How much KBr should be added to 1 L of 0.05 M AgNO_3 solution just to start precipitation of AgBr? K_(sp) of AgBr = 5 x 10^-13.

Jan 30, 2016

${K}_{s p}$ means that $\left[A {g}^{+}\right] \cdot \left[B {r}^{-}\right] = 5 \cdot {10}^{-} 13$,
because the 'sp' suffix stands for 'solution product'.

Explanation:

Concentration of $A {g}^{+} = 0.05 m o l / L$
(one mole of $A {g}^{+}$ per mole of $A g N {O}_{3}$)
So $\left[A {g}^{+}\right] = 5 \cdot {10}^{-} 2$

Now we plug in what we know:
$\left(5 \cdot {10}^{-} 2\right) \cdot \left[B {r}^{-}\right] = 5 \cdot {10}^{-} 13$
$\left[B {r}^{-}\right] = \frac{\cancel{5} \cdot {10}^{-} 13}{\cancel{5} \cdot {10}^{-} 2} = 1 \cdot {10}^{-} 11 m o l / L$

If you have the $K B r$ in a not very diluted form, you may assume that you'll only need a drop or two, and that the $1 L$ won't change significantly.
So you'll need $1 \cdot {10}^{-} 11 m o l$ of $K B r$ 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.