# Question d03db

Nov 22, 2017

Yes, a precipitate would form.

#### Explanation:

You know that silver nitrate and sodium chloride solutions can be mixed to produce silver chloride, $\text{AgCl}$, an insoluble ionic compound that can precipitate out of the solution if the concentrations of the silver(I) cations and of the chloride anions are high enough.

You also know that for

${\text{AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

you have a solubility product constant, ${K}_{s p}$, equal to

${K}_{s p} = \left[{\text{Ag"^(+)] * ["Cl}}^{-}\right] = 1.8 \cdot {10}^{- 10}$

In order for a precipitate to form you need to have

$\left[{\text{Ag"^(+)] * ["Cl}}^{-}\right] \ge {K}_{s p}$

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

${V}_{\text{total" = "45 mL + 85 mL = 130 mL}}$

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

$0.1558 \cdot 0.008827 \text{ >> } 1.8 \cdot {10}^{- 10}$

So all you have to do now is to figure out which of the two reactants acts as a limiting reagent.

${\text{Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-) -> "AgCl}}_{\left(s\right)}$

Since you have a $1 : 1$ mole ratio between the silver(I) cations and the chloride anions, and since you have fewer moles of chloride anions present in the solution, you can say that the chloride anions will act as the limiting reagent here.

The reaction will consume all the moles of chloride anions present in the solution and leave you with

${\text{0.02025 moles " - " 0.0011475 moles" = "0.0191 moles Ag}}^{+}$

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

${\text{AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

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 $\to$ think the common ion effect here.

If you take $s$ to be the molar solubility of the salt in the resulting solution, you can say that you have

$1.8 \cdot {10}^{- 10} = \left(0.15 + s\right) \cdot s$

$s = 1.2 \cdot {10}^{- 9}$

which means that the resulting solution has

$\left[{\text{Cl}}^{-}\right] = 1.2 \cdot {10}^{- 9}$ $\text{M}$

Once again, the answer is rounded to two sig figs.

Nov 22, 2017

Yes, a precipitate would form. ["Ag"^"+"] = "0.147 mol/L". ["Cl"^"-"] = 1.2 × 10^"-9" "mol/L"

#### Explanation:

The equation for the reaction is

$\text{Ag"^"+" + "Cl"^"-" → "AgCl}$

Step 1. Calculate the initial moles of $\text{Ag"^"+}$ and $\text{Cl"^"-}$

$\text{Moles of Ag"^"+" = 0.045 color(red)(cancel(color(black)("L Ag"^"+"))) × "0.45 mol Ag"^"+"/(1 color(red)(cancel(color(black)("L Ag"^"+")))) = " 0.0202 mol Ag"^"+}$

$\text{Moles of Cl"^"-" = 0.085 color(red)(cancel(color(black)("L Cl"^"-"))) × "0.0135 mol Cl"^"-"/(1 color(red)(cancel(color(black)("L Cl"^"-")))) = "0.001 15 mol Cl"^"-}$

Step 2. Calculate the amounts of each ion that will react

$\textcolor{w h i t e}{m m m m m m} \text{Ag"^"+" +color(white)(mll) "Cl"^"-" color(white)(m)→ "AgCl}$
$\text{I/mol} : \textcolor{w h i t e}{m m} 0.0202 \textcolor{w h i t e}{m l l} 0.001 15$
$\text{C/mol": color(white)(m)"-0.00115"color(white)(m)"-0.00115}$
$\text{E/mol} : \textcolor{w h i t e}{m l} 0.0191 \textcolor{w h i t e}{m m} 0$

Step 3. Calculate the concentration of $\text{Ag"^"+}$ after mixing

$\text{Total volume = 45 mL + 85 mL = 130 mL = 0.130 L}$

["Ag"^"+"] = "0.0191 mol"/"0.130 L" = "0.147 mol/L"

So, we have a precipitate of $\text{AgCl}$ in equilibrium with 0.147 mol/L $\text{Ag"^"+}$.

Step 4. Calculate the equilibrium concentration of $\text{Cl"^"-}$

$\textcolor{w h i t e}{m m m m m m} \text{AgCl" ⇌ "Ag"^"+" + "Cl"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m m m l l} 0.147 \textcolor{w h i t e}{m m} 0$
$\text{C/mol·L"^"-1":color(white)(mmmmm)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1":color(white)(mmmll)"0.147+} x \textcolor{w h i t e}{m l} x$

K_text(sp) = ["Ag"^"+"]["Cl"^"-"] = x(0.147+x) = 1.8 × 10^"-10"

Because ${K}_{\textrm{s p}}$ is so small, x ≪ 0.147.

Then,

0.147x = 1.8 × 10^"-10"

x = (1.8 × 10^"-10") /0.147 = 1.2 × 10^"-9"

["Cl"^"-"] = 1.2 × 10^"-9" "mol/L"#