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  • Need explanation for: MY and NY3 two neartly insoluble salts, have te same Ksp values of 6.2 x 10-13 at room temperature. Which statement would be true in regard of MYand NY3?

MY and NY3 two neartly insoluble salts, have te same Ksp values of 6.2 x 10-13 at room temperature. Which statement would be true in regard of MY and NY3?

  • Option 1)

    The molar solubilities of MY and NY3 in water are indetical.

  • Option 2)

    The molar solubility of MY in water is less than that of NY3.

  • Option 3)

    The salts MY and NY3 are more soluble in 0.5 M KY than in pure water.

  • Option 4)

    The addition of the salt of KY to solution of MY and NY3 will have no effect on their solubilities.

 

Answers (1)

best_answer

As we learnt in

Solubility product constant -

e.g.

BaSO_{4}\rightleftharpoons Ba^{2+}(aq)+SO_{4}^{2-}(aq)


K=\frac{[Ba^{2+}]\:[SO_{4}^{2-}]}{[BaSO_{4}]}

- wherein

For a pure solid substance the concentration remain constant

\therefore \:K_{sp}=K[BaSO_{4}]
 

               =[Ba^{2+}]\:[SO_{4}^{2-}]
 

    K_{sp}=solubility\:product

 

 MY\rightleftharpoons M^{+}+Y^{-}                                    K_{sp}=6.2 \times 10^{-13}

1-s              s             s

\Rightarrow \frac{s^{2}}{1-s}=6.2 \times 10^{-13}

s << 1

\therefore s = \sqrt{6.2\times 10^{-13}}=7.87 \times 10^{-7}

Also 

\\NY_{3}\rightleftharpoons N^{3+}+3Y^{-},\ \: \: \: \: K_{sp}=6.2 \times 10^{-13}\\1-s'\: \: \: \: \: \: \: s' \: \: \: \: 3s'

\Rightarrow K_{sp}=\frac{(3s') (s)}{(1-s')}

\Rightarrow 27 s'^{4}=6.2 \times 10^{-13}

\Rightarrow s'=3.89 \times 10^{-4}

Clearly s' > s

 

 


Option 1)

The molar solubilities of MY and NY3 in water are indetical.

Incorrect

Option 2)

The molar solubility of MY in water is less than that of NY3.

Correct

Option 3)

The salts MY and NY3 are more soluble in 0.5 M KY than in pure water.

Incorrect

Option 4)

The addition of the salt of KY to solution of MY and NY3 will have no effect on their solubilities.

Incorrect

Posted by

Aadil

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