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1) Will a precipiate of PbCl 2 (s) form when various concentrations and volumes

ID: 1018575 • Letter: 1

Question

1) Will a precipiate of PbCl2(s) form when various concentrations and volumes of Pb(NO3)2(aq) and HCl(aq) are mixed?

PbCl2 Ksp = 1.2 x 105

- 100 mL of 5.0 x 10^-5 M of Pb(NO3)2 and 100 mL of 5.0 x 10^-5 M HCl A. No

- 100 mL of 5.0 x 10^-1 M of Pb(NO3)2 and 100 mL of 5.0 x 10^-1 M HCl B. Yes

- 500 mL of 5.0 x 10^-1 M of Pb(NO3)2 and 100 mL of 5.0 x 10^-1 M HCl

- 500 mL of 5.0 x 10^-1 M of Pb(NO3)2 and 500 mL of 5.0 x 10^-1 M HCl

2) Will a precipiate of BaSO4(s) form when various concentrations and volumes of BaCl2(aq) and Na2SO4(aq) are mixed?

BaSO4 Ksp = 1.1 x 1010

- 100 mL of 4.0 x 10^-3 M of BaCl2 and 300 mL of 6.0 x 10^-4 M Na2SO4 A. Yes

- 500 mL of 4.0 x 10^-6 M of BaCl2 and 500 mL of 6.0 x 10^-6 M Na2SO4 B. No

- 200 mL of 4.0 x 10^-2 M of BaCl2 and 500 mL of 6.0 x 10^-6 M Na2SO4

- 250 mL of 4.0 x 10^-5 M of BaCl2 and 250 mL of 6.0 x 10^-10 M Na2SO4

3) What is the Q (IP) when 290 mL of 0.073 M solution of Na2CO3(aq) are mixed with 243 mL of 0.052 M solution of BaCl2(aq). The solid product is BaCO3. Express your answer in scientific notation.

Explanation / Answer


Ksp of PbCl2 = 1.2 * 10^-5 M and Kip = [Pb+2][Cl-]^2
Kip of PbCl2 = (100*5*10^-5)(100*5*10^-5)^2 = 1.25*10^-7
Kip < Ksp hence no precipitation occurs.
Kip of PbCl2 = (100*5*10^-1)(100*5*10^-1)^2 = 1.25*10^5
Kip > Ksp hence precipitation occurs.
Kip of PbCl2 = (500*5*10^-1)(100*5*10^-1)^2 = 6.25*10^5
Kip > Ksp hence precipitation occurs.
Kip of PbCl2 = (500*5*10^-1)(500*5*10^-1)^2 = 1.56*10^7
Kip > Ksp hence precipitation occurs.
Ksp of BaSO4 = 1.1*10^-10 M and Kip = [Ba+2][SO4-2]
Kip of BaSO4 = (100*4*10^-3)(300*6*10^-4) = 7.2*10^-2
Kip > Ksp hence precipitation occurs
Kip of BaSO4 = (500*4*10^-3)(500*6*10^-4) = 0.6
Kip > Ksp hence precipitation occurs
Kip of BaSO4 = (200*4*10^-3)(500*6*10^-4) = 0.24
Kip > Ksp hence precipitation occurs
Kip of BaSO4 = (250*4*10^-3)(250*6*10^-4) = 0.15
Kip > Ksp hence precipitation occurs
Kip = [Ba+2][CO3 2-] = (243*0.052)(290*0.073)
Kip = 2.675*10^2