posted by Bob .
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)
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?
Ksp= (CaSO4)(KCl)^2 / (CaCl2)(K2SO4)
Ksp= 2.4 x 10^ -5
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.
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
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.
(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.