Animated Solution for Chemistry - Ionic Equilibrium: The molar solubility of ${\text{Zn(OH)}}_2$ in $0.1\text{ M NaOH}$ solution is $x\times 10^{-18}\text{ M}$. The value of $x$ is ......... (Nearest integer) (Given; The solubility product of ${\text{Zn(OH)}}_2$ is $2\times 10^{-20}$).

Unraveling the Solubility of Zinc Hydroxide

Imagine a beaker where solid zinc hydroxide, ${\text{Zn(OH)}}_2$, is trying to dissolve. It sets up a delicate equilibrium, releasing zinc ions and hydroxide ions into the solution. We can represent this process with the equation:

Let's call its molar solubility $S$. This means for every mole of ${\text{Zn(OH)}}_2$ that dissolves, it produces $S$ moles of ${\text{Zn}}^{2+}$ and $2S$ moles of ${\text{OH}}^{-}$.

But wait, our beaker isn't just filled with pure water. It already contains a $0.1\text{ M}$ solution of sodium hydroxide ($\text{NaOH}$). Sodium hydroxide is a strong base, which means it completely dissociates into sodium ions (${\text{Na}}^{+}$) and hydroxide ions (${\text{OH}}^{-}$).

This floods the solution with hydroxide ions, creating a classic scenario known as the Common Ion Effect. According to Le Chatelier's principle, this excess of ${\text{OH}}^{-}$ ions will push the equilibrium of the sparingly soluble ${\text{Zn(OH)}}_2$ heavily to the left, drastically reducing its solubility.

Now, what is the total concentration of hydroxide ions in our beaker? It's the sum of the ions from both sources:

Here is where we make a crucial, time-saving approximation. Because ${\text{Zn(OH)}}_2$ is sparingly soluble (its $K_{sp}$ is a tiny $2\times 10^{-20}$), the value of $S$ is incredibly small. Therefore, $2S$ is practically negligible compared to $0.1$. We can safely approximate:

Don't make the silly mistake of keeping the $2S$ in your calculations; it will only lead to a complex quadratic or cubic equation that isn't necessary!

Let's bring in the solubility product constant, $K_{sp}$. The formula is the concentration of zinc ions times the square of the hydroxide ion concentration:

Substituting our known values and our approximated hydroxide concentration, we get:

Now, it's just simple algebra. $(0.1{)}^2$ is $0.01$, or $10^{-2}$. Dividing the $K_{sp}$ by this gives us the solubility $S$:

The problem states that the solubility is $x\times 10^{-18}\text{ M}$. Comparing this with our calculated value, we can clearly see that the value of $x$ is exactly 2.