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6-2.Equilibrium-II (Ionic Equilibrium)
easy
Solubility product of silver bromide is $5.0 \,\,\times \,\,10^{-13}$ . The quantity of potassium
bromide (molar mass taken as $120\,\,g\,mol^{-1}$ ) to be added to $1\,L$ of $0.05\,\,M$ solution of silver nitrate to start the precipitation of $AgBr$ is
A
$1.2\, \times \,{10^{ - 10}}\,g$
B
$1.2\, \times \,{10^{ - 9}}\,g$
C
$6.2\, \times \,{10^{ - 5}}\,g$
D
$5.0\, \times \,{10^{ - 8}}\,g$
Solution
$5.0 \times 10^{-13}=\left[\mathrm{Br}^{-}\right][0.05]$
$\left[\mathrm{Br}^{-}\right]=10^{-11}\, \mathrm{M}$
Amount of $\mathrm{AgBr}=10^{-11} \times 120 \,\mathrm{g}$
Standard 11
Chemistry
