Will a precipitate of CaSO4 form if 40.0mL of 2.0 x 10^3M CaCl2 is mixed with 60.0mL of 3.0 x 10^-2 Na2SO4?
You want to calculate Qsp and compare with Ksp.
..........CaSO4 ==> Ca^2+ + SO4^2-
Qsp has the same format as Ksp; i.e., Qsp = (Ca^2+)(SO4^2-). Substitute the following.
(Ca^2+) = 2E3M x 40/100 = ? (Note: I wonder if you meant 2.0E-3???
(SO4^2-)= 3.0E-2 x 60/100 = ?
Calculate Qsp. If Qsp > Ksp, a ppt will form. If Qsp < Ksp, a ppt will not form. You will need to look up the Ksp value in your text for CaSO4.
To determine if a precipitate of CaSO4 will form when 40.0 mL of 2.0 x 10^3 M CaCl2 is mixed with 60.0 mL of 3.0 x 10^-2 Na2SO4, we need to consider if a reaction will occur between the two solutions and if the resulting product is insoluble (and thus forms a precipitate).
First, we need to write out the balanced chemical equation for the reaction between CaCl2 and Na2SO4:
CaCl2 + Na2SO4 → CaSO4 + 2 NaCl
Next, we need to determine the solubility of CaSO4. By referring to a solubility table or doing some research, we can find that CaSO4 is sparingly soluble in water.
Based on the solubility rules, we know that sulfates (SO4^2-) are generally soluble except when they are paired with certain cations, including calcium (Ca^2+). Therefore, CaSO4 is expected to be insoluble and form a precipitate.
Finally, to confirm whether a precipitate will indeed form, we need to calculate the concentrations of Ca^2+ and SO4^2- ions in the resulting solution after mixing the two solutions.
1. Calculate the moles of CaCl2:
Moles of CaCl2 = volume (in L) × concentration (in mol/L)
Moles of CaCl2 = 40.0 mL × (1 L / 1000 mL) × 2.0 x 10^3 mol/L
Moles of CaCl2 = 0.080 mol
2. Calculate the moles of Na2SO4:
Moles of Na2SO4 = volume (in L) × concentration (in mol/L)
Moles of Na2SO4 = 60.0 mL × (1 L / 1000 mL) × 3.0 x 10^-2 mol/L
Moles of Na2SO4 = 0.0018 mol
3. Determine the limiting reactant by comparing the mole ratios from the balanced chemical equation:
From the balanced equation, we can see that the molar ratio between CaCl2 and CaSO4 is 1:1. Therefore, the limiting reactant is CaCl2 since it has a lower number of moles.
4. Based on the limiting reactant, calculate the maximum number of moles of CaSO4 that can form:
Moles of CaSO4 = Moles of CaCl2
Moles of CaSO4 = 0.080 mol
5. Calculate the concentration (M) of CaSO4 in the final solution:
Volume of final solution = Volume of CaCl2 + Volume of Na2SO4
Volume of final solution = 40.0 mL + 60.0 mL = 100.0 mL = 0.100 L
Concentration of CaSO4 = moles of CaSO4 / volume of final solution
Concentration of CaSO4 = 0.080 mol / 0.100 L
Concentration of CaSO4 = 0.80 M
Since the resulting concentration of CaSO4 exceeds its solubility (0.80 M > sparingly soluble), a precipitate of CaSO4 will form when 40.0 mL of 2.0 x 10^3 M CaCl2 is mixed with 60.0 mL of 3.0 x 10^-2 Na2SO4.