# 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.

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 ($PbCl_2$)

$PbCl_2\rightleftharpoons Pb^{2+}+2Cl^-$
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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