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7.68     The solubility product constant of Ag2CrO4 and AgBr is 1.1 × 10-12 and 5.0 × 10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.

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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
 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}

Ionization of Silver bromide (AgBr)
AgBr\rightleftharpoons Ag^++Br^-

 K_{sp} of AgBr = 5\times 10^{-13}

[Ag^+] = s' 
[Br^-] = s'

\\\Rightarrow 5\times 10^{-13} = s'.s'\\ =5\times 10^{-13} =s'^2

s' = \sqrt{\frac{5\times 10^{-13}}{1}}=\sqrt{0.5\times 10^{-12}}
     =7.07 \times 10^{-7}

Now, the ratio of solubilities 
\Rightarrow \frac{s}{s'}=\frac{6.5\times 10^{-5}}{1.1\times 10^{-12}}
            = 9.91

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manish

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