Question
What did you observe after the addition of dilute HCl to the saturated lead chloride solution? Explain observations using relevant chemical reaction.
I observed that a white precipitate was formed.
Not sure how to explain why it formed
I observed that a white precipitate was formed.
Not sure how to explain why it formed
Answers
You had a saturated solution of PbCl2 which means that (Pb^+)(Cl^-)^2 was just equal to Ksp.
When you added dilute HCl, you increased the (Cl^-) [the common ion effect] and this DECREASED the solubility of PbCl2; therefore, some of the PbCl2 pptd and that's what you saw.
PbCl2 ==> Pb^2+ + 2Cl^-
(Pb^2+)(Cl^-)2 = Ksp.
So note that if (Cl^-) is added to the above equation, the Cl^- is increased, Le Chatelier's principle tells you that will force the equilibrium to the left and that DECREASES solubility which means some of the PbCl2 from the saturated solution must come out of solution. You observed that as a ppt.
When you added dilute HCl, you increased the (Cl^-) [the common ion effect] and this DECREASED the solubility of PbCl2; therefore, some of the PbCl2 pptd and that's what you saw.
PbCl2 ==> Pb^2+ + 2Cl^-
(Pb^2+)(Cl^-)2 = Ksp.
So note that if (Cl^-) is added to the above equation, the Cl^- is increased, Le Chatelier's principle tells you that will force the equilibrium to the left and that DECREASES solubility which means some of the PbCl2 from the saturated solution must come out of solution. You observed that as a ppt.
Thank you!
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