Q. 674.1( 7 Votes )

# Determine the sol

Answer :

To determine the solubility, we apply the formula

Ksp = [A+][B-]

Where Ksp is the solubility product which is equal to the product of ionic concentrations in a saturated solution.

a) Silver chromate

ionization of silver chromate:

Ag2CrO4 2Ag+ + CrO42-

As we know that,

Ksp = [A+] [B-]

Where A and B are the ions dissolved

In the above reaction,

[A+] = [Ag+]

[B-] = [CrO42-]

Ksp = [Ag+]2 + [CrO42-]

As Ksp of Ag2CrO4 =1.1 × 10-12 (given)

Let ‘s’ be the solubility of Ag2CrO4

[Ag+] = 2s

[CrO42-] =s

1.1 × 10–12 = (2s)2 s

1.1 × 10-12 = 4(s)3

0.275 × 10–12 = s3

s = s = 0.65 × 10-4

Thus, Molarity of Ag+ = 2s= 2× 0.65× 10-4 = 1.30× 10-4 M

Molarity of CrO42- = s = 0.65 × 10-4 M

b) Barium chromate

ionization of barium chromate:

BaCrO4 Ba2+ + CrO42-

As we know that,

Ksp = [A+] [B-]

Where A and B are the ions dissolved

In the above reaction,

[A+] = [Ba2+]

[B-] = [CrO42-]

Ksp = [Ba2+] + [CrO42-]

As Ksp of BaCrO4 =1.2 × 10-10 (given)

Let ‘s’ be the solubility of BaCrO4

[Ba2+] =s

[CrO42-] =s

1.2 × 10–10 = (s× s)

1.2 × 10-10 = (s)2

s = s = 1.09 × 10-5

Thus, Molarity of Ba2+ and CrO42- = s= 1.09× 10-5M

c) Ferric hydroxide

ionization of ferric hydroxide

Fe(OH)3 Fe3+ + 3OH-

As we know that,

Ksp = [A+] [B-]

Where A and B are the ions dissolved

In the above reaction,

[A+] = [Fe3+]

[B-] = [OH-]

Ksp = [Fe3+] + [OH-]

As Ksp of Fe(OH)3 =1.0 × 10-38 (given)

Let ‘s’ be the solubility of Fe(OH)3

[Fe3+] =s

[3OH-] = 3s

1.0× 10–38 = (s)(3s)3

1.0 × 10-38 = (27s)4

0.37 × 10–38 = s3

s = s = 1.39 × 10-10

Thus, Molarity of Fe3+ = s= 1.39× 10-10 M

Molarity of OH- = 3s = 4.17 × 10-10M

d) Lead chloride

ionization of lead chloride:

PbCl2 Pb2+ + 2Cl-

As we know that,

Ksp = [A+] [B-]

Where A and B are the ions dissolved

In the above reaction,

[A+] = [Pb2+]

[B-] = [Cl-]

Ksp = [Pb2+] + [Cl-]2

As Ksp of PbCl2 =1.6 × 10-5 (given)

Let ‘s’ be the solubility of PbCl2

[Pb2+] =s

[Cl-] =2 s

1.6 × 10–5 = (s)(2s)2

1.6 × 10-5 = (4s)3

0.4 × 10–5 = s3

s = s = 1.58 × 10-2

Thus, Molarity of Pb2+ = s= 1.58× 10-2 M

Molarity of Cl- = 2s = 3.16 × 10-2 M

e) Mercurous Iodide

ionization of mercurous iodide:

Hg2I2 Hg22+ + 2I-

As we know that,

Ksp = [A+] [B-]

Where A and B are the ions dissolved

In the above reaction,

[A+] = [Hg22+]

[B-] = [I-]2

Ksp = [Hg22+] + [I-]2

As Ksp of Hg2I2 = 4.5 × 10–29

Let ‘s’ be the solubility of Hg2I2

[Hg22+] =s

[I-] 2s

4.5 × 10–29 = (s)(2s)2

4.5 × 10–29 = (4s)3

4.5 × 10–29 = s3

s = s = 2.24× 10-10

Thus, Molarity of Hg22+ = s= 2.24× 10-10 M

Molarity of I- = 2s = 4.48 × 10-10 M

Note: Ksp increases with increase in temperature.

In a saturated solution, Ksp = [A+][B-]

In an unsaturated solution of AB, Ksp> [A+][B-] means more solute can be dissolved.

In a super saturated solution of AB, Ksp< [A+][B-] means precipitation will start to occur

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