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  • Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table . Determine also the molarities of individual ions.

7.67     Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table. Determine also the molarities of individual ions.

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Solubility product is the product of ionic concentrations in a saturated solution.
K_{sp}=[A^+][B^-]
(i) silver chromate (Ag_2CrO_{4})
Ionization of silver chromate

Ag_2CrO_{4}\rightleftharpoons 2Ag^++CrO_{4}^{2-}
Let "s" be the solubility of Ag_2CrO_{4}
[Ag^+] = 2s 
[CrO_{4}^{2-}] = s
According to the table K_{sp} of Ag_2CrO_{4} = 1.1\times 10^{-12}

\\\Rightarrow 1.1\times 10^{-12} = (2s)^2.s\\ =1.1\times 10^{-12} =2s^3

s = \sqrt[3]{\frac{1.1\times 10^{-12}}{4}}
     =0.65 \times 10^{-4}

(ii) Barium chromate (BaCrO_{4})
Ionization of silver chromate

BaCrO_{4}\rightleftharpoons Ba^{2+}+CrO_{4}^{2-}
Let "s" be the solubility of BaCrO_{4}
[Ba^{2+}] = s 
[CrO_{4}^{2-}] = s
According to the table K_{sp} of BaCrO_{4} = 1.2\times 10^{-12}

\\\Rightarrow 1.2\times 10^{-10} = s.s\\ =1.1\times 10^{-10} =s^2

s = \sqrt{\frac{1.2\times 10^{-10}}{1}}
     =1.09 \times 10^{-5}

(iii) Ferric hydroxide (Fe(OH)_{3})
Ionization of Ferric hydroxide 

Fe(OH)_3\rightleftharpoons Fe^{3+}+ 3OH^-
Let "s" be the solubility of Fe(OH)_{3}
[Fe^{3+}] = s 
[OH^-] = 3s
According to the table K_{sp} of Fe(OH)_{3} = 1.0\times 10^{-38}

\\\Rightarrow 1.0\times 10^{-38} = s.(3s)^3\\ =1.0\times 10^{-38} =27s^4

s = \sqrt[4]{\frac{1.0\times 10^{-38}}{27}}
     =1.39 \times 10^{-10}

(iv) 

Lead chloride (PbCl_2)
Ionization of Lead chloride

PbCl_2\rightleftharpoons Pb^{2+}+2Cl^-
Let "s" be the solubility of PbCl_2
[Pb^{2+}] = s 
[Cl^-] = 2s
According to the table K_{sp} of PbCl_2 = 1.6\times 10^{-5}

\\\Rightarrow 1.6\times 10^{-5} = s.(2s)^2\\ =1.6\times 10^{-5} =4s^3

s = \sqrt[3]{\frac{1.6\times 10^{-5}}{4}}
     =1.58 \times 10^{-2}

So molarity of Pb^{2+}=1.58 \times 10^{-2}M and molarity of Cl^{-}=3.16 \times 10^{-2}M

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manish

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