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At 25°C, the solubility product of Mg \left ( OH \right )_{2} is 1.0 × 10-11. At which pH, will Mg^{2+} ions start precipitating in the form of Mg \left ( OH \right )_{2} from a solution of 0.001 M Mg2+ ions ?

  • Option 1)

    8

  • Option 2)

    9

  • Option 3)

    10

  • Option 4)

    11

 

Answers (1)

best_answer

As we have learned

General expression of solubility product -

M_{x}X_{y}\rightleftharpoons xM^{p+}(aq)+yX^{2-}(aq)


(x.p^{+}=y.\bar{q})

- wherein

Its solubility product is 
 

K_{sp}=[M^{p+}]^{x}[X^{q-}]^{y}

 

 K_{sp}  of Mg(OH)_2= [ Mg ^{2+}][OH^-]^2

[OH^-]= \sqrt{\frac{K_{sp}}{[Mg^{2+}]}}= \sqrt{\frac{1 \times 10^{-11}}{10^{-3}}} = 10^{-4 }M

p OH = 4

p H = 14-pOH= 14-4 = 10

 

 

 

 


Option 1)

8

Option 2)

9

Option 3)

10

Option 4)

11

Posted by

SudhirSol

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