# Question b716d

Mar 27, 2016

$\text{446 mL}$

#### Explanation:

Start by taking a look at the balanced chemical equation that describes this double replacement reaction

$\textcolor{red}{2} {\text{NaI"_text((aq]) + "Hg"("NO"_3)_text(2(aq]) -> "HgI"_text(2(s]) darr + 2"NaNO}}_{\textrm{3 \left(a q\right]}}$

Sodium iodide, $\text{NaI}$, and mercury(II) nitrate, "Hg"("NO"_3)_2, are soluble ionic compounds that dissociate completely in aqueous solution to form cations and anions.

This means that you can rewrite the balanced chemical equation as

color(red)(2) xx overbrace(("Na"_text((aq]) + "I"_text((aq])))^(color(blue)("NaI"_text((aq]))) + overbrace("Hg"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-))^(color(purple)("Hg"("NO"_3)_text(2(aq]))) -> "HgI"_text(2(s]) $\downarrow$ +2 xx overbrace(("Na"_text((aq]) + "NO"_text(3(aq])))^(color(black)("NaNO"_text(3(aq]))

The net ionic equation, which you get by eliminating spectator ions, i.e. ions that are present on both sides of the equation, looks like this

$\textcolor{red}{2} {\text{I"_text((aq])^(-) + "Hg"_text((aq])^(2+) -> "HgI}}_{\textrm{2 \left(s\right]}} \downarrow$

So, in order to precipitate one mole of mercury(II) cations, ${\text{Hg}}^{2 +}$, you need $\textcolor{red}{2}$ moles of iodide anions, ${\text{I}}^{-}$.

Use the molarity and volume of the mercury(II) nitrate solution to determine how many moles of mercury(II) nitrate were used to make the solution

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

Plug in your values to get - do not forget to convert the volume of the solution from milliliters to liters

${n}_{H g {\left(N {O}_{3}\right)}_{2}} = \text{1.68 mol" color(red)(cancel(color(black)("L"^(-1)))) * 183 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.30744 moles}$

Since mercury(II) nitrate dissociates in a $1 : 1$ mole ratio to form mercury(II) cations, you know that

${n}_{H {g}^{2 +}} = \text{0.30744 moles}$

Use the $\textcolor{red}{2} : 1$ mole ratio that exists between the two ions to find how many moles of iodide anions must be delivered to the solution

0.30744color(red)(cancel(color(black)("moles Hg"^(2+)))) * (color(red)(2)color(white)(a)"moles I"^(-))/(1color(red)(cancel(color(black)("mole Hg"^(2+))))) = "0.6149 moles I"^(-)

Sodium iodide dissociates in a $1 : 1$ mole ratio to form iodide anions, so you know that

${n}_{N a I} = \text{0.6149 moles}$

Now use the molarity of the sodium iodide solution to determine what volume would contain this many moles

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

You will have

V_(NaI) = (0.6149 color(red)(cancel(color(black)("moles"))))/(1.38color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.4456 L"#

Rounded to three sig figs and expressed in milliliters, the answer will be

${V}_{N a I} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{446 mL} \textcolor{w h i t e}{\frac{a}{a}} |}}}$