Question #b716d

1 Answer
Mar 27, 2016

Answer:

#"446 mL"#

Explanation:

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

#color(red)(2)"NaI"_text((aq]) + "Hg"("NO"_3)_text(2(aq]) -> "HgI"_text(2(s]) darr + 2"NaNO"_text(3(aq])#

Sodium iodide, #"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])# #darr# #+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

#color(red)(2)"I"_text((aq])^(-) + "Hg"_text((aq])^(2+) -> "HgI"_text(2(s]) darr#

So, in order to precipitate one mole of mercury(II) cations, #"Hg"^(2+)#, you need #color(red)(2)# moles of iodide anions, #"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

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

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

#n_(Hg(NO_3)_2) = "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_(Hg^(2+)) = "0.30744 moles"#

Use the #color(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_(NaI) = "0.6149 moles"#

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(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_(NaI) = color(green)(|bar(ul(color(white)(a/a)"446 mL"color(white)(a/a)|)))#