a solution contains Cr3+ ion and Mg2+ ion. The addition of 1.00 L of 1.55 M NaF solution is required to cause the complete precipitation of these ions as CrF3(s) and MgF2(s). The total mass of the precipitate is 49.8 g.
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To find the number of moles of Cr3+ and Mg2+ ions in the original solution, we can use the equation Molarity (M) = Moles (mol) / Volume (L).
1. Calculate the number of moles of NaF in the solution:
Molarity (M) = 1.55 M
Volume (L) = 1.00 L
Moles of NaF = Molarity × Volume = 1.55 M × 1.00 L = 1.55 mol
Since NaF dissociates to produce one Na+ ion and one F- ion, the concentration of F- ions is also 1.55 M.
2. Calculate the number of moles of F- ions required to precipitate the Cr3+ and Mg2+ ions:
Using the balanced chemical equation for the formation of the precipitate:
3 NaF + CrCl3 → CrF3(s) + 3 NaCl
The stoichiometric ratio between NaF and CrF3 is 3:1. Therefore, 1.55 mol of F- ions will react with (1.55 / 3) mol of Cr3+ ions.
3. Calculate the number of moles of Cr3+ ions:
Moles of Cr3+ ions = (1.55 mol of F- ions) / (3 mol of F- ions per 1 mol of Cr3+ ions)
Moles of Cr3+ ions = 1.55 mol / 3 = 0.52 mol
4. Calculate the number of moles of Mg2+ ions:
Since MgF2 has a 1:2 stoichiometric ratio with respect to F- ions, the number of moles of Mg2+ ions is twice that of F- ions.
Moles of Mg2+ ions = 2 × (1.55 mol of F- ions) = 2.00 mol
5. Calculate the mass of the precipitate:
The molar mass of CrF3 is 101.99 g/mol.
The molar mass of MgF2 is 62.32 g/mol.
Mass of CrF3 = Moles of Cr3+ ions × Molar mass of CrF3 = 0.52 mol × 101.99 g/mol = 53.04 g
Mass of MgF2 = Moles of Mg2+ ions × Molar mass of MgF2 = 2.00 mol × 62.32 g/mol = 124.64 g
Total mass of precipitate = Mass of CrF3 + Mass of MgF2 = 53.04 g + 124.64 g = 177.68 g
Since the given total mass of precipitate is 49.8 g, there seems to be an error in the provided information or calculations need to be reviewed.
To solve this problem, we need to determine the number of moles of Cr3+ and Mg2+ ions in the solution and then calculate the moles of fluoride ions required to completely precipitate these ions. Finally, we can calculate the mass of the precipitate formed.
1. Calculate the moles of Cr3+ ions in the solution:
We know that the concentration of the NaF solution is 1.55 M, which means it contains 1.55 moles of fluoride ions per liter of solution. Since the stoichiometric ratio between CrF3 and Cr3+ ions is 1:3, we can calculate the moles of Cr3+ ions as follows:
Moles of Cr3+ = 1.55 M (concentration of NaF) × 1.00 L (volume of NaF solution) × 3 (stoichiometric ratio)
Moles of Cr3+ = 4.65 moles
2. Calculate the moles of Mg2+ ions in the solution:
The stoichiometric ratio between MgF2 and Mg2+ ions is 1:1, so the moles of Mg2+ ions are equal to the moles of MgF2 precipitated. Since the total mass of the precipitate is given as 49.8 g, we can calculate the moles of Mg2+ ions:
Moles of Mg2+ = Mass of precipitate (g) / molar mass of MgF2 (g/mol)
Moles of Mg2+ = 49.8 g / (24.31 g/mol + 2 * 19.00 g/mol)
Moles of Mg2+ = 49.8 g / 62.31 g/mol
Moles of Mg2+ = 0.797 moles
3. Calculate the moles of fluoride ions required:
We already know the stoichiometric ratio between Cr3+ and fluoride ions is 1:3. Since we have 4.65 moles of Cr3+ ions, the moles of fluoride ions required can be calculated as follows:
Moles of fluoride ions = Moles of Cr3+ × 3
Moles of fluoride ions = 4.65 moles × 3
Moles of fluoride ions = 13.95 moles
4. Calculate the molarity of the NaF solution:
We have the moles of fluoride ions (13.95 moles) and the volume of the NaF solution (1.00 L). The molarity is calculated as moles/volume:
Molarity of NaF = Moles of fluoride ions / Volume of NaF solution
Molarity of NaF = 13.95 moles / 1.00 L
Molarity of NaF = 13.95 M
So, the molarity of the NaF solution required to completely precipitate the Cr3+ and Mg2+ ions is 13.95 M.