Questions: Question 4 2 Points Calculate the volume of 2.09 M Nal that would be needed to precipitate all of the Hg^+2 ion from 272 mL of a 2.02 M Hg(NO3)2. The equation for the reaction is 2 Nal(aq) + Hg^2(NO3)2(aq) -> Hgl2(s) + 2 NaNO3(aq)

Solution Steps

First, calculate the moles of \(\mathrm{Hg}^{2+}\) ions present in the solution. Use the formula:

\[ \text{moles of } \mathrm{Hg}^{2+} = \text{Molarity} \times \text{Volume} \]

Given:

• Molarity of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) = 2.02 M
• Volume of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) = 272 mL = 0.272 L

\[ \text{moles of } \mathrm{Hg}^{2+} = 2.02 \, \text{M} \times 0.272 \, \text{L} = 0.5494 \, \text{moles} \]

From the balanced chemical equation:

\[ 2 \mathrm{NaI} + \mathrm{Hg}^{2+} \rightarrow \mathrm{HgI}_{2} + 2 \mathrm{NaNO}_{3} \]

It is clear that 2 moles of \(\mathrm{NaI}\) are required for every mole of \(\mathrm{Hg}^{2+}\). Therefore, the moles of \(\mathrm{NaI}\) needed are:

\[ \text{moles of } \mathrm{NaI} = 2 \times 0.5494 = 1.0988 \, \text{moles} \]

Now, calculate the volume of 2.09 M \(\mathrm{NaI}\) solution needed to provide 1.0988 moles of \(\mathrm{NaI}\). Use the formula:

\[ \text{Volume} = \frac{\text{moles}}{\text{Molarity}} \]

\[ \text{Volume of } \mathrm{NaI} = \frac{1.0988 \, \text{moles}}{2.09 \, \text{M}} = 0.5258 \, \text{L} \]

Convert the volume from liters to milliliters:

\[ 0.5258 \, \text{L} = 525.8 \, \text{mL} \]

Final Answer