Predict whether a precipitate of $PbI_2$ will be formed or not on mixing $20 \ mL$ of $3 \times 10^{-3} \ M$ $Pb(NO_3)_2$ solution with $80 \ mL$ of $2 \times 10^{-3} \ M$ $NaI$ solution. $K_{sp}$ for lead iodide $(PbI_2)$ is $6.0 \times 10^{-9}$.
(D) Step $1$: Calculate the total volume of the mixture: $V_{total} = 20 \ mL + 80 \ mL = 100 \ mL = 0.1 \ L$.
Step $2$: Calculate the concentration of $Pb^{2+}$ ions: $[Pb^{2+}] = \frac{20 \ mL \times 3 \times 10^{-3} \ M}{100 \ mL} = 6 \times 10^{-4} \ M$.
Step $3$: Calculate the concentration of $I^-$ ions: $[I^-] = \frac{80 \ mL \times 2 \times 10^{-3} \ M}{100 \ mL} = 1.6 \times 10^{-3} \ M$.
Step $4$: Calculate the ionic product $(Q_{sp})$ for $PbI_2$: $Q_{sp} = [Pb^{2+}][I^-]^2 = (6 \times 10^{-4}) \times (1.6 \times 10^{-3})^2 = 6 \times 10^{-4} \times 2.56 \times 10^{-6} = 1.536 \times 10^{-9}$.
Step $5$: Compare $Q_{sp}$ with $K_{sp}$: Since $Q_{sp} (1.536 \times 10^{-9}) < K_{sp} (6.0 \times 10^{-9})$,the solution is unsaturated and no precipitate of $PbI_2$ will be formed.
Step $2$: Calculate the concentration of $Pb^{2+}$ ions: $[Pb^{2+}] = \frac{20 \ mL \times 3 \times 10^{-3} \ M}{100 \ mL} = 6 \times 10^{-4} \ M$.
Step $3$: Calculate the concentration of $I^-$ ions: $[I^-] = \frac{80 \ mL \times 2 \times 10^{-3} \ M}{100 \ mL} = 1.6 \times 10^{-3} \ M$.
Step $4$: Calculate the ionic product $(Q_{sp})$ for $PbI_2$: $Q_{sp} = [Pb^{2+}][I^-]^2 = (6 \times 10^{-4}) \times (1.6 \times 10^{-3})^2 = 6 \times 10^{-4} \times 2.56 \times 10^{-6} = 1.536 \times 10^{-9}$.
Step $5$: Compare $Q_{sp}$ with $K_{sp}$: Since $Q_{sp} (1.536 \times 10^{-9}) < K_{sp} (6.0 \times 10^{-9})$,the solution is unsaturated and no precipitate of $PbI_2$ will be formed.
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