Required amount of KBr (molar mass = 119 ) in gram to start precipitation of AgBr in 500, mL solutio

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6-2.Equilibrium-II (Ionic Equilibrium)

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The required amount of $KBr$ (molar mass $= 119$) in gram to start the precipitation of $AgBr$ in $500\, mL$ solution of $0.05\, M\, AgNO_3$ will be ($K_{SP}$ of $AgBr = 5 \times 10^{-13}$)

A

$1.19 \times {10^{ - 9}}\,g$

B

$4 \times {10^{ - 11}}\,g$

C

$5.95 \times {10^{ - 10}}\,g$

D

$2.97 \times {10^{ - 10}}\,g$

Solution

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Standard 11

Chemistry

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