Solubility Comparison Question

[kfcman289]

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I took a practice MCAT and there was a question where it said the following:

Since PbI2(s) reacts with Na2CO3(aq) to fully yield PbCO3(s), then PbCO3(s) is less soluble than PbI2(s). Can you assume a solubility relationship just off a simple reaction like this?

 

[AwayFromReality]

Yeah you can definitely tell a solubility relationship with the given info.

To start off you have an aqueous solution of Na2CO3 which means you have Na+ and CO32- ions in water already with no visible solid.

Now you add PbI2 solid into this aqueous solution. The solid begins to disassociate into Pb2+ and I- ions. This means the solution now contains two cations (Na+, Pb2+) and two anions (CO32- and I-)

When a precipitate is formed from a solution it means that the solid that precipitated is not very soluble. So now Pb2+ (a cation) has a choice between two anions. It can choose to bond with CO32- and form a solid or it can choose I- to form a solid. Considering the fact that the Pb2+ decided to bond with CO32- instead of with the original I-, we can safely conclude that PbCO3(s) is less soluble than PbI2(s).

 

[kfcman289]

Yeah you can definitely tell a solubility relationship with the given info.

To start off you have an aqueous solution of Na2CO3 which means you have Na+ and CO32- ions in water already with no visible solid.

Now you add PbI2 solid into this aqueous solution. The solid beings to disassociate into Pb2+ and I- ions. This means the solution now contains two cations (Na+, Pb2+) and two anions (CO32- and I-)

When a precipitate is formed from a solution it means that the solid that precipitated is not very soluble. So now Pb2+ (a cation) has a choice between two anions. It can choose to bond with CO32- and form a solid or it can choose I- to form a solid. Considering the fact that the Pb2+ decided to bond with CO32- instead of with the original I-, we can safely conclude that PbCO3(s) is less soluble than PbI2(s).

Ok thanks got it