# Calculate the reaction quotient, Q, for mixing 100ml of 0.00028M "Pb"("NO"_3)_2 with 200ml of 0.0012M "NaCl". The K_(sp) for "PbCl"_2 is 1.9 xx10^(-4). Does a precipitate form?

## Calculate the ion product(a.k.a reaction quotient, Q) for mixing 100ml of 0.00028M Pb(NO3)2 with 200ml of 0.0012M Nacl. Ksp for PbCl2 is 1.9x10^-4. Does a precipitate form?

Jun 3, 2016

No, a precipitate does not form.

#### Explanation:

Your goal here is to determine if the concentrations of lead(II) cations, ${\text{Pb}}^{2 +}$, and chloride anions, ${\text{Cl}}^{-}$, are high enough to allow for the formation of lead(II) chloride, ${\text{PbCl}}_{2}$, an insoluble solid that precipitates out of solution.

Lead(II) nitrate, "Pb"("NO"_3)_2, dissociates in a $1 : 1$ mole ratio to form lead(II) cations

${\text{Pb"("NO"_ 3)_ (color(red)(2)(aq)) -> "Pb"_ ((aq))^(2+) + color(red)(2)"NO}}_{3 \left(a q\right)}^{-}$

For every mole of lead(II) nitrate present in the solution, you will have one mole of lead(II) cations. Use the molarity and volume of the solution to determine how many moles of lead(II) cations you have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

n_(Pb^(2+)) = "0.00028 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{P {b}^{2 +}} = 2.8 \cdot {10}^{- 5} {\text{moles Pb}}^{2 +}$

Do the same for the sodium chloride solution, $\text{NaCl}$

${\text{NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

You will have

n_(Cl^(-)) = "0.0012 mol" color(red)(cancel(color(black)("L"^(-1)))) * 200 * 10^(-3)color(red)(cancel(color(black)("L")))

${n}_{C {l}^{-}} = 2.4 \cdot {10}^{- 4} {\text{moles Cl}}^{-}$

Now, the volume of the resulting solution (assuming volumes are additive) will be

${V}_{\text{total}} = {V}_{P b {\left(N {O}_{3}\right)}_{2}} + {V}_{N a C l}$

${V}_{\text{total" = "100 mL" + "200 mL" = "300 mL}}$

The concentrations of the lead(II) cations and of the chloride anions (not yet incorporating ${\text{PbCl}}_{2}$) in the resulting solution will be

["Pb"^(2+)] = (2.8 * 10^(-5)"moles")/(300 * 10^(-3)"L") = 9.33 * 10^(-5)"M"

["Cl"^(-)] = (2.4 * 10^(-4)"moles")/(300 * 10^(-3)"L") = 8.0 * 10^(-4)"M"

Now, lead(II) chloride dissociates according to the following equilibrium reaction

${\text{PbCl"_ (color(red)(2)(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + color(red)(2)"Cl}}_{\left(a q\right)}^{-}$

Its solubility product constant, ${K}_{s p}$, is equal to

${K}_{s p} = {\left[{\text{Pb"^(2+)] * ["Cl}}^{-}\right]}^{\textcolor{red}{2}}$

Keep in mind that the solubility product constant is calculated using equilibrium concentrations. In order to determine whether or not a precipitate will form, you need to use the initial concentrations of the two ions.

This is what the reaction quotient, ${Q}_{s p}$, is used for

Q_(sp) = ["Pb"^(2+)]_"initial" * ["Cl"^(-)]_"initial"^color(red)(2)

A precipitate will form if

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{Q}_{s p} > {K}_{s p}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to find

${Q}_{s p} = 9.33 \cdot {10}^{- 5} \cdot {\left(8.0 \cdot {10}^{- 4}\right)}^{\textcolor{red}{2}}$

${Q}_{s p} = 6.0 \cdot {10}^{- 11}$ $\text{ << }$ $\stackrel{{K}_{s p}}{\overbrace{1.9 \times {10}^{- 4}}}$

Since ${Q}_{s p} < {K}_{s p}$, and actually quite significantly smaller, a precipitate will not be formed when these two solutions are mixed.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{Q}_{s p} < {K}_{s p} \implies \text{NO precipitate}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$