A sample containing 0.0500 mole of Fe2(SO4)3 is dissolved in enough water to make 1.00 L of solu

step 1 I think I've written everything down here. We are asked to consider this equation that I've writ

A sample containing 0.0500 mole of Fe2(SO4)3 is dissolved in...

A sample containing 0.0500 mole of $\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$ is dissolved in enough water to make 1.00 L of solution. This solution contains hydrated $\mathrm{SO}_4{ }^{2-}$ and $\mathrm{Fe}^{3+}$ ions. The latter behaves as an acid: $$ \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6{ }^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q) $$ a. Calculate the expected osmotic pressure of this solution at $25^{\circ} \mathrm{C}$ if the above dissociation is negligible. b. The actual osmotic pressure of the solution is 6.73 atm at $25^{\circ} \mathrm{C}$. Calculate $K_\alpha$ for the dissociation reaction of $\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6{ }^{3+}$. (To do this calculation, you must assume that none of the ions go through the semipermeable membrane. Actually, this is not a great assumption for the tiny $\mathrm{H}^{+}$ion.)

A sample containing 0.0500 mole of $\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$ is dissolved in enough water to make 1.00 L of solution. This solution contains hydrated $\mathrm{SO}_4{ }^{2-}$ and $\mathrm{Fe}^{3+}$ ions. The latter behaves as an acid:
$$
\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6{ }^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q)
$$
a. Calculate the expected osmotic pressure of this solution at $25^{\circ} \mathrm{C}$ if the above dissociation is negligible.
b. The actual osmotic pressure of the solution is 6.73 atm at $25^{\circ} \mathrm{C}$. Calculate $K_\alpha$ for the dissociation reaction of $\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6{ }^{3+}$. (To do this calculation, you must assume that none of the ions go through the semipermeable membrane. Actually, this is not a great assumption for the tiny $\mathrm{H}^{+}$ion.)

Key Concepts

Equilibrium Constant

The equilibrium constant (K? in the context of partial dissociation) is a measure of the position of equilibrium in a chemical reaction. It represents the ratio of the concentrations of the products to the reactants once the reaction has reached equilibrium. In the discussion of an acid that undergoes partial dissociation, K? helps determine the fraction of the original species that ionizes, which in turn affects the overall colligative properties of the solution.

van ’t Hoff Factor

The van ’t Hoff factor (i) quantifies the effective number of particles into which a solute dissociates in solution. It is used in colligative property calculations to account for the increase in particle number due to dissociation or ionization. When a compound dissociates or undergoes a reaction in solution, the actual number of particles produced can be determined by multiplying the initial concentration by the van ’t Hoff factor.

Acid Dissociation Equilibrium

Acid dissociation equilibrium refers to the dynamic process where a weak acid partially ionizes in solution, establishing an equilibrium between the non-dissociated acid and its ions. The equilibrium is characterized by an equilibrium constant, which quantifies the extent of dissociation. This concept is applied when evaluating how the species in the solution, such as a hydrated metal ion acting as an acid, dissociate and alter the number of solute particles.

Osmotic Pressure

Osmotic pressure is a colligative property that depends solely on the number of solute particles in a solution rather than their identity. It is defined by the relationship ? = MRT, where M is the molarity of the solute, R is the gas constant, and T is the absolute temperature. This property is critical for understanding how solute concentration affects the movement of solvent through a semipermeable membrane.

Colligative Properties

Colligative properties are characteristics of solutions that depend on the number of dissolved particles rather than their chemical nature. These include osmotic pressure, boiling point elevation, freezing point depression, and vapor pressure lowering. In the context of this problem, the focus is on osmotic pressure, which illustrates how dissociation into more particles can enhance the observed effect.

Transcript

00:01 I think i've written everything down here.

00:02 We are asked to consider this equation that i've written right here.

00:06 Let me highlight that quick.

00:08 And we are asked to calculate the osmotic pressure, which is calculated as follows.

00:17 That will equal i, our van't hoff, times our molarity, times r, times t.

00:26 So my van't hoff will equal five on this one.

00:35 It's one, two, three, four, five.

00:38 So i'm going to have my osmotic pressure will equal five times the molarity, 0 .500 moles per liter, times 0 .08206 latm over k mole, times 25 degrees c plus 273.

01:07 And make a note this is 298.

01:11 So this will be, on my calculator, five times 0 .5 times 0 .08206 times 298.

01:22 And i get 61 .1 atmospheres.

01:34 There's my answer for a.

01:38 Now what's b ask us? move up here.

01:44 B, the actual osmotic pressure is 6 .73 atmospheres.

02:07 And again, at 25 degrees celsius.

02:13 Calculate the ka for the dissociation.

02:22 This is hard.

02:27 Reaction of, that's given, of your feh2o63 plus.

02:44 And we're told to make some assumptions that aren't great assumptions, but we can handle this.

02:52 Okay.

03:02 Let's calculate our van't hoff factor, excuse me, by taking our osmotic pressure, which is 6.

03:13 Oh, we've got to change this to bars.

03:18 Okay.

03:28 So to convert that to bars, i'm going to take my 6 .73 times 1 .01325 bars per atmosphere.

03:51 And this will equal 6 .73 times 1 .01325.

04:00 Whoopsie.

04:03 6 .73 times 1 .01325.

04:11 That is 6 .82.

04:17 Okay.

04:18 So i have 6 .82 bar.

04:24 Okay.

04:24 So far, so good.

04:26 And now i'm going to divide that by my molarity, which is 0 .500 moles per liter times, let's do 8 .3145 times 10 to the minus 2 l bar per k mole times 298 kelvin.

05:02 Okay.

05:05 So, okay.

05:17 So i have 6 .82 divided by that's an l not a 2 .5 divided by 8 .3145 times 10 to the minus 2 divided by 298.

05:49 I think that should be, yeah, that should be 10 to the minus 3.

05:54 My bad.

05:55 So i will be equal to 5 .51.

06:01 Okay.

06:02 So three of those, three of the 5 .51 are sulfate ions and 2 .51 are hydrated iron three plus ions and they act like an acid.

06:31 Okay...