To find the final concentration of mercury in the solution, we will first need to determine the amount of mercury contributed by both mercury chloride (HgCl₂) and mercury sulfate (HgSO₄) in terms of moles, then convert into molarity and percent concentration.
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Calculate the Molar Masses:
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Mercury Chloride (HgCl₂):
- Molar mass of Hg = 200.59 g/mol
- Molar mass of Cl = 35.45 g/mol (2 Cl)
- Molar mass of HgCl₂ = 200.59 g/mol + 2 × 35.45 g/mol = 200.59 + 70.90 = 271.49 g/mol
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Mercury Sulfate (HgSO₄):
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol (4 O)
- Molar mass of HgSO₄ = 200.59 g/mol + 32.07 g/mol + 4 × 16.00 g/mol = 200.59 + 32.07 + 64.00 = 296.66 g/mol
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Calculate the moles of each compound:
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Moles of HgCl₂:
Each mole of HgCl₂ provides 1 mole of Hg, so moles of Hg from HgCl₂ = 0.00903 mol. -
Moles of HgSO₄:
Each mole of HgSO₄ provides 1 mole of Hg, so moles of Hg from HgSO₄ = 0.01575 mol.
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Total moles of mercury:
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Calculate the final concentration in Molarity: The final volume of the solution is 50 mL, which is 0.050 L.
Rounded to 4 significant figures:
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Calculate % (w/v):
- % (w/v) is calculated as:
The total mass of mercury from both sources: - From HgCl₂:
- Mass of Hg from HgCl₂ =
- Mass of Hg from HgCl₂ =
- From HgSO₄:
- Mass of Hg from HgSO₄ =
- Mass of Hg from HgSO₄ =
The total mass of Mercury is:
Now calculate the % (w/v):
Rounded to 3 significant figures:
- % (w/v) is calculated as:
Final Results:
- Concentration in Molarity: 0.4956 M
- Concentration in % (w/v): 9.94%