Question

K$_{sp}$ of Mg(OH)$_2$ is \(1\times10^{-12}\). 0.01 M MgCl$_2$ solution will precipitate when limiting pH is

• A. 8
• B. 9
• C. 10
• D. 12

Hint:

Use solubility product comparison with ionic product.

Answer 1

The Correct Option is C

Step 1: Ionic product \[ IP=[Mg^{2+}][OH^-]^2. \]
Step 2: Precipitation when \(IP>1\times10^{-12}\). Given \([Mg^{2+}]=0.01\). \[ [OH^-]^2>1\times10^{-10} \Rightarrow[OH^-]>1\times10^{-5}. \] 
Step 3: \[ pOH<5 \Rightarrow pH>9. \] Nearest option 10.