100. mL of 0.020 M CaCl2(aq) is added to 200. mL of 0.20 M K2SO4(aq). Calculate the equilibrium concentrations of Ca^2+ (aq) and SO4^2-(aq)
Setting up an ICE chart is a great move; however, I don't get the answers you have (perhaps I just can't red the spacing since the board doesn't recognize anything over one space). At any rate, the equation is right but the Ksp is wrong.
CaCl2 + K2SO4 ==> CaSO4
Since I can't use spaces, I do it this way but you can transfer it to paper and make it look the normal ICE chart way.
CaCl2 + K2SO4 ==> CaSO4 + 2KCl.
initial:
moles CaCl2 = M x L = 0.02M x 0.1 L = 0.002 moles.
moles K2SO4 = 0.2M x 0.200 L = 0.04 moles.
CaSO4 = 0
change: obviously, the K2SO4 is in excess and CaCl2 is the limiting reagent.
CaCl2 = -0.002
K2SO4 = -0.002
CaSO4 = +0.002
equilibrium:
CaCl2 = 0
K2SO4 = 0.04-0.002 = 0.038
CaSO4 = 0+0.002 = 0.002
So we form 0.002 moles CaSO4 which will ppt (if Ksp is exceeded) and the Ca^+2 is determined by the solubility of CaSO4. What is the K2SO4 concn? It is 0.038 moles/0.300 L = 0.127 M
Set up Ksp and solve for Ca^+2.
That is
(Ca^+2)(SO4^-2) = 2.4 x 10^-5.
(Ca^+2)(0.127) = 2.4 x 10^-5. So solve for Ca^+2, you have SO4^-2 which is what the problem asked. You COULD calculate the CaSO4 precipitated if you take the (Ca^+2), convert to grams Ca^+, then convert to g CaSO4, finally convert for 300 mL and not 1 L.
Write the Ksp expression, calculate the moles CaCl2 and moles K2SO4. Determine which is in excess (if either), the excess one will be a common ion for the pptn of CaSO4. Post your work if you get stuck.
I decided to set up an ICE chart but I am a little confused. This is what I did:
CaCl2 + K2SO4 = CaSO4 + 2 KCl
I 0.020M 0.20M 0 0
C -X -X +X +2X
E 0.020-x 0.20-x 0+x 0+2x
Am I right in saying this?
Additional information:
Ksp= (CaSO4)(KCl)^2 / (CaCl2)(K2SO4)
Ksp= 2.4 x 10^ -5
K2SO4+CaCl2
To calculate the equilibrium concentrations of Ca^2+ (aq) and SO4^2- (aq), we need to first determine the moles of CaCl2 and K2SO4 that are present in the solution after they react.
Step 1: Calculate the moles of CaCl2:
moles of CaCl2 = (volume in liters) x (molarity)
= 0.100 L x 0.020 mol/L
= 0.002 mol
Step 2: Calculate the moles of K2SO4:
moles of K2SO4 = (volume in liters) x (molarity)
= 0.200 L x 0.20 mol/L
= 0.040 mol
Step 3: Determine the stoichiometry of the reaction.
The balanced chemical equation for the reaction between CaCl2 and K2SO4 is:
CaCl2(aq) + K2SO4(aq) → 2KCl(aq) + CaSO4(s)
From the equation, we can see that 1 mole of CaCl2 reacts with 1 mole of K2SO4 to produce 1 mole of CaSO4.
Step 4: Determine the limiting reactant.
The limiting reactant is the one that is consumed completely in the reaction and limits the amount of product that can be formed. To determine the limiting reactant, we compare the moles of each reactant to their stoichiometric ratios.
moles of CaCl2 = 0.002 mol
moles of K2SO4 = 0.040 mol
From the stoichiometry of the reaction, we can see that 0.040 mol of K2SO4 would require 0.040 mol of CaCl2 to completely react. Since we have only 0.002 mol of CaCl2, it is the limiting reactant.
Step 5: Calculate the moles of CaSO4 formed.
Since 1 mole of CaCl2 reacts with 1 mole of CaSO4, and CaCl2 is the limiting reactant, we can conclude that 0.002 mol of CaSO4 is formed.
Step 6: Calculate the equilibrium concentrations.
The initial volume of the solution is the sum of the volumes of CaCl2 and K2SO4:
volume of initial solution = 100 mL + 200 mL = 300 mL = 0.300 L
Using the balanced equation, we can determine that the stoichiometric ratio of CaSO4 to Ca^2+ is 1:1, and the stoichiometric ratio of CaSO4 to SO4^2- is also 1:1.
Therefore, the equilibrium concentration of Ca^2+ (aq) is equal to the moles of CaSO4 formed divided by the total volume of the solution:
concentration of Ca^2+ (aq) = (0.002 mol) / (0.300 L) = 0.0067 M
The equilibrium concentration of SO4^2- (aq) is also equal to the moles of CaSO4 formed divided by the total volume of the solution:
concentration of SO4^2- (aq) = (0.002 mol) / (0.300 L) = 0.0067 M
So, the equilibrium concentrations of Ca^2+ (aq) and SO4^2- (aq) are both 0.0067 M.