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How to solve this problem- - Equilibrium - JEE Main

Let the solubility of an aqueous solution of Mg(OH)be x then its K_{sp}  is

  • Option 1)

    4x^{3}

  • Option 2)

    108x^{5}

  • Option 3)

    27x^{4}

  • Option 4)

    9x

 
Answers (1)
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As we learnt in 

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^{2-}]^{y}

 

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

                               X               2X

K_{sp}=\left [ Mg^{2+} \right ] \left [ OH \right^{-} ]^{2}

K_{sp}=x\times (2x)^{2}=4x^{3}


Option 1)

4x^{3}

This option is correct 

Option 2)

108x^{5}

This option is incorrect 

Option 3)

27x^{4}

This option is incorrect 

Option 4)

9x

This option is incorrect 

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