Will a precipiate of BaSO 4 (s) form when various concentrations and volumes of

ID: 503862 • Letter: W

Question

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

Remember to account for the total volume in determinting the new concentration of each ion when the solutions are mixed.

100 mL of 4.0 x 103 of BaCl2 and 300 mL of 6.0 x 104 Na2SO4

500 mL of 4.0 x 106 of BaCl2 and 500 mL of 6.0 x 106 Na2SO4

200 mL of 4.0 x 102 of BaCl2 and 500 mL of 6.0 x 106 Na2SO4

250 mL of 4.0 x 105 of BaCl2 and 250 mL of 6.0 x 1010 Na2SO4

Yes.

- A. B.

100 mL of 4.0 x 103 of BaCl2 and 300 mL of 6.0 x 104 Na2SO4

- A. B.

500 mL of 4.0 x 106 of BaCl2 and 500 mL of 6.0 x 106 Na2SO4

- A. B.

200 mL of 4.0 x 102 of BaCl2 and 500 mL of 6.0 x 106 Na2SO4

- A. B.

250 mL of 4.0 x 105 of BaCl2 and 250 mL of 6.0 x 1010 Na2SO4

Yes.

1) [Ba+2] = 100x4x10-3/400 =1x10-3 M and [SO4] = 300x6.0x10-4 /400= 4.5x10-4 M

Thus Ionic product = (1x10-3 M ) ( 4.5x10-4 )

= 4.5x10-7

since ionic product > Ksp . precipitation occurs.

[Ba+2] = 500x4x10-6/1000 =2x10-6 M and [SO4] = 500x6.0x10-6/1000= 3.0x10-6 M

Thus Ionic product = (2x10-6 M ) ( 3x10-6 )

= 6x10-12

since IP < ksp no precipitation occurs.

[Ba+2] = 200x4x10-2/700 =1.14x10-2M and [SO4] = 500x6.0x10-6 /400= 7.5x10-6M

Thus Ionic product = (1.14x10-2 M ) ( 7.5x10-6 )

= 8.55x10-8

since IP > Ksp precipitation occurs.

[Ba+2] = 250x4x10-5/500 =2x10-5 M and [SO4] = 250x6.0x10-10 /500= 3x10-10M

Thus Ionic product = (2x10-5M ) ( 3x10-10)

= 6x10-15

since IP < Ksp no precipitation occurs.

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