Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# For $Cu{{(OH)}_{2}}$ , ${{K}_{sp}}=1.6\times {{10}^{-19}}$ ,the molar solubility of $Cu{{(OH)}_{2}}$ is:(A) $3.4\times {{10}^{-7}}M$ (B) $5.4\times {{10}^{-7}}M$ (C) $2.7\times {{10}^{-11}}M$ (D) $5.1\times {{10}^{-10}}M$

Last updated date: 22nd Jul 2024
Total views: 348k
Views today: 5.48k
Verified
348k+ views
Hint :We know that the Molar solubility is the property of solute to get dissolved in the solvent to form a solution. The solubility product is the measure of the number of solute dissolves. The equilibrium constant of the solubility is proportional to the concentration of ions of solute in a solution. Here by using the concept of solubility product constant. Write down the equation and then let the solubility of the aqueous part and put in the equation.

Solubility is defined as the property of a substance (solute) to get dissolved in a solvent in order to form a solution. The solubility product constant is the equilibrium constant for the dissolution of a solid solute in a solvent to form a solution. It is denoted by the symbol $Ksp$ . We can form an idea about the dissolution of an ionic compound by the value of its solubility product.
We are given that the ${{K}_{sp}}$ of $Cu{{(OH)}_{2}}$ is $1.6\times {{10}^{-19}}$ and we have to find the molar solubility of $Cu{{(OH)}_{2}}$ , so firstly we will write down the equation for $Cu{{(OH)}_{2}}$ ,
$Cu{{(OH)}_{2}}(s)\text{ }\rightleftharpoons \text{ C}{{\text{u}}^{2+}}(aq)\text{ + 2O}{{\text{H}}^{-}}(aq)$
After dissolving in a suitable solvent the ions will be produced as follows,
Now the next step will be using the formula of solubility product constant which will be as follows:
${{K}_{sp}}=\dfrac{[C{{u}^{2+}}(aq)]\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}(aq){{]}^{2}}}{[Cu{{(OH)}_{2}}(s)]}$
Now as we know that solid do not contribute in concentration so we will neglect $[Cu{{(OH)}_{2}}(s)]$ in our equation, here we have to note one thing that ${{K}_{sp}}$ is the solubility product constant so we will do one thing let us assume that the molar solubility be S so solubility of $[C{{u}^{2+}}(aq)]$ and $\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}(aq)]$ be S,
Now putting all things in equation,
${{K}_{sp}}=[C{{u}^{2+}}(aq)]\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}(aq)]$
On Substituting we get;
$1.6\times {{10}^{-19}}=\text{ S}\times {{\text{S}}^{2}}$
On the further solving we get;
$1.6\times {{10}^{-19}}={{S}^{3}}$
Now taking cube root both side our S will come,
$S=5.4\times {{10}^{-7}}mol\text{ }{{\text{L}}^{-1}}$
So the molar solubility of $Cu{{(OH)}_{2}}$ is $5.4\times {{10}^{-7}}mol\text{ }{{\text{L}}^{-1}}$
Therefore, the correct answer is Option B.

Note :
Remember that the molar solubility (S) is very low as compared to the concentration of the solution. Therefore neglect the term S for simplification. The solubility product is not the same as that of molar solubility. Solubility is the amount dissolved in solution but molar solubility is no solute dissolved per liter of solution. Also keep in mind that the solid concentration remains constant throughout the equilibrium. And keep note of the coefficients of the ions produced because they will come in powers which is necessary for finding the solubility.