Solubility Question

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meliora27

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I was under the impression that all chlorides, bromides, and iodides are soluble except those containing Ag+, Pb2+, or Hg_2 2+. That being said, how is it possible in a rxn b/w PbCO3 (s) and 2 HCl (aq) to produce PbCl2 (aq). Perhaps I'm just exhausted from studying all day but if anyone wouldn't mine explaining this or correcting my (perhaps faulty) knowledge of solubility rules I'd be quite appreciative. Thanks!

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I was under the impression that all chlorides, bromides, and iodides are soluble except those containing Ag+, Pb2+, or Hg_2 2+. That being said, how is it possible in a rxn b/w PbCO3 (s) and 2 HCl (aq) to produce PbCl2 (aq). Perhaps I'm just exhausted from studying all day but if anyone wouldn't mine explaining this or correcting my (perhaps faulty) knowledge of solubility rules I'd be quite appreciative. Thanks!

HCl is a strong acid. It will form H+ and Cl- ions. C032- is a base and it is more favorable for it to react with H+ to form H20 and C02 gas. The Cl- then forms with Pb as you stated.
 
Don't think you really answered his question.

OP - Is solubility even relevant to the question asked? It could just be a typo.
 
HCl is a strong acid. It will form H+ and Cl- ions. C032- is a base and it is more favorable for it to react with H+ to form H20 and C02 gas. The Cl- then forms with Pb as you stated.

Yes, typo. Sorry, thinking of something else originally.
 
The solubility of PbCl2 in water is low (9.9 g/L at 20 °C) and for practical purposes it is considered insoluble. Its solubility-product constant (Ksp) at 25 °C (298 K) is 1.6 x 10-5. It is one of only three commonly insoluble chlorides, the other two being silver chloride (AgCl) with Ksp = 1.8 x 10-10 and mercury (I) chloride (Hg2Cl2) with Ksp = 1.3 x 10-18.
Your solubility rule knowledge is perfectly right. PbCl2 is insoluble but if the concentrations of Pb2+ and Cl - do not exceed Ksp then it remains in aqueous medium.PbCl2 (aq) means very very dilute solution. http://en.wikipedia.org/wiki/Cotunnite#cite_note-0
 
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