Search any question & find its solution
Question:
Answered & Verified by Expert
The solubility product of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ is $32 \times 10^{-}$ 12. What is the concentration of $\mathrm{CrO}_{4}^{-}$ions in that solution?
Options:
Solution:
1176 Upvotes
Verified Answer
The correct answer is:
$2 \times 10^{-4} \mathrm{M}$
$\mathrm{Ag}_{2} \mathrm{CrO}_{4} \longrightarrow 2 \mathrm{Ag}^{+}+\mathrm{Cr} \mathrm{O}_{4}^{2-}$
$\mathrm{S}$
$\mathrm{K}_{\mathrm{SP}}=(2 \mathrm{~s})^{2} \mathrm{~s}=4 \mathrm{~s}^{2} \cdot \mathrm{s}=4 \mathrm{~s}^{3}$
$\mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{Sp}}}{4}\right)^{1 / 3} \Rightarrow\left(\frac{32 \times 10^{-12}}{4}\right)^{1 / 3}$
$=2 \times 10^{-4} \mathrm{M}$
$\mathrm{S}$
$\mathrm{K}_{\mathrm{SP}}=(2 \mathrm{~s})^{2} \mathrm{~s}=4 \mathrm{~s}^{2} \cdot \mathrm{s}=4 \mathrm{~s}^{3}$
$\mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{Sp}}}{4}\right)^{1 / 3} \Rightarrow\left(\frac{32 \times 10^{-12}}{4}\right)^{1 / 3}$
$=2 \times 10^{-4} \mathrm{M}$
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.