Chp8_KspConsid

Chp8_KspConsid - Some Ksp Considerations Consider the...

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Some Ksp Considerations Consider the mixing of 10.0 mL of 0.50 M Pb(NO3)2 with 15mL of 0.50 M KI. The nitrate and potassium are “spectator” ions and the “net ionic equation” from CH 101 would be: 2 1 2 2 PbI I Pb + - + which describes the formation of the sparingly soluble lead iodide from the ions. Note that this reaction is the reverse of the solubility product (Ksp) reaction for lead iodide (which has the value 6.5e–9). The equilibrium constant for the formation of PbI2 is thus the inverse of this, 1.54e+8. Suppose we are asked to calculate the concentration of both ions in the solution after mixing. Consider two approaches: (a) Recognize that the equilibrium constant for the formation of the solid is “very large”. Thus, we assume that the reaction is extensive to the right and that “all” of the limiting reagent is consumed. (n.b. if we write the Ksp reaction, we would have the reaction run extensively to the left). Thus (in mmol): Pb^2+ 2I^1– = PbI2 (i) 5.00 7.50 0 ( ) –3.75 –7.50 +3.50 (f) 1.25 ~0 3.50 Be certain you understand which is the limiting reagent. [Pb
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Chp8_KspConsid - Some Ksp Considerations Consider the...

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