When 50.0 mL of a 0.010 M CuNO3 solution is mixed with 150.0 mL of a 0.010 M NaCl solution, a precipitate forms (Ksp = 1.9 x 10^-7).
a) Determine all final ion concentrations
b) Determine the mass of the precipitate formed.
How do i do this?
To determine the final ion concentrations and the mass of the precipitate formed, you need to understand the concept of solubility product constant and how it relates to the equilibrium of a solution.
1. Calculate the moles of CuNO3 and NaCl:
- Moles of CuNO3 = Volume of CuNO3 solution (in L) × Concentration of CuNO3 (in M)
- Moles of NaCl = Volume of NaCl solution (in L) × Concentration of NaCl (in M)
In this case:
- Moles of CuNO3 = 0.050 L × 0.010 M = 0.0005 mol
- Moles of NaCl = 0.150 L × 0.010 M = 0.0015 mol
2. Determine the ions present in the final solution:
- CuNO3 dissociates into Cu²⁺ and NO₃⁻ ions.
- NaCl dissociates into Na⁺ and Cl⁻ ions.
3. Use the stoichiometry of the reaction to determine the final concentrations of these ions:
- Since the reaction is between Cu²⁺ and Cl⁻ ions, upon mixing, they will react to form a precipitate of CuCl₂.
- The stoichiometric ratio of Cu²⁺ to Cl⁻ is 1:2.
- Final concentration of Cu²⁺: Initially, there were 0.0005 mol of Cu²⁺ from CuNO3. Assuming all of it reacts, the final concentration of Cu²⁺ would be (0.0005 mol / (0.050 L + 0.150 L)) = 0.002 M.
- Final concentration of Cl⁻: Initially, there were 0.0015 mol of Cl⁻ from NaCl. Assuming all of it reacts, the final concentration of Cl⁻ would be (2 × 0.0015 mol / (0.050 L + 0.150 L)) = 0.006 M.
4. To determine if a precipitate forms, compare the ion product (Cu²⁺ × Cl⁻) to the solubility product constant (Ksp = 1.9 x 10^-7):
- Ion product (IP) = Cu²⁺ × Cl⁻ = 0.002 M × 0.006 M = 1.2 x 10^-5
- If the ion product (IP) is greater than the solubility product constant (Ksp), a precipitate will form.
- In this case, IP (1.2 x 10^-5) is greater than Ksp (1.9 x 10^-7), so a precipitate will form.
5. Determine the mass of the precipitate formed:
- To calculate the mass of the precipitate (CuCl₂), you need to know the limiting reactant.
- In this case, the limiting reactant is Cu²⁺ since the stoichiometric ratio of Cu²⁺ to Cl⁻ is 1:2.
- The moles of CuCl₂ formed will be half the moles of Cu²⁺ since the stoichiometric ratio is 1:2.
- Moles of CuCl₂ = 0.0005 mol / 2 = 0.00025 mol
- To find the mass of the precipitate formed, you need to know the molar mass of CuCl₂. Assume it is 134.45 g/mol.
- Mass of precipitate formed = Moles of CuCl₂ × Molar mass of CuCl₂
- Mass of precipitate formed = 0.00025 mol × 134.45 g/mol = 0.0336 g (rounded to three significant figures)
Therefore, the final ion concentrations are 0.002 M for Cu²⁺ and 0.006 M for Cl⁻. The mass of the precipitate formed is approximately 0.0336 grams.