Solubility Rules - Why is Lead Iodide insoluble? Watch

Jessinoch
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In chemistry right now we're learning solubility rules, so I know whether nitrates, chlorides, sulfates, oxides, hydroxides and carbonates are soluble or insoluble.

In class today we did precipitation reactions where we had to predict if the reaction between Lead (II) Nitrate and Potassium Iodide would form a precipitate or not? I know that the two products I could get from this reaction would be Lead Iodide and Potassium Nitrate. However, the solubility rules that I was given as a handout did not say anything about compounds like Lead Iodide, it only told me that Potassium Nitrate was soluble in water.

So with this, I presumed that there was going to be no precipitate, however, when I did the experiment I found that a yellow precipitate was formed which must be the Lead Iodide. But my question is, how would I know that Lead Iodide is insoluble in water? Is there a rule for compounds like this as like I said before I only know about nitrates, chlorides, sulfates, oxides, hydroxides and carbonates.

Thanks for your help, and sorry if this was so long lol
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TSR Jessica
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Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.


Just quoting in Fox Corner so she can move the thread if needed :wizard:
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MexicanKeith
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(Original post by Jessinoch)
In chemistry right now we're learning solubility rules, so I know whether nitrates, chlorides, sulfates, oxides, hydroxides and carbonates are soluble or insoluble.

In class today we did precipitation reactions where we had to predict if the reaction between Lead (II) Nitrate and Potassium Iodide would form a precipitate or not? I know that the two products I could get from this reaction would be Lead Iodide and Potassium Nitrate. However, the solubility rules that I was given as a handout did not say anything about compounds like Lead Iodide, it only told me that Potassium Nitrate was soluble in water.

So with this, I presumed that there was going to be no precipitate, however, when I did the experiment I found that a yellow precipitate was formed which must be the Lead Iodide. But my question is, how would I know that Lead Iodide is insoluble in water? Is there a rule for compounds like this as like I said before I only know about nitrates, chlorides, sulfates, oxides, hydroxides and carbonates.

Thanks for your help, and sorry if this was so long lol
The rules like "sulphates are soluble" often have exceptions (eg Calcium Sulphate)

Halogen compounds are no different, many are soluble, but some are not! You may have come across the Silver Nitrate test for Halogen compounds, this works because AgCl, AgBr and AgI are insoluble, similarly the lead compounds PbCl PbBr and PbI are insoluble. Just something you have to know at GCSE/A level chemistry.

The next bit is a bit more of an advanced explanation (hopefully comprehensible though!):

Solubility depends on two main factors, how strong ionic attractions are in the solid and how well solvated each ion is when it's dissolved in water. Using just the ionic model, this gives rise (for group 1 halides atleast and so is probably at least representative of these compounds) to a pattern whereby ions of similar size form insoluble solids, whereas ions of different size form more soluble compounds.

Halide anions have an addtional electron, and so are always quite large (Cl-: 167 pm, Br-: 182pm, I-: 206pm), so their least soluble compounds would be expected with large cations, Pb2+ is largeish (133pm) being low in the periodic table, and Ag+ is largeish too (115pm) because it too is quite low and has only lost 1 outer electron. This is a possible explanation as to why these compounds are insoluble whilst things like NaCl are very soluble (because Na+ is a small cation (102pm).

These numbers don't marry up completely with solubilities, this is probably because silver and lead ions are much more polarisable and so, in the solid form they have additional covalent interactions holding them together, making them even less soluble!
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