A clumsy chemist accidentally pours 100 mL of a 0.004 M Na2SO4 solution into a beaker containing 500 mL of 0.050 M CaCl2 solution. Will a precipitate be formed?
To determine if a precipitate will be formed when the two solutions are mixed, we need to consider if a reaction will occur between the ions in the solutions that results in the formation of an insoluble compound.
1. Write the balanced equation for the possible reaction between Na2SO4 and CaCl2:
Na2SO4 + CaCl2 → 2NaCl + CaSO4
2. Calculate the number of moles of Na2SO4:
Molarity (M) = moles (mol) / volume (L)
0.004 M = moles Na2SO4 / 0.1 L
Moles Na2SO4 = 0.004 M * 0.1 L = 0.0004 mol
3. Calculate the number of moles of CaCl2:
Molarity (M) = moles (mol) / volume (L)
0.050 M = moles CaCl2 / 0.5 L
Moles CaCl2 = 0.050 M * 0.5 L = 0.025 mol
4. Determine the limiting reactant, which is the reactant that will be completely consumed in the reaction. We compare the moles of Na2SO4 and CaCl2 to see which one is less.
The moles of Na2SO4 (0.0004 mol) is much less than the moles of CaCl2 (0.025 mol).
5. Since we have excess CaCl2, only a portion of Na2SO4 will react to form a precipitate. Let's determine the amount of CaSO4 that can potentially form:
From the balanced equation, we know that for every 1 mole of Na2SO4, 1 mole of CaSO4 is formed.
The moles of CaSO4 formed = 0.0004 mol
6. Calculate the final volume of the solution after mixing:
The final volume is the sum of the volumes of the two solutions: 100 mL + 500 mL = 600 mL = 0.6 L
7. Calculate the final concentration of Na2SO4 and CaCl2 in the mixed solution:
Final concentration of Na2SO4 = moles of Na2SO4 / final volume
= 0.0004 mol / 0.6 L
= 0.00067 M
Final concentration of CaCl2 = moles of CaCl2 / final volume
= 0.025 mol / 0.6 L
= 0.042 M
8. Compare the solubility product constants (Ksp) of NaCl and CaSO4:
The Ksp of NaCl is very high, indicating that it is highly soluble in water and will not form a precipitate.
The Ksp of CaSO4 is relatively low, suggesting that it is not very soluble in water and can potentially form a precipitate.
9. Compare the final concentrations of Na2SO4 (0.00067 M) and CaCl2 (0.042 M) in the mixed solution to their respective solubility product constants (Ksp):
The concentration of CaCl2 (0.042 M) is less than the solubility product constant of CaSO4, implying that a precipitate of CaSO4 may form.
Therefore, based on the calculations and solubility product constants, it is likely that a precipitat
To determine whether a precipitate will be formed when the two solutions are mixed, we need to compare the solubility of the resulting compound with the concentration of the ions in solution.
First, let's write out the balanced chemical equation for the reaction between Na2SO4 and CaCl2:
Na2SO4 + CaCl2 -> 2 NaCl + CaSO4
From the balanced equation, we see that calcium sulfate (CaSO4) is formed. The solubility of calcium sulfate in water at room temperature is relatively low.
To determine if a precipitate will form, we need to compare the solubility product constant (Ksp) of calcium sulfate with the concentrations of the ions in solution.
The solubility product constant (Ksp) expression for calcium sulfate (CaSO4) is:
Ksp = [Ca2+][SO4 2-]
Given that the concentration of CaCl2 is 0.050 M and the volume is 500 mL, we can calculate the concentration of Ca2+ ions:
[Ca2+] = concentration of Ca2+ = 0.050 M
Now, let's determine the concentration of SO4 2- ions in solution. Since Na2SO4 is a strong electrolyte, it dissociates completely in water:
Na2SO4 -> 2 Na+ + SO4 2-
The concentration of SO4 2- ions in the mixture will be half of the concentration of Na2SO4. Given that the concentration of Na2SO4 is 0.004 M and the volume is 100 mL, we can calculate the concentration of SO4 2- ions:
[SO4 2-] = (1/2) * (0.004 M) = 0.002 M
Now, we can substitute the values into the Ksp expression:
Ksp = (0.050 M) * (0.002 M) = 0.0001
If the calculated Ksp value is greater than the solubility product constant (Ksp) of calcium sulfate, a precipitate will form. If it is less, no precipitate will form.
The Ksp value for calcium sulfate is approximately 9.1 x 10^-6.
Since the calculated Ksp value (0.0001) is greater than the solubility product constant (9.1 x 10^-6) of calcium sulfate, a precipitate will likely form when the two solutions are mixed.
Na2SO4 + CaCl2 ==> 2NaCl + CaSO4
I looked up on the internet to find Ksp for CaSO4 to be 4.93E-5. You should have posted Ksp from your notes/text. Use the Ksp from your notes and not the one I found.
What is the molar concentration of each component for the final solution?
M CaCl2 = 0.050 x (500/600) = 0.0417
M Na2SO4 = 0.004 x (100/600) = 0.000667
Qsp CaSO4 = (Ca^2+)(SO4^2-) =(0.000667)(0.0417) = 2.78E-5
A ppt will occur IF Ksp is exceeded. Since 2.78E-5 < 4.93E-5, no ppt will occur.