If you were to take a quantity, of PbCI_2, a sparingly soluble salt - maybe 10g
ID: 498917 • Letter: I
Question
If you were to take a quantity, of PbCI_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: PbCI_2 (s) Pb^2+ (aq) + 2CI^- (aq). What will the ratio of the concentrations of Pb^2+ to CI^- be? In other words, wat will be [Pb^2+]/[CI^-]? Equal volumes of 1.0 M Pb(NO_3)_2j and 0.001 M NaI 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? 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