The concentration of Ag+ ions is 1x10^-6 M in AgBr solution. Find minimum concentration of Br- ions necessary to cause precipitation of AgBr. ( Ksp of AgBr= 4x10^-13 M^2)?

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The concentration of Ag+ ions is 1x10^-6 M in AgBr solution. Find minimum concentration of Br- ions necessary to cause precipitation of AgBr. ( Ksp of AgBr= 4x10^-13 M^2)?

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Answer 1 · 2017-05-07T10:35:26

We interrogate the equilibrium reaction:

$A g^{+} + B r^{-} \to A g B r (s) ⏐ ↓$ $Ag^(+) + Br^(-) rarr AgBr(s)darr$

Explanation:

We have been given the thermodynamic equilibrium constant for this reaction, i.e.:

$K_{sp} = [A g^{+}] [B r^{-}] = 4 \times 10^{- 13}$ $K_"sp"=[Ag^+][Br^-]=4xx10^(-13)$ .

$K_{sp}$ $K_"sp"$ is the so-called $solubility product$ $"solubility product"$ , which have been measured for a host of inorganic salts. And so we simply substitute in $[A g^{+}]$ $[Ag^+]$ , and solve for $[B r^{-}]$ $[Br^-]$ ............

$[B r^{-}] = \frac{K_{sp}}{[A g^{+}]} = \frac{4 \times 10^{- 13}}{1 \times 10^{- 6}} = 4.0 \times 10^{- 7} \cdot m o l \cdot L^{- 1}$ $[Br^-]=K_"sp"/([Ag^+])=(4xx10^-13)/(1xx10^-6)=4.0xx10^-7*mol*L^-1$

And thus, should $[A g^{+}] > 4.0 \times 10^{- 7} \cdot m o l \cdot L^{- 1}$ $[Ag^+]>4.0xx10^-7*mol*L^-1$ , a precipitate of creamy yellow $A g B r$ $AgBr$ salt will be observed.

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The concentration of Ag+ ions is 1x10^-6 M in AgBr solution. Find minimum concentration of Br- ions necessary to cause precipitation of AgBr. ( Ksp of AgBr= 4x10^-13 M^2)?

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