Answer :

To determine the solubility, we apply the formula

K_{sp} = [A^{+}][B^{-}]

Where K_{sp} is the solubility product which is equal to the product of ionic concentrations in a saturated solution.

a) Silver chromate

ionization of silver chromate:

Ag_{2}CrO_{4} ↔ 2Ag^{+} + CrO_{4}^{2-}

As we know that,

Ks_{p} = [A^{+}] [B^{-}]

Where A and B are the ions dissolved

In the above reaction,

⇒[A^{+}] = [Ag^{+}]

⇒[B^{-}] = [CrO_{4}^{2}^{-}]

∴ K_{sp} = [Ag^{+}]^{2} + [CrO_{4}^{2-}]

As K_{sp} of Ag_{2}CrO_{4} =1.1 × 10^{-12} (given)

Let ‘s’ be the solubility of Ag_{2}CrO_{4}

[Ag^{+}] = 2s

[CrO_{4}^{2-}] =s

∴ 1.1 × 10^{–12} = (2s)^{2} s

⇒1.1 × 10^{-12} = 4(s)^{3}

^{⇒}0.275 × 10^{–12} = s^{3}

⇒s =

⇒s = 0.65 × 10^{-4}

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

Molarity of CrO_{4}^{2}- = s = 0.65 × 10^{-4} M

b) Barium chromate

ionization of barium chromate:

BaCrO_{4} ↔ Ba^{2+} + CrO_{4}^{2-}

As we know that,

Ks_{p} = [A^{+}] [B^{-}]

Where A and B are the ions dissolved

In the above reaction,

⇒[A^{+}] = [Ba^{2+}]

[B^{-}] = [CrO_{4}^{2-}]

∴ K_{sp} = [Ba^{2+}] + [CrO_{4}^{2-}]

As K_{sp} of BaCrO_{4} =1.2 × 10^{-10} (given)

Let ‘s’ be the solubility of BaCrO_{4}

[Ba^{2+}] =s

[CrO_{4}^{2}-] =s

∴ 1.2 × 10^{–10} = (s× s)

^{⇒}1.2 × 10^{-10} = (s)^{2}

⇒s =

⇒s = 1.09 × 10^{-5}

Thus, Molarity of Ba^{2+} and CrO_{4}^{2-} = s= 1.09× 10^{-5}M

c) Ferric hydroxide

ionization of ferric hydroxide

Fe(OH)_{3} ↔ Fe^{3+} + 3OH^{-}

As we know that,

Ks_{p} = [A^{+}] [B^{-}]

Where A and B are the ions dissolved

In the above reaction,

⇒[A^{+}] = [Fe^{3+}]

⇒[B^{-}] = [OH^{-}]

∴ K_{sp} = [Fe^{3+}] + [OH^{-}]

As K_{sp} of Fe(OH)_{3} =1.0 × 10^{-38} (given)

Let ‘s’ be the solubility of Fe(OH)_{3}

[Fe^{3+}] =s

[3OH^{-}] = 3s

∴ 1.0× 10^{–38} = (s)(3s)^{3}

^{⇒}1.0 × 10^{-38} = (27s)^{4}

^{⇒}0.37 × 10^{–38} = s^{3}

⇒s =

⇒s = 1.39 × 10^{-10}

Thus, Molarity of Fe^{3+} = s= 1.39× 10^{-10} M

Molarity of OH^{-} = 3s = 4.17 × 10^{-10}M

d) Lead chloride

ionization of lead chloride:

PbCl_{2} ↔ Pb^{2+} + 2Cl^{-}

As we know that,

Ks_{p} = [A^{+}] [B^{-}]

Where A and B are the ions dissolved

In the above reaction,

⇒[A^{+}] = [Pb^{2+}]

⇒[B^{-}] = [Cl^{-}]

∴ K_{sp} = [Pb^{2+}] + [Cl^{-}]^{2}

As K_{sp} of PbCl_{2} =1.6 × 10^{-5} (given)

Let ‘s’ be the solubility of PbCl_{2}

[Pb^{2+}] =s

[Cl^{-}] =2 s

∴ 1.6 × 10^{–5} = (s)(2s)^{2}

^{⇒}1.6 × 10^{-5} = (4s)^{3}

^{⇒}0.4 × 10^{–5} = s^{3}

⇒s =

⇒s = 1.58 × 10^{-2}

Thus, Molarity of Pb^{2+} = s= 1.58× 10^{-2} M

Molarity of Cl^{-} = 2s = 3.16 × 10^{-2} M

e) Mercurous Iodide

ionization of mercurous iodide:

Hg_{2}I_{2} ↔ Hg_{2}^{2+} + 2I^{-}

As we know that,

Ks_{p} = [A^{+}] [B^{-}]

Where A and B are the ions dissolved

In the above reaction,

⇒[A^{+}] = [Hg_{2}^{2+}]

⇒[B^{-}] = [I^{-}]^{2}

∴ K_{sp} = [Hg_{2}^{2+}] + [I^{-}]^{2}

As K_{sp} of Hg_{2}I_{2} = 4.5 × 10^{–29}

Let ‘s’ be the solubility of Hg_{2}I_{2}

[Hg_{2}^{2+}] =s

[I^{-}] 2s

∴ 4.5 × 10^{–29} = (s)(2s)^{2}

⇒4.5 × 10^{–29} = (4s)^{3}

⇒4.5 × 10^{–29} = s^{3}

⇒s =

⇒s = 2.24× 10^{-10}

Thus, Molarity of Hg_{2}^{2+} = s= 2.24× 10^{-10} M

Molarity of I- = 2s = 4.48 × 10^{-10} M

Note: K_{sp} increases with increase in temperature.

⇒In a saturated solution, K_{sp} = [A^{+}][B^{-}]

⇒In an unsaturated solution of AB, K_{sp}> [A^{+}][B^{-}] means more solute can be dissolved.

⇒In a super saturated solution of AB, K_{sp}< [A^{+}][B^{-}] means precipitation will start to occur

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