Question

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

BaSO₄ K_sp = 1.1 x 10⁻¹⁰ Μ

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

-       A.       B.    100 mL of 4.0 x 10−3 Μ of BaCl2 and 300 mL of 6.0 x 10−4 Μ Na2SO4       -       A.       B.    500 mL of 4.0 x 10−6 Μ of BaCl2 and 500 mL of 6.0 x 10−6 Μ Na2SO4       -       A.       B.    200 mL of 4.0 x 10−2 Μ of BaCl2 and 500 mL of 6.0 x 10−6 Μ Na2SO4       -       A.       B.    250 mL of 4.0 x 10−5 Μ of BaCl2 and 250 mL of 6.0 x 10−10 Μ Na2SO4

A. Yes. B. No

Answer 1

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.

2)

[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.

3)

[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.

4)

[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.