# 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)?

May 7, 2017

We interrogate the equilibrium reaction:

$A {g}^{+} + B {r}^{-} \rightarrow A g B r \left(s\right) \downarrow$

#### Explanation:

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

${K}_{\text{sp}} = \left[A {g}^{+}\right] \left[B {r}^{-}\right] = 4 \times {10}^{- 13}$.

${K}_{\text{sp}}$ is the so-called $\text{solubility product}$, which have been measured for a host of inorganic salts. And so we simply substitute in $\left[A {g}^{+}\right]$, and solve for $\left[B {r}^{-}\right]$............

$\left[B {r}^{-}\right] = {K}_{\text{sp}} / \left(\left[A {g}^{+}\right]\right) = \frac{4 \times {10}^{-} 13}{1 \times {10}^{-} 6} = 4.0 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$

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