Question #9c1c3

1 Answer
Mar 20, 2016

Answer:

Here's what I got.

Explanation:

!! LONG ANSWER !!

The idea here is that the magnesium hydroxide, #"Mg"("OH")_2#, will react with the glacial acetic acid, #"CH"_3"COOH"#, to form magnesium acetate, #("CH"_3"COO")_2"Mg"#, and water.

This neutralization reaction will actually consume some of the acetic acid, which means that the resulting solution will contain less solvent than what you start with. Keep this in mind.

The balanced chemical equation for this reaction looks like this

#"Mg"("OH")_text(2(s]) + color(red)(2)"CH"_3"COOH"_text((l]) -> 2"CH"_3"COO"^(-) + "Mg"^(2+) + 2"H"_2"O"_text((l])#

Now, the freezing point of a solution depends on the concentration of particles of solute present in solution.

In your case, the particles of solute will include the acetate anion, #"CH"_3"COO"^(-)#, the magnesium cation, #"Mg"^(2+)#, and water.

In essence, you're looking for any molecule or ion that is not acetic acid.

Mathematically, you can express the freezing-point depression of a solution by using the equation

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "#, where

#DeltaT_f# - the freezing-point depression;
#i# - the van't Hoff factor
#K_f# - the cryoscopic constant of the solvent;
#b# - the molality of the solution.

The van't Hoff factor tells you how many moles of particles of solute you get per mole of solute dissolved in the solvent.

In your case, one mole of magnesium hydroxide will produce

  • two moles of acetate anions, #2 xx "CH"_3"COO"^(-)#
  • one mole of magnesium cations, #1 xx "Mg"^(2+)#
  • two moles of water, #2 xx "H"_2"O"#

Since you get a total of five moles of particles for every one mole of sodium hydroxide, the van't Hoff factor will be #i=color(red)(5)#.

Use the molar masses of magnesium hydroxide and acetic acid (I'll use #"HAc"# for simplicity) to calculate how many moles of each you're mixing

#2.3 color(red)(cancel(color(black)("g"))) * ("1 mole Mg"("OH")_2)/(58.32color(red)(cancel(color(black)("g")))) = "0.03944 moles Mg"("OH")_2#

#25.0 color(red)(cancel(color(black)("g"))) * ("1 mole HAc")/(60.05color(red)(cancel(color(black)("g")))) = "0.4163 moles HAc"#

The reaction consumes #color(red)(2)# moles of acetic acid for every #1# mole of magnesium hydroxide, so use this mole ratio to determine how many moles of acetic acid will be consumed by the reaction

#0.03944color(red)(cancel(color(black)("moles Mg"("OH")_2))) * (color(red)(2)color(white)(a)"moles HAc")/(1color(red)(cancel(color(black)("mole Mg"("OH")_2)))) = "0.07888 moles HAc"#

This means that the resulting solution will only contain

#n_(CH_3COOH) = 0.4163 - 0.07888 = "0.3374 moles HAc"#

This will be equivalent to

#0.3374color(red)(cancel(color(black)("moles HAc"))) * "60.05 g"/(1color(red)(cancel(color(black)("mole HAc")))) = "20.26 g"#

To get the molality of the solution, use the number of moles of magnesium hydroxide and the mass of acetic acid that remains in the resulting solution.

As you know, to get a solution's molality you need to use number of moles of solute and the mass of the solvent expressed in kilograms.

#color(blue)(|bar(ul(color(white)(a/a)"molality" = "number of moles of solute"/"kilograms of solution" xx 100color(white)(a/a)|)))#

In your case, you will have

#b = "0.03944 moles"/(20.26 * 10^(-3)"kg") = "1.947 mol kg"^(-1)#

Plug your values into the equation for freezing-point depression

#DeltaT_f = color(red)(5) * 3.90""^@"C" color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 1.947color(red)(cancel(color(black)("mol")))color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_f = 37.97^@"C"#

The freezing-point depression is defined as

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "#, where

#T_f^@# - the freezing point of the pure solvent
#T_"f sol"# - the freezing point of the solution

The freezing point of the solution will thus be

#T_"f sol" = T_f^@ - DeltaT_f#

#T_"f sol" = 16.6^@"C" - 37.97^@"C" = color(green)(|bar(ul(color(white)(a/a)-21.4^@"C"color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.