Question

The solubility of Cd(OH)2, at pH = 13 at 25°C is
(Ksp of Cd(OH)2 = 2.5 * 10-14)

(1) 2.7 x 10-13
(2) 1.09 x 10-10
(3) 2.5 x 10-12
(4) 1.6 x 10-14​

Answer 1

Answer:

Ksp of Cd(OH)2 is 2.5*$10^{-14}$

pH is 13

Ksp=[Cd+2][OH^-1]^2

     =(1.5*10^-5)(3*10^-5)^2

     =1.5*9*10^-15

=15.5*10^-15

=1.6*10^-14

so option 4 is correct

Answer 2

The solubility of Cd(OH)2, at pH = 13 at $25^{o}C$ is  (Ksp of $Cd(OH)_{2} = 2.5 \times 10^{-14}$) is $2.5 \times 10^{-12}$.

Explanation:

The reaction equation for dissociation of $Cd(OH)_{2}$ is as follows.

               $Cd(OH)_{2} \rightarrow Cd^{2+} + 2OH^{-}$

Let us assume that "s" moles of $Cd(OH)_{2}$ gives "s" moles of $Cd^{2+}$ and 2s moles of $OH^{-}$.

As it is given that pH of the given reaction is 13 so the value of pOH is as follows.

                        pH + pOH = 14

                        13 + pOH = 14

                                pOH = 1

Therefore,      $[OH^{-}] = 10^{-1}$

 Therefore, the expression for $K_{sp}$ of this reaction is as follows.

                   $K_{sp} = [s][(2s+0.1)]^{2}$

                   $2.5 \times 10^{-14} = [s][(2s+0.1)]^{2}$

                        s = $2.5 \times 10^{-12}$

Therefore, solubility of $Cd(OH)_{2}$ is $2.5 \times 10^{-12}$.

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