MCQ
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
    $1.2\, \times \,{10^{ - 10}}\,g$
  • $1.2\, \times \,{10^{ - 9}}\,g$
  • C
    $6.2\, \times \,{10^{ - 5}}\,g$
  • D
    $5.0\, \times \,{10^{ - 8}}\,g$

Answer

Correct option: B.
$1.2\, \times \,{10^{ - 9}}\,g$
b
$5.0 \times 10^{-13}=\left[\mathrm{Br}^{-}\right][0.05]$

$\left[\mathrm{Br}^{-}\right]=10^{-11}\, \mathrm{M}$

Amount of $\mathrm{AgBr}=10^{-11} \times 120 \,\mathrm{g}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free