# Question #52bf5

Jun 23, 2015

If you look at the concentrations, you will see that $B {r}^{-}$ is greatly in excess, about a factor of 1 to 1000.

#### Explanation:

So the tiny bit of $B {r}^{-}$ that reacts with the $A {g}^{+}$ will hardly make a dent in the concentration (less than 0,1% = third decimal).
Of course the concentration is halved (twice the amount of liquid).
We may take [Br^-] to be $1.0 \cdot {10}^{-} 3$ after the reaction.

If we now set up the solubility product:

${K}_{s p} = \left[A {g}^{+}\right] \cdot \left[B {r}^{-}\right]$

Solving:
$\left[A {g}^{+}\right] \cdot \left[1.0 \cdot {10}^{-} 3\right] = 5.0 \cdot {10}^{-} 13 \to$

$\left[A {g}^{+}\right] = \frac{5.0 \cdot {10}^{-} 13}{1.0 \cdot {10}^{-} 3} = 5.0 \cdot {10}^{-} 10 M$