Question

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

Answer 1

To calculate the final concentration of mercury (Hg) in the solution, we need to determine the total amount of mercury from both mercury chloride (HgCl2) and mercury sulfate (HgSO4), and then express this concentration in molarity (M) and percent weight/volume (% w/v).

Step 1: Calculate the moles of Mercury from HgCl2 and HgSO4

1. Mercury Chloride (HgCl2)

1. Molar Mass of HgCl2:

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

2. Moles of HgCl2: \[ \text{Moles of HgCl2} = \frac{2.45 , \text{g}}{271.49 , \text{g/mol}} \approx 0.00903 , \text{mol} \]

3. Moles of Mercury from HgCl2: Since every mole of HgCl2 gives one mole of Hg: \[ \text{Moles of Hg from HgCl2} = 0.00903 , \text{mol} \]

2. Mercury Sulfate (HgSO4)

1. Molar Mass of HgSO4:

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

2. Moles of HgSO4: \[ \text{Moles of HgSO4} = \frac{4.68 , \text{g}}{296.66 , \text{g/mol}} \approx 0.01575 , \text{mol} \]

3. Moles of Mercury from HgSO4: Since every mole of HgSO4 gives one mole of Hg: \[ \text{Moles of Hg from HgSO4} = 0.01575 , \text{mol} \]

Step 2: Total Moles of Mercury

Add the moles of mercury obtained from both compounds: \[ \text{Total moles of Hg} = 0.00903 , \text{mol} + 0.01575 , \text{mol} = 0.02478 , \text{mol} \]

Step 3: Calculate the Concentration in Molarity (M)

Molarity (M) is defined as moles of solute per liter of solution. The final volume of the solution is 500 mL, which is 0.500 L.

\[ \text{Concentration (M)} = \frac{\text{Total moles of Hg}}{\text{Volume in L}} = \frac{0.02478 , \text{mol}}{0.500 , \text{L}} = 0.04956 , \text{M} \]

Step 4: Calculate the Concentration in % (w/v)

Percent weight/volume (% w/v) is calculated using the formula: \[ \text{% (w/v)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 \]

The total mass of mercury can be calculated by using the molar mass of mercury (Hg = 200.59 g/mol): \[ \text{Mass of Hg} = \text{Total moles of Hg} \times \text{molar mass of Hg} = 0.02478 , \text{mol} \times 200.59 , \text{g/mol} \approx 4.973 , \text{g} \]

Now, calculating the % (w/v): \[ \text{% (w/v)} = \frac{4.973 , \text{g}}{500 , \text{mL}} \times 100 = 0.9946 , % \]

Final Results

1. Concentration in Molarity (M): \[ 0.04956 , \text{M} \quad (\text{to 4 significant figures}) \]

2. Concentration in % (w/v): \[ 0.995 % \quad (\text{to 3 significant figures}) \]

Thus, the final concentration of mercury (Hg) in the solution is approximately 0.04956 M and 0.995% (w/v).