If you were to take a quantity, of PbCl_2, a sparingly soluble salt- maybe 10g-a

Question

If you were to take a quantity, of PbCl_2, a sparingly soluble salt- maybe 10g-and put it into water, some of it would dissolve based on its K_sp. The equation for this reaction is PbCl_2(s). D Pb^2+_(aq) + 2Cl^-_(aq). What will the ratio of the concentrations of Pb^2+ to Cl^- be? In other words, what will be [Pb^2+]/[Cl]? Equal volumes of 1.0 M Pb(NO_3)_2, and 0.001 M NaI are mixed in a beaker and PbI_a precipitates out as the solution reached equilibrium, what will the ratio of the concentrations of Pb^2+ to I^- be in the beaker at equilibrium? Is this the same ratio as you found in problem #1? Why not? In this lab, you will be working with potassium chromate (find the SDS here). Please list at list three specific hazards associated with potassium chromate.

Explanation / Answer

1) PbCl2 (s) -----------> Pb+2 + 2 Cl-

s 0 0 initial concnetrations

- s 2s equilibrium concnetrations

Thus the ratio of [Pb+2]/[Cl-] = s /2s = 0.5

2) Pb(NO3)2 + 2 NaI ------------> PbI2 (s) + 2 NaNO3

1.0 0.001 0 0 initial concentrations

1.0 - 0.0005 0 0.0005 -

= 0.9995 /2 0.0005 /2 - equilibrium concentration

NOw as PbI2 is a sparingly soluble salt all the iodide concentration comes from it only.

PbI2 -------------> Pb+2 + 2 I-

s 0.9995/2 s

ksp of pBI2 = 4.41x10-9

= [Pb+2] [I-]2

  

4.41x10-9 = ( 0.9995/2 ) s2

thus s = 9.39x10-5 M

Thus the ratio of [Pb+2] /[I-] = (0.9995/2) / 9.39 x10-5

= 5322 :1

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