If pH of a saturated solution of Ba\left ( OH \right )_2  is 12, the value of its K_{\left ( sp \right )} is

  • Option 1)

    4.00\times 10^{-6}M^{3}

  • Option 2)

    4.00\times 10^{-7}M^{3}

  • Option 3)

    5.00\times 10^{-6}M^{3}

  • Option 4)

    5.00\times 10^{-7}M^{3}

 

Answers (1)

As 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}

 

 We know that, p\left ( H \right ) + p\left (OH \right ) =14

\Rightarrow 12+p\left ( OH \right ) = 14

p\left ( OH \right ) = 2\\\ \\\Rightarrow \left [ OH^-} \right ] = 10^{-2}

Ba\left ( OH \right )_{2}\rightleftharpoons Ba^{2+}+2\ {OH^-}

                                S                 2S

2S = \left [ \ {OH^-} \right ] = 10^{-2}

S = \frac{1}{2}\times 10^{-2}

\Rightarrow K_{sp} = S\cdot \left ( 2S \right )^{2} = 4S^{3} = 4\times \left ( \frac{10}{2}^{-2} \right )^{3}

                = 5\times 10^{-7}


Option 1)

4.00\times 10^{-6}M^{3}

This option is incorrect

Option 2)

4.00\times 10^{-7}M^{3}

This option is incorrect

Option 3)

5.00\times 10^{-6}M^{3}

This option is incorrect

Option 4)

5.00\times 10^{-7}M^{3}

This option is correct

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