For part (a) No, a precipitate will not form.

I will show you how to do point (a), and you solve point (b) as practice.

So, you’re mixing two that contain soluble and you want to figure out if a precipitate will form.

Silver nitrate and potassium chloride will dissociate completely in aqueous solution to produce silver and potassium cations, on one hand, and nitrate and chloride anions, on the other.

If and only if the concentrations of the silver cations and chloride anions are high enough, silver chloride, an insoluble ionic compound, will precipitate out of solution.

Now, you know that silver chloride has a equal to ##1.6 * 10^(-10##. The dissociation of silver chloride in aqueous solution is represented by an equilibrium

##”AgCl”_text((s]) rightleftharpoons “Ag”_text((aq])^(+) + “Cl”_text((aq])^(-)##

The nitrate anions and potassium ations are spectator ions, so you can ignore them altogether.

By definition, ##K_(sp)## is equal to

##K_(sp) = [“Ag”^(+)] * [“Cl”^(-)]##

This tells you that if the concentration of silver cations multiplied by the concentration of the chloride anions is at least equal to ##K_(sp)##, a precipitate will be formed.

If these two concentrations do not satisfy this inequality

##K_(sp) <= [“Ag”^(+)] * [“Cl”^(-)]##

then a precipitate will not be formed.

So, start by calculating the total volume of the solution

##V_”total” = V_(AgNO_3) + V_(KCl)##

##V_”total” = “200 mL” + “900 mL” = “1100 mL”##

Use the molarities and the initial volumes of the two solutions to determine the number of moles of each compound

##color(blue)(C = n/V implies n = C * V)##

##n = 1.0 * 10^(-4)”M” * 200 * 10^(-3)”L” = 2.0 * 10^(-5)”moles Ag”^(+)##

and

##n = 1.0 * 10^(-6)”M” * 900 * 10^(-3)”L” = 9.0 * 10^(-7)”moles Cl”^(-)##

The molarities of the silver cations and chloride anions in the final solution will be

##[“Ag”^(+)] = (2.0 * 10^(-5)”moles”)/(1100 * 10^(-3)”L”) = 1.82 * 10^(-5)”M”##

and

##[“Cl”^(-)] = (9.0 * 10^(-7)”moles”)/(1100 * 10^(-3)”L”) = 8.18 * 10^(-7)”M”##

This means that you have

##[“Ag”^(+)] * [“Cl”^(-)] = 1.82 * 10^(-5) * 8.18 * 10^(-7) = 1.5 * 10^(-11)##

Since this value is smaller than ##K_(sp) = 1.6 * 10^(-10)##, a precipitate will not form in the final solution.

Our website has a team of professional writers who can help you write any of your homework. They will write your papers from scratch. We also have a team of editors just to make sure all papers are of HIGH QUALITY & PLAGIARISM FREE. To make an Order you only need to click Ask A Question and we will direct you to our Order Page at WriteDemy. Then fill Our Order Form with all your assignment instructions. Select your deadline and pay for your paper. You will get it few hours before your set deadline.