way to give out Sodium Stannite which is soluble and there is a formation of 2 molecules of water along with it. The reaction of the same is given as:
$Sn{{(OH)}_{2}}+NaOH\to N{{a}_{2}}Sn{{O}_{2}}+2{{H}_{2}}O$, here Sodium Stannite is formed which is soluble in water.
Therefore, \[Al{{(OH)}_{3}}\]\[Zn{{(OH)}_{2}}\]\[Sn{{(OH)}_{2}}\] are soluble in excess NaOH.
Therefore, the correct options are \[Al{{(OH)}_{3}}\]\[Zn{{(OH)}_{2}}\]\[Sn{{(OH)}_{2}}\]

Note: Since, it is asked to find out the compounds which are soluble in excess of NaOH. We should know that NaOH is a polar molecule so we know that like dissolves like and hence only polar molecules would be soluble in excess of NaOH. Another thing to keep in mind is to find the most acidic compound among the following because NaOH is itself a very strong basic compound.