Calculate ΔG at 25°C for the precipitation of lead(II) iodide from mixing 100. mL of 0.10 M lead(II) nitrate solution with 100. mL of 0.10 M sodium iodide solution. The ΔG° for this precipitation reaction at 25°C = −46.2 kJ/mol.

Answer 1

dG = dGo + RTlnQ
Write the equation. Determine the limiting reagent. Q = (NaNO3)^2/[Pb(NO2)]
Determine the concentrations for Q and calculate. Post your work if you get stuck.