How much $K B r$ $KBr$ should be added to $1$ $1$ $L$ $L$ of $0.05$ $0.05$ $M$ $M$ $A g N O_{3}$ $AgNO_3$ solution just to start precipitation of $A g B r$ $AgBr$ ? $K_{s p}$ $K_(sp)$ of $A g B r$ $AgBr$ = $5$ $5$ x $10^{- 13}$ $10^-13$ .

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How much $K B r$ $KBr$ should be added to $1$ $1$ $L$ $L$ of $0.05$ $0.05$ $M$ $M$ $A g N O_{3}$ $AgNO_3$ solution just to start precipitation of $A g B r$ $AgBr$ ? $K_{s p}$ $K_(sp)$ of $A g B r$ $AgBr$ = $5$ $5$ x $10^{- 13}$ $10^-13$ .

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Answer 1 · 2016-01-30T23:15:22

$K_{s p}$ $K_(sp)$ means that $[A g^{+}] \cdot [B r^{-}] = 5 \cdot 10^{- 13}$ $[Ag^+]*[Br^-]=5*10^-13$ ,
because the 'sp' suffix stands for 'solution product'.

Explanation:

Concentration of $A g^{+} = 0.05 m o l / L$ $Ag^+=0.05mol//L$
(one mole of $A g^{+}$ $Ag^+$ per mole of $A g N O_{3}$ $AgNO_3$ )
So $[A g^{+}] = 5 \cdot 10^{- 2}$ $[Ag^+]=5*10^-2$

Now we plug in what we know:
$(5 \cdot 10^{- 2}) \cdot [B r^{-}] = 5 \cdot 10^{- 13}$ $(5*10^-2)*[Br^-]=5*10^-13$
$[Br^-]=(cancel5*10^-13)/(cancel5*10^-2)=1*10^-11mol//L$

If you have the $KBr$ in a not very diluted form, you may assume that you'll only need a drop or two, and that the $1L$ won't change significantly.
So you'll need $1*10^-11mol$ of $KBr$ and this can easily be converted to grams (or rather milligrams).

Extra :
One drop more and the solution will turn cloudy white, and turn dark on exposure to sunlight -- that's what classic photography is based upon.

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How much $K B r$ $KBr$ should be added to $1$ $1$ $L$ $L$ of $0.05$ $0.05$ $M$ $M$ $A g N O_{3}$ $AgNO_3$ solution just to start precipitation of $A g B r$ $AgBr$ ? $K_{s p}$ $K_(sp)$ of $A g B r$ $AgBr$ = $5$ $5$ x $10^{- 13}$ $10^-13$ .

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Explanation:

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How much KBrKBrKBr should be added to 111 LLL of 0.050.050.05 MMM AgNO_3AgNO3AgNO_3 solution just to start precipitation of AgBrAgBrAgBr? K_(sp)KspK_(sp) of AgBrAgBrAgBr = 555 x 10^-1310−1310^-13.

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How much $K B r$ $KBr$ should be added to $1$ $1$ $L$ $L$ of $0.05$ $0.05$ $M$ $M$ $A g N O_{3}$ $AgNO_3$ solution just to start precipitation of $A g B r$ $AgBr$ ? $K_{s p}$ $K_(sp)$ of $A g B r$ $AgBr$ = $5$ $5$ x $10^{- 13}$ $10^-13$ .