TPRHSW Stoichiometry Problem

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deleted388502

Hi,

TPRHSW had a problem and its solution confused me a bit.
"What is the molar concentration of Ca2+ in a solution that is prepared by mixing 15mL of a 0.02M CaCl2 solution with 10mL of a 0.04M CaSO4 solution?"

A. 0.014M
B. 0.020M
C. 0.08M
D. 0.035M

I set this up with (0.02)(15)(2) + (0.04)(10)/25, but the solution does not multiply the 0.02 by 2 and I'm confused as to why/

Thanks!

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Hi,

TPRHSW had a problem and its solution confused me a bit.
"What is the molar concentration of Ca2+ in a solution that is prepared by mixing 15mL of a 0.02M CaCl2 solution with 10mL of a 0.04M CaSO4 solution?"

A. 0.014M
B. 0.020M
C. 0.08M
D. 0.035M

I set this up with (0.02)(15)(2) + (0.04)(10)/25, but the solution does not multiply the 0.02 by 2 and I'm confused as to why/

Thanks!
moles CaSO4 = (0.01L)(0.04M) = (1/100)(4/100) = 4/10000 = 0.0004 moles
moles CaCl2 = (0.015L)(0.02M) = (15/100)(2/100) = 30/10000 = 0.0030 moles

Both of these are ionic spefcies that will fully dissociate in water to form the calcium ions.

Total Calcium ions: 0.003 + 0.0004 = 0.0034 moles
Total Volume (after mixing): 15mL + 10mL = 25mL = 0.025L

Molar Concentration: 0.0034 moles Ca2+/0.025L = 3.4e-3/2.5e-2 = 3.4/2.5 e-1 ~~~ approx. 0.34M (choice d). Probably easier to ignore converting to L and stick with mL.
 
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LOL I totally just realized that in the 25 minutes I stared at this problem I was looking at CaCl2 and thinking that it had 2x Ca and not Cl2. Wow.

Thanks for your help on an incredibly simple problem I should've known the first time. Thanks!
 
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moles CaSO4 = (0.01L)(0.04M) = (1/100)(4/100) = 4/10000 = 0.0004 moles
moles CaCl2 = (0.015L)(0.02M) = (15/100)(2/100) = 30/10000 = 0.0030 moles

Both of these are ionic spefcies that will fully dissociate in water to form the calcium ions.

You sure about that? That's flat out wrong. You know, you can clean drywall with water. Gypsum therefore probably isn't that soluble. Its Ksp (off the top my head) is 9.1*10^-6.
 
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