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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) 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 $[Ag^+] = 0.05 \, M$ and $K_{sp} = 5 	imes 10^{-13}$.$[Br^-] = \frac{K_{sp}}{[Ag^+]} = \frac{5 	imes 10^{-13}}{0.05} = 10^{-11} \, M$.Number of moles of $Br^-$ required $= Molarity 	imes Volume(L) = 10^{-11} \, mol/L 	imes 0.5 \, L = 0.5 	imes 10^{-11} \, mol$.Since $KBr$ dissociates as $KBr ightarrow K^+ + Br^-$,moles of $KBr$ required $= 0.5 	imes 10^{-11} \, mol$.Mass of $KBr = moles 	imes molar \, mass = 0.5 	imes 10^{-11} 	imes 119 = 59.5 	imes 10^{-11} = 5.95 	imes 10^{-10} \, g$.

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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