Question #d03db
2 Answers
Yes, a precipitate would form.
Explanation:
You know that silver nitrate and sodium chloride solutions can be mixed to produce silver chloride,
You also know that for
#"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#
you have a solubility product constant,
#K_(sp) = ["Ag"^(+)] * ["Cl"^(-)] = 1.8 * 10^(-10)#
In order for a precipitate to form you need to have
#["Ag"^(+)] * ["Cl"^(-)] >= K_(sp)#
So, start by calculating the number of moles of each ion. For the silver(I) cations, you will have
#45 color(red)(cancel(color(black)("mL solution"))) * "0.45 moles Ag"^(+)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.02025 moles Ag"^(+)#
For the chloride anions, you have
#85 color(red)(cancel(color(black)("mL solution"))) * "0.0135 moles Cl"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0011475 moles Cl"^(-)#
After you mix the two solutions, the total volume of the resulting solution will be
#V_"total" = "45 mL + 85 mL = 130 mL"#
This means that the concentrations of the two ions in the resulting solution will be
#["Ag"^(+)] = "0.02025 moles"/(130 * 10^(-3)color(white)(.)"L") = "0.1558 mol L"^(-1)#
#["Cl"^(-)] = "0.0011475 moles"/(130 * 10^(-3)color(white)(.)"L") = "0.008827 mol L"^(-1)#
At this point, it should be clear that precipitation does occur because you have
#0.1558 * 0.008827 " >> " 1.8 * 10^(-10)#
So all you have to do now is to figure out which of the two reactants acts as a limiting reagent.
#"Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-) -> "AgCl"_ ((s))#
Since you have a
The reaction will consume all the moles of chloride anions present in the solution and leave you with
#"0.02025 moles " - " 0.0011475 moles" = "0.0191 moles Ag"^(+)#
The concentration of the silver(I) cations will be
#["Ag"^(-)] = "0.0191 moles"/(130 * 10^(-3)color(white)(.)"L") = "0.15 mol L"^(-1)#
The answer is rounded to two sig figs.
You could calculate the equilibrium concentration of chloride anions in the resulting solution by using the dissociation equilibrium of the silver chloride
#"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#
but the fact that the solution contains excess silver(I) cations means that the concentration of the chloride anions in the solution is very, very low
If you take
#1.8 * 10^(-10) = (0.15 + s) * s#
This quadratic equation produces
#s = 1.2 * 10^(-9)#
which means that the resulting solution has
#["Cl"^(-)] = 1.2 * 10^(-9)# #"M"#
Once again, the answer is rounded to two sig figs.
Yes, a precipitate would form.
Explanation:
The equation for the reaction is
#"Ag"^"+" + "Cl"^"-" → "AgCl"#
Step 1. Calculate the initial moles of
Step 2. Calculate the amounts of each ion that will react
Step 3. Calculate the concentration of
So, we have a precipitate of
Step 4. Calculate the equilibrium concentration of
Because
Then,
∴