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Chemistry Question on Equilibrium

Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compounds, would they use? Ksp(AgCl)=1.8×10−10K_{s p}( AgCl )=1.8 \times 10^{-10} Ksp(AgBr)=5.0×10−13K_{s p}( AgBr )=5.0 \times 10^{-13} Ksp(Ag2CrO4)=2.4×10−12K_{s p}\left( Ag _{2} CrO _{4}\right)=2.4 \times 10^{-12}

A

AgClAgCl

B

AgBrAgBr

C

Ag2CrO4A{{g}_{2}}Cr{{O}_{4}}

D

None of these

Answer

Ag2CrO4A{{g}_{2}}Cr{{O}_{4}}

Explanation

Solution

For binary salts (like AgCl,AgBrAgCl , AgBr )
s=Ksps =\sqrt{K_{s p}}
∴\therefore Solubility of AgCl=1.8×10−10AgCl =\sqrt{1.8 \times 10^{-10}}
=1.35×10−7 mol/L=1.35 \times 10^{-7} \,mol / L
Solubility of AgBr=5.0×10−13AgBr =\sqrt{5.0 \times 10^{-13}}
=7.1×10−7mol/L=7.1 \times 10^{-7} mol / L
For, Ag2CrO4,Ksp=4s3 Ag _{2} CrO _{4}, K_{s p} =4 s^{3}
∴\therefore Solubility of Ag2CrO4Ag _{2} CrO _{4}
=Ksp43=2.4×10−1243=\sqrt[3]{\frac{K_{s p}}{4}}=\sqrt[3]{\frac{2.4 \times 10^{-12}}{4}}
=600×10−153=\sqrt[3]{600 \times 10^{-15}}
=8.44×10−5mol/L=8.44 \times 10^{-5} mol / L
As Ag2CrO4Ag _{2} CrO _{4} has maximum solubility, it will give maximum Ag+Ag ^{+}ions in solution. Hence, it will be used.