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The molar solubility of Cd(OH)_{2} is 1.84\times 10^{-5}M in water. The expected solubility of Cd(OH)_{2} in a buffer solution of pH = 12 is :

 

  • Option 1)

    1.84\times 10^{-9}M

     

     

     

  • Option 2)

    \frac{2.49}{1.84}\times 10^{-9}M

  • Option 3)

    6.23\times 10^{-11}M

  • Option 4)

    2.49\times 10^{-10}M

Answers (1)

best_answer

Ksp of C\alpha (OH)^{3}=4S^{3}

Ksp = 45^{3}=4\times (1.84\times 10^{-5})

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

                                S               2S+ 10^{-2}

  2S - > this OH comes from dissociation of Cd(OH)_{2}

    10^{-2} - > this OH comes from a buffer                                      

 Given P^{H}=12

            P^{OH}=2

            [OH] = 10^{-2}

Ksp= Kea

45^{3}=s[25+10^{-2}]^{2}

4\times (1.84\times 10^{-5})^{3}=s\times 10^{-4}

S=2.49\times 10^{-10}M


Option 1)

1.84\times 10^{-9}M

 

 

 

Option 2)

\frac{2.49}{1.84}\times 10^{-9}M

Option 3)

6.23\times 10^{-11}M

Option 4)

2.49\times 10^{-10}M
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