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Answer 1 · 2017-05-23T15:25:44

Yes.

At equilibrium we have:

$sf(AgCl_((s))rightleftharpoonsAg_((aq))^++Cl_((aq))^-)$

For which:

$sf(K_(sp)=[Ag_((aq))^+][Cl_((aq))^-]=1.8xx10^(-10)color(white)(x)"mol"^2."l"^(-2))$

Where the concentrations are those at equilibrium

The reaction quotient Q is given by:

$sf(Q=[Ag_((aq))^+][Cl_((aq))^-])$

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

$sf(Q=4.0xx10^(-5)xx1.0xx10^(-5)=4.0xx10^(-10)color(white)(x)"mol"^2."l"^(-2))$

You can see that $sf(Q>K_(sp))$ . 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.