1 times 10^{-5} S M AgNO_3 is added to 1 L | vedclass

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(C) The concentration of $Ag^{+}$ ions from $AgNO_3$ is $[Ag^{+}] = 10^{-5} \ M$.The concentration of $NO_3^{-}$ ions is $[NO_3^{-}] = 10^{-5} \ M$.Du

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) The concentration of $Ag^{+}$ ions from $AgNO_3$ is $[Ag^{+}] = 10^{-5} \ M$.The concentration of $NO_3^{-}$ ions is $[NO_3^{-}] = 10^{-5} \ M$.Due to the common ion effect,the concentration of $Br^{-}$ ions is determined by the solubility product constant: $[Br^{-}] = \frac{K_{sp}}{[Ag^{+}]} = \frac{4.9 	imes 10^{-13}}{10^{-5}} = 4.9 	imes 10^{-8} \ M$.The conductivity $\kappa$ is given by $\kappa = \sum C_i \lambda_i$.For $Ag^{+}$: $\kappa_{Ag^{+}} = 10^{-5} 	imes 6 	imes 10^{-3} = 6 	imes 10^{-8} \ S \ m^{-1}$.For $Br^{-}$: $\kappa_{Br^{-}} = 4.9 	imes 10^{-8} 	imes 8 	imes 10^{-3} = 39.2 	imes 10^{-11} \ S \ m^{-1} = 0.392 	imes 10^{-8} \ S \ m^{-1}$.For $NO_3^{-}$: $\kappa_{NO_3^{-}} = 10^{-5} 	imes 7 	imes 10^{-3} = 7 	imes 10^{-8} \ S \ m^{-1}$.Total conductivity $\kappa = (6 + 0.392 + 7) 	imes 10^{-8} \ S \ m^{-1} = 13.392 	imes 10^{-8} \ S \ m^{-1}$.Rounding to the nearest integer,we get $13 	imes 10^{-8} \ S \ m^{-1}$.

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