# Will a precipitate form if 20 mL of 1 x 10^-7 M AgNO_3 is mixed with 20 mL of 2 x 10^9 M NaCl at 25°C? For AgCl, K_(sp) = 1.8 x 10^-10

Mar 16, 2017

$\text{Non...........}$

#### Explanation:

We assess the equilibrium reaction:

$A {g}^{+} + C {l}^{-} r i g h t \le f t h a r p \infty n s A g C l \left(s\right) \downarrow$, ${K}_{\text{sp}} = 1.8 \times {10}^{-} 10$

Now ${K}_{\text{sp}} = \left[A {g}^{+}\right] \left[N {O}_{3}^{-}\right]$, and initially, we have the following concentrations:

$\left[A {g}^{+}\right] = \frac{20 \times {10}^{-} 3 L \times 1.0 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1}{40 \times {10}^{-} 3 \cdot L} = 5 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$

$\left[C {l}^{-}\right] = \frac{20 \times {10}^{-} 3 L \times 2.0 \times {10}^{-} 9 \cdot m o l \cdot {L}^{-} 1}{40 \times {10}^{-} 3 \cdot L} = 1 \times {10}^{-} 9 \cdot m o l \cdot {L}^{-} 1$

Note that I assume you meant that $\left[N a C l\right] = 2 \times {10}^{-} 9 \cdot m o l \cdot {L}^{-} 1$, the given concentration is unreasonably high.

Now the ion product, ${Q}_{\text{rxn}} = \left[A {g}^{+}\right] \left[C {l}^{-}\right] = 5 \times {10}^{-} 8 \times 1 \times {10}^{-} 9 = 5 \times {10}^{-} 17$.

And since ${Q}_{\text{rxn}}$ $<$ ${K}_{\text{sp}}$ precipitation of $A g C l$ should not occur.