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(B) First,calculate the final concentrations of the ions in the mixture (total volume = $300 \ mL + 100 \ mL = 400 \ mL$):$[Pb^{2+}] = \frac{300 \ mL

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$Q > K_{sp}$

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(B) First,calculate the final concentrations of the ions in the mixture (total volume = $300 \ mL + 100 \ mL = 400 \ mL$):$[Pb^{2+}] = \frac{300 \ mL 	imes 0.134 \ M}{400 \ mL} = 0.1005 \ M$$[Cl^-] = \frac{100 \ mL 	imes 0.4 \ M}{400 \ mL} = 0.1 \ M$Next,calculate the reaction quotient $Q$ for the dissociation $PbCl_{2(s)} ightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$:$Q = [Pb^{2+}][Cl^-]^2$$Q = (0.1005) 	imes (0.1)^2$$Q = 0.1005 	imes 0.01 = 1.005 	imes 10^{-3}$Comparing $Q$ with $K_{sp}$ $(1.6 	imes 10^{-5})$:$1.005 	imes 10^{-3} > 1.6 	imes 10^{-5}$Therefore,$Q > K_{sp}$.

The $K_{sp}$ for the following dissociation is $1.6 \times 10^{-5}$:
$PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$
Which of the following choices is correct for a mixture of $300 \ mL$ of $0.134 \ M \ Pb(NO_3)_2$ and $100 \ mL$ of $0.4 \ M \ NaCl$?

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The $K_{sp}$ for the following dissociation is $1.6 \times 10^{-5}$:$PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$Which of the following choices is correct for a mixture of $300 \ mL$ of $0.134 \ M \ Pb(NO_3)_2$ and $100 \ mL$ of $0.4 \ M \ NaCl$?