# Question #a3ed1

May 23, 2017

Yes.

#### Explanation:

At equilibrium we have:

$\textsf{A g C {l}_{\left(s\right)} r i g h t \le f t h a r p \infty n s A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}}$

For which:

$\textsf{{K}_{s p} = \left[A {g}_{\left(a q\right)}^{+}\right] \left[C {l}_{\left(a q\right)}^{-}\right] = 1.8 \times {10}^{- 10} \textcolor{w h i t e}{x} {\text{mol"^2."l}}^{- 2}}$

Where the concentrations are those at equilibrium

The reaction quotient Q is given by:

$\textsf{Q = \left[A {g}_{\left(a q\right)}^{+}\right] \left[C {l}_{\left(a q\right)}^{-}\right]}$

Where the concentrations are the initial ones given. Putting in the numbers:

$\textsf{Q = 4.0 \times {10}^{- 5} \times 1.0 \times {10}^{- 5} = 4.0 \times {10}^{- 10} \textcolor{w h i t e}{x} {\text{mol"^2."l}}^{- 2}}$

You can see that $\textsf{Q > {K}_{s p}}$. When this is the case this means that the position of equilibrium will be driven to the left.

This means that we would expect the precipitate of AgCl to form.