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

Question

If you were to take a quantity, of PbCl_2, a sparingly soluble salt - maybe log -and put it into water, some of it would dissolve based on its K_sp. The equation for this reaction is: PbCl _2(s) Pb^2+(aq) + 2Cl^- (aq). What will the ratio of the concentrations of Pb^2+ to Cl^- be? In other words, wat will be [Pb^2+]/[Cl^-]? Equal volumes of 1.0 M Pb(NO_3)_2 and 0.001 M Nail are mixed in a beaker and PbI_2 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?

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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