What volume of 0.89 molar NaI solution is necessary to precipitate all the Hg²⁺ from 51 mL of a 0.65 molar Hg(NO₃)₂ solution as HgI₂?

Question

Quizplus · Accepted Answer

The balanced equation for the reaction is: Hg(NO₃)₂ + 2 NaI → HgI₂ + 2 NaNO3 From the equation, it can be seen that 1 mole of Hg(NO₃)₂ reacts with 2 moles of NaI. Number of moles of Hg(NO₃)₂ in 51 mL of 0.65 M solution = 0.033 mol Number of moles of NaI required for complete reaction = 2 × 0.033 mol = 0.066 mol Concentration of NaI solution = 0.89 M Volume of NaI solution required = (0.066 mol) / (0.89 mol/L) = 0.074 L or 74 mL Therefore, the answer is B.

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