Question

Question #a3ed1

Answer 1

Yes.

Explanation:

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.