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  • Give answer! Solubility product of silver bromide is 5.0 × 10-13 . The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05M solution of silver nitrate to start the precipitation of AgBr is

Solubility product of silver bromide is 5.0 × 10-13 . The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05M solution of silver nitrate to start the precipitation of AgBr is

  • Option 1)

    5.0 × 10-8 g

  • Option 2)

    1.2 x 10-10 g

  • Option 3)

    1.2 x 10-9 g

  • Option 4)

    6.2 x 10-5 g

 

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}

 

 AgBr \rightleftharpoons Ag ^++ Br^-

K_{sp} \; \; of \; \; AgBr = [ Ag^+][Br^-]

5 \times 10 ^{-13 }= 0.05 [Br ^- ]

[Br ^-] = \frac{5 \times 10^{13}}{0.05}= 1 \times 10^{-11}M

Moles of KBr = = 1 \times 1 0^{-11}

Weight of KBr = = 1.2 \times 1 0^{-9}g 

 

 

 

 


Option 1)

5.0 × 10-8 g

Option 2)

1.2 x 10-10 g

Option 3)

1.2 x 10-9 g

Option 4)

6.2 x 10-5 g

Posted by

SudhirSol

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