Solid sodium carbonate is slowly added to 50.0 mL of a 0.0310 M calcium acetate solution. The concentration of carbonate ion required to just initiate precipitation is
Ksp = (Ca^2+)(CO3^2-)
You know Ksp, you know (Ca^2+---it's 0.0310M), solve for (CO3^2-). The 50 mL is not used.
To determine the concentration of carbonate ion required to just initiate precipitation, we need to consider the balanced chemical equation for the reaction between sodium carbonate and calcium acetate:
Na2CO3 + Ca(C2H3O2)2 → 2NaC2H3O2 + CaCO3
In this reaction, sodium carbonate (Na2CO3) reacts with calcium acetate (Ca(C2H3O2)2) to form sodium acetate (NaC2H3O2) and calcium carbonate (CaCO3).
The key concept here is that a precipitate of calcium carbonate will form when the concentration of carbonate ion (CO3^2-) in solution exceeds the solubility product constant (Ksp) of calcium carbonate.
Here's how we can determine the concentration of carbonate ion required to just initiate precipitation:
1. Calculate the moles of calcium acetate:
Molarity (M) = moles (mol) / volume (L)
Rearrange the equation to solve for moles:
Moles = Molarity * volume
Moles of calcium acetate = 0.0310 M * 0.0500 L
2. Convert the moles of calcium acetate to moles of carbonate ion:
From the balanced equation, the stoichiometric ratio between calcium acetate and carbonate ion is 1:1.
Therefore, the moles of carbonate ion = moles of calcium acetate.
3. Convert the moles of carbonate ion to concentration:
Concentration (M) = moles (mol) / volume (L)
The volume of the solution did not change, so the concentration of carbonate ion is the same as the moles of carbonate ion.
Finally, the concentration of carbonate ion required to just initiate precipitation is the concentration calculated in step 3.