Question

A 500 mL solution of mercury (Hg) solution was prepared by mixing 2.45 g of mercury chloride (HgCl2) and 4.68 g of mercury sulfate (HgSO4) in one container then dissolve and complete to volume with distilled water. What is the final concentration Mercury (Hg) in final solution? Express concentration in Molarity to 4 significant figures and %(w/v) to 3 significant figures.

Answer 1

To calculate the concentration of mercury (Hg) in the final solution, we first need to convert the masses of the mercury compounds (HgCl2 and HgSO4) into moles of mercury (Hg) and then find the final concentration in molarity (M) and weight/volume percentage (% w/v).

Step 1: Calculate moles of mercury from each compound

Molar mass of compounds:

1. Mercury Chloride (HgCl2):

   • Hg = 200.59 g/mol
   • Cl = 35.45 g/mol (2 Cl atoms)
   • Molar mass of HgCl2 = 200.59 + (2 × 35.45) = 200.59 + 70.90 = 271.49 g/mol

2. Mercury Sulfate (HgSO4):

   • Hg = 200.59 g/mol
   • S = 32.07 g/mol
   • O = 16.00 g/mol (4 O atoms)
   • Molar mass of HgSO4 = 200.59 + 32.07 + (4 × 16.00)
   • = 200.59 + 32.07 + 64.00 = 296.66 g/mol

Step 2: Calculate moles of Hg from each component

Moles of Hg from HgCl2:

\[ \text{Moles of HgCl2} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.45 \text{ g}}{271.49 \text{ g/mol}} \approx 0.00903 \text{ moles} \]

There is 1 mole of Hg in 1 mole of HgCl2: \[ \text{Moles of Hg from HgCl2} = 0.00903 \]

Moles of Hg from HgSO4:

\[ \text{Moles of HgSO4} = \frac{4.68 \text{ g}}{296.66 \text{ g/mol}} \approx 0.01577 \text{ moles} \]

There is also 1 mole of Hg in 1 mole of HgSO4: \[ \text{Moles of Hg from HgSO4} = 0.01577 \]

Step 3: Total moles of mercury (Hg)

\[ \text{Total moles of Hg} = 0.00903 + 0.01577 \approx 0.02480 \text{ moles} \]

Step 4: Calculate Molarity of Hg in the solution

The final volume is 500 mL or 0.500 L. Molarity (M) is defined as: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.02480 \text{ moles}}{0.500 \text{ L}} = 0.04960 \text{ M} \]

Step 5: Calculate % (w/v) concentration

The weight/volume percentage (%) is calculated as: \[ % (w/v) = \left( \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \right) \times 100 \]

The total mass of Hg: \[ \text{Total mass of Hg} = (0.00903 \text{ moles} + 0.01577 \text{ moles}) \times 200.59 \text{ g/mol} = 0.02480 \text{ moles} \times 200.59 \text{ g/mol} \approx 4.97358 \text{ g} \]

Now, calculate % (w/v): \[ % (w/v) = \left( \frac{4.97358 \text{ g}}{500 \text{ mL}} \right) \times 100 \approx 0.994716% \approx 0.995 % \]

Final Results

• Molarity of mercury (Hg) = 0.04960 M (to 4 significant figures)
• % (w/v) = 0.995% (to 3 significant figures)