This is a solubility product problem with CaCO3 in which there is a common ion (the Ca in Ca(NO3)2)
CaCO3 ==> Ca^+2 + CO3^=
Ksp = (Ca^+2)(CO3^=) = ??
Look up the Ksp for CaCO3.
Ca(NO3)2 ==> Ca^+2 + 2NO3^- (100% ionized)
[Ca(NO3)2] = 0.05 M
Let y = solubility of CaCO3, then y = concn of Ca^+2 from CaCO3 and y is concn of CO3^=. Total (Ca^+2) = y+0.05 (that's Ca from CaCO3 and Ca from Ca(NO3)2).
Solve for y, the units will be mols/L CaCO3, then convert from mols/L to mols/3.0 x 10^-2 mL, and from there to grams.
0.05 M = (Ca^+2) from Ca(NO3)2
Hi. I'm not sure if my strategy for this problem is correct:
How many grams of CaCO3 will dissolve in 3.0x10^-2 mL of 0.050M Ca(NO3)2?
I think I'm supposed to convert molarity of Ca(NO3)2 into moles, but then i don't know how to get from mols of that to mols of CaCO3..is it a 1 to 1 ratio? from then on do i just convert to grams?
Thanks.
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