Search any question & find its solution
Question:
Answered & Verified by Expert
The solubility of $\mathrm{CuBr}$ is $2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$ at $25^{\circ} \mathrm{C}$. The $K_{s p}$ value for $\mathrm{CuBr}$ is
Options:
Solution:
2235 Upvotes
Verified Answer
The correct answer is:
$4 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2}$
$\mathrm{CuBr} \rightarrow \mathrm{Cu}^{+}+\mathrm{Br}^{-}$
Solubility of $\mathrm{CuBr}$ is $2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
Therefore, solubility of $\mathrm{Cu}^{+}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
$\Rightarrow$ Solubility product $=2 \times 10^{-4} \times 2 \times 10^{-4}$
$=4 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2}$
Solubility of $\mathrm{CuBr}$ is $2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
Therefore, solubility of $\mathrm{Cu}^{+}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
$\Rightarrow$ Solubility product $=2 \times 10^{-4} \times 2 \times 10^{-4}$
$=4 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.