The required amount of KBr (molar mass = 119 | vedclass

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(C) The precipitation of $AgBr$ starts when the ionic product exceeds the solubility product $(K_{SP})$.$AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq

Vedclass · Accepted Answer

$5.95 	imes 10^{-10} \ g$

Vedclass · Answer

(C) The precipitation of $AgBr$ starts when the ionic product exceeds the solubility product $(K_{SP})$.$AgBr(s) ightleftharpoons Ag^+(aq) + Br^-(aq)$$K_{SP} = [Ag^+][Br^-]$Given $[Ag^+] = 0.05 \ M = 5 	imes 10^{-2} \ M$.$5 	imes 10^{-13} = (5 	imes 10^{-2}) 	imes [Br^-]$$[Br^-] = \frac{5 	imes 10^{-13}}{5 	imes 10^{-2}} = 10^{-11} \ M$.Number of moles of $Br^-$ required $= 	ext{Molarity} 	imes 	ext{Volume in Liters} = 10^{-11} \ mol/L 	imes 0.5 \ L = 0.5 	imes 10^{-11} \ mol$.Since $KBr$ dissociates as $KBr ightarrow K^+ + Br^-$,moles of $KBr$ required $= 0.5 	imes 10^{-11} \ mol$.Mass of $KBr = 	ext{moles} 	imes 	ext{molar mass} = 0.5 	imes 10^{-11} \ mol 	imes 119 \ g/mol = 59.5 	imes 10^{-11} \ g = 5.95 	imes 10^{-10} \ g$.

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

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