Question

The K sp for barium sulfate is 1.1 x 10 -10 . Calculate the molar solubility of
barium sulfate.

Answer 1

Answer:

BaSO

4

(s)⟶Ba

2+

(aq)+SO

4

2−

(aq)

∴K

sp

=[Ba

2+

][SO

4

2−

]=x

Then, 1.5×10

−9

=x×x;x

2

=15×10

−10

or3.87×10

−5

Then, solubility of BaSO

4

in pure water is 3.87×10

−5

.

Let the solubility of BaSO

4

in 0.1MBaCl

2

be

′

s

′

Allsolid

BaSO

4

(s)

⇒

0.1M

Ba

2+

(aq)

+SO

4

−

(aq)

Initial (from BaCl

2

) 0

At equilibrium (0.1M+s) s

Hence, 1.5×10

−9

=(s+0.1)×s=s×0.1(Since s<<1)

s=1.5×10

−8

Thus the solubility of BaSO

4

in presence of 0.1M BaCl

2

is 1.5×10

−8