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Featured 4 months ago

Because polarity is a continuum, and solubility lies somewhere on that continuum.......

Ethanol is miscible in water in all concentrations; and so is methanol. On the other hand petroleum ether, hexanes, is completely miscible in ethanol, whereas, believe it or not, methanol is IMMISCIBLE in hexanes. By the same token, should we have an aqueous solution of brine; addition of ethanol would precipitate the brine.

Why? Well, clearly ethanol has polar and non-polar functionality. The hydroxyl group promotes water solubility, whereas the hydrocarbyl tail allows some solubility in a non-polar solvent such as hexanes. The effect of the hydroxyl group is quite startling with respect to volatility. Ethanol has a normal boiling point of

Throw in some polarity (but not hydrogen bonding), and we find that

As regards the use of ethanol as a solvent for non-polar solutes, it is one of the most useful solvents in the laboratory, and it is probably the first solvent we would turn to for recrystallization of organic solutes. Organic solutes tend to have some solubility in HOT ethanol, whereas, upon cooling, the solutes tend to crystallize out - which is arguably the effect of the hydroxyl group (and thus ethanol is a preferred solvent to recrystallize such solutes).

Ethanol is also none too flammable; certainly less so than hexanes; I would happily use a heat gun on an ethanolic solution, whereas I would think twice about using it on a hexanes solution. As an important bonus, ethanol smells nice, and is not too hard on your hands.

Just to add for clarity. The ethanol we use in a lab is often known supplied as industrial methylated spirit. Why? Because when it is supplied it is adulterated with so-called denaturants such as phenol, or methanol (hence

Featured 4 months ago

It depends on how concentrated your solution is and what it contains. It becomes simplest for ideally-dilute solutions containing strong electrolytes wherein we ignore ion pairing.

If we do that, then the **ionic strength** in terms of *molality* is given as follows (*Physical Chemistry, Levine, pg. 312*):

#I_m = 1/2sum_i z_i^2m_i# where for a strong electrolyte represented by

#M_(nu_(+))X_(nu_(-))(s) stackrel(H_2O(l)" ")(->) nu_(+)M^(z_(+))(aq) + nu_(-)X^(z_(-))(aq)# ,

we have:

#m_i# is themolalityof theentire#i# th strong electrolyte.For instance, a cation within

#"HCl"# would have a molality of#m_+ = nu_+m_i# , and an anion within#"HCl"# would have a molality of#m_(-) = nu_(-)m_i# . The#m_i# would then correspond to#"HCl"# .In other words, the stoichiometry of the ion gives its contribution factor.

#z_i# is its charge.For instance, a cation would have a charge

#z_(+)# and anion would have a charge of#z_(-)# . This charge has both magnitude and sign, but the sign goes away by squaring.

**EXAMPLE: HCl**

Then for simplicity, first consider a solution containing only **one** strong electrolyte

Its **ionic strength** is given by:

#I_m = 1/2 (z_(H^(+))^2m_(H^(+)) + z_(Cl^(-))^2m_(Cl^(-)))#

#= 1/2(z_(H^(+))^2nu_(H^(+))m_(HCl) + z_(Cl^(-))^2nu_(Cl^(-))m_(HCl))#

But since they are the same charge magnitudes,

#=> 1/2(|z_(H^(+))z_(Cl^(-))|nu_(H^(+))m_(HCl) + |z_(H^(+))z_(Cl^(-))|nu_(Cl^(-))m_(HCl))#

#= 1/2|z_(H^(+))z_(Cl^(-))|(nu_(H^(+))m_(HCl) + nu_(Cl^(-))m_(HCl))#

#= 1/2|z_(H^(+))z_(Cl^(-))|(nu_(H^(+)) + nu_(Cl^(-)))m_(HCl)#

And so,

#color(blue)(I_m) = 1/2|1 cdot -1| cdot (1 + 1) cdot "0.01 m" = color(blue)("0.01 mol solute/kg solvent")#

*That should be no surprise. A 1:1 electrolyte should have the same molality in solution as it would before it dissociates.*

**EXAMPLE: THREE-ELECTROLYTE SOLUTION**

Now let's say you had a more complicated solution. Let's say we had **three** strong electrolytes:

#"0.01 m"# #"NaCl"# #"0.02 m"# #"HCl"# #"0.03 m"# #"BaCl"_2#

Again, ignoring ion pairing. We instead get for our initial **ionic strength** expression:

#I_m = 1/2sum_i z_i^2 m_i#

#= 1/2[z_(Na^(+))^2m_(Na^(+)) + z_(H^(+))^2m_(H^(+)) + z_(Ba^(2+))^2m_(Ba^(2+)) + z_(Cl^(-))^2m_(Cl^(-))]#

We should note that the chloride comes from all three electrolytes, so

#"NaCl"(aq) -> "Na"^(+)(aq) + "Cl"^(-)#

#"HCl"(aq) -> "H"^(+)(aq) + "Cl"^(-)(aq)#

#"BaCl"_2(aq) -> "Ba"^(2+)(aq) + 2"Cl"^(-)(aq)#

This means:

#I_m = 1/2[|z_(+)z_(-)|m_(Na^(+)) + |z_(+)z_(-)|m_(H^(+)) + z_(Ba^(2+))^2m_(Ba^(2+)) + |z_(+)z_(-)|m_(Cl^(-))]#

#= 1/2[|z_(+)z_(-)|nu_(Na^(+))m_(NaCl) + |z_(+)z_(-)|nu_(H^(+))m_(HCl) + z_(Ba^(2+))^2nu_(Ba^(2+))m_(BaCl_2) + |z_(+)z_(-)|(nu_(Cl^(-))m_(NaCl) + nu_(Cl^(-))m_(HCl) + nu_(Cl^(-))m_(BaCl_2))]#

From this we then get:

#color(blue)(I_m) = 1/2[|1 cdot -1|cdot 1 cdot "0.01 m NaCl" + |1 cdot -1|cdot 1 cdot "0.02 m HCl" + 2^2 cdot 1 cdot "0.03 m BaCl"_2 + |1 cdot -1|(1 cdot "0.01 m NaCl" + 1 cdot "0.02 m HCl" + 2 cdot "0.03 m BaCl"_2)]#

#=# #color(blue)("0.12 mol solutes/kg solvent")#

That's the rigorous way to do it. You could also just assign the concentration of the electrolyte to the ion, and multiply by its charge squared, then add it all up based on the concentration each electrolyte contributes.

#color(blue)(I_m) = 1/2[1^2 cdot 1 cdot "0.01 m Na"^(+) + {1^2 cdot 1 cdot "0.01 m Cl"^(-) + 1^2 cdot 1 cdot "0.02 m Cl"^(-) + 1^2 cdot 2 cdot "0.03 m Cl"^(-)} + 1^2 cdot 1 cdot "0.02 m H"^(+) + 2^2 cdot 1 cdot "0.03 m Ba"^(2+)]#

#=# #color(blue)("0.12 mol solutes/kg solvent")#

Featured 3 months ago

There is none. You need the wave function for the particular orbital, which is not readily available unless the atom is hydrogen...

In the case of **hydrogenic atoms**, i.e. one-electron atoms, the wave functions are available in most physical chemistry textbooks up through

As a simple but not undercomplicated example, consider one of the

The hydrogenic atom wave function for the

#psi_(2pz) = R_(21)(r)Y_(1)^(0)(theta,phi)#

#= 1/(4sqrt(2pi)) (Z/a_0)^(3//2) sigmae^(-sigma//2)costheta# ,where:

#sigma = Zr//a_0# .#Z# is the atomic number.#a_0 = "0.0529177 nm"# is the Bohr radius.#r# is the radial distance away from the nucleus.#R_(nl)(r)# is the radial component of the wave function.#Y_(l)^(m_l)(theta,phi)# is the angular component of the wave function.

The radial nodes are where

Since nonzero constants are clearly never zero, we just have to pick out the functions that have

#R_(21)(r) prop sigmae^(-sigma//2)#

#Y_(1)^(0)(theta,phi) prop costheta#

**RADIAL NODES?**

So, solving for the radial nodes, with

#(Zr)/(a_0)e^(-Zr//2a_0) = 0#

This exponential never goes to zero except at

#=> (Zr)/(a_0) = 0#

From here we can realize that there are **no radial nodes** in the

The only solution to this is

**ANGULAR NODES?**

Solving for the angular nodes,

#costheta = 0#

And this only nonredundantly applies for

So, in this domain,

Furthermore, there is no

#"Probability Density of 2p"_z#

#= int_(0)^(2pi) int_(0)^(pi) int_(0)^(oo) psi_(2pz)^"*"psi_(2pz)r^2dr sinthetad theta d phi#

#= color(red)(1/(4sqrt(2pi)) (Z/a_0)^(5) int_(0)^(2pi) d phi) int_(0)^(pi) cos^2theta sintheta d theta int_(0)^(oo) r^2e^(-Zr//a_0)dr#

#= color(red)((2pi)/(4sqrt(2pi))(Z/a_0)^(5)) int_(0)^(pi) cos^2theta sintheta d theta int_(0)^(oo) r^2e^(-Zr//a_0)dr#

Having no dependence on the value of

Therefore, the **angular node** is a plane perpendicular to the *plane*, just like it shows in the above diagram.

Featured 2 months ago

Here's what's going on here.

You're actually dealing with a **disproportionation reaction** here. In a disproportionation reaction, the *same element* undergoes **both** oxidation and reduction.

In this case, manganese(VI) is **reduced** to manganese(IV) and **oxidized** to manganese(VII).

#stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + stackrel(color(blue)(+4))("Mn")"O"_ (2(s))#

The **reduction half-reaction** looks like this

#stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s))#

Here each atom of manganese takes in **electrons**, which is why the oxidation number of manganese goes from

To balance the atoms of oxygen, use the fact that this reaction takes place in an *acidic medium* and add water molecules to the side that needs oxygen and protons,

#4"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l))#

Notice that the half-reaction is balanced in terms of **charge** because you have

#4 xx (1+) + (2-) + 2 xx (1-) = 0 + 0#

The **oxidation half-reaction** looks like this

#stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + "e"^(-)#

This time, each atom of manganese loses **electron**, which is why the oxidation number of manganese goes from

The atoms of oxygen are already balanced, so you don't need to use water molecules and protons. Once again, the half-reaction is balanced in terms of **charge** because you have

#(2-) = (1-) + (1-)#

So, you know that the balanced half-reactions look like this

#{(4"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l))), (color(white)(aaaaaaaaaaaaaa)stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + "e"^(-)) :}#

In every redox reaction, the number of electrons lost in the oxidation half-reaction must be **equal** to the number of electrons gained in the reduction half-reaction, so multiply the oxidation half-reaction by

#{(4"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l))), (color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + 2"e"^(-)) :}#

Add the two half-reactions to find the balanced chemical equation that describes this disproportionation reaction.

#{(4"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l))), (color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + 2"e"^(-)) :}#

#color(white)(a)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#

#4"H"_ ((aq))^(+) + ["MnO"_ (4(aq))^(2-) + 2"MnO"_ (4(aq))^(2-)] + color(red)(cancel(color(black)(2"e"^(-)))) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + color(red)(cancel(color(black)(2"e"^(-)))) + 2"H"_ 2"O"_ ((l))#

You will end up with

#4"H"_ ((aq))^(+) + 3"MnO"_ (4(aq))^(2-) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + 2"H"_ 2"O"_ ((l))#

Featured 2 weeks ago

Depends on the system... but if we are telling the whole truth, then it would be a zero minimum energy that isn't possible.

Even though the rigid rotator has zero energy in the ground state, it is not violating the Uncertainty Principle because it also translates in space and its total energy is unstated.

- An
**electron in a one-dimensional box**of length#a# CANNOT have zero energy:

#E_n = (n^2pi^2ℏ^2)/(2ma^2)# ,#n = 1, 2, 3, . . . #

If it did, that would require that it have zero velocity, which means it has well-known momentum, which means it has POORLY known position, as per Heisenberg's Uncertainty Principle. But that means it ISN'T trapped in the box of length

#a# that it IS (by construction) trapped in.So we have concluded that zero energy

#-># the particle doesn't exist in the box, i.e. the system does not exist.

- A
**harmonic oscillator**CANNOT have zero energy:

#E_upsilon = ℏomega(upsilon + 1/2)# ,#upsilon = 0, 1, 2, . . . # It has a zero-point energy of

#E_0 = 1/2ℏomega# , which is positive.

Zero energy would mean it has zero frequency, and thus that the oscillator is not moving. But again, that would suggest that its momentum is well-known, which would suggest its position is NOT, as per Heisenberg's Uncertainty Principle... even when it is not moving and its position IS well-known.

Thus, we have a contradiction and the system cannot exist if it is to have

#E = 0# .

- A
**rigid rotator**CAN have zero energy (but this is a straight-up lie), and does NOT violate the Heisenberg Uncertainty Principle as we can only calculatefor this system (hence, nothing is stated about the translational kinetic energy, which can be nonzero, so that momentum is nonzero):*rotational energy*

#E_J = hcBJ(J+1)# ,#J = 0, 1, 2, 3, . . . #

This is confidently stated here in this book. I believe it, as again, the translational energy from the Schrodinger equation of this system has to be stated by the person who constructed the system.

At a rotational quantum level

#J = 0# , the ground-stateenergy isrotational#E_0 = hcBJ(J+1) = 0# .However, the wave function at

#J = 0# is a constant (zero angular momentum), and being only a function of rotation angles, does not give information about the linear momentum (i.e. the translational kinetic energy is unstated).This is because in solving the rigid rotator system, we separate out the total wave function into an arbitrary

translationalpart#psi(r)# (which just gives a phase factor to the total wave function#psi(r,theta,phi# )) and a solvablerotationalpart#psi(theta,phi)# .Hence, we are

notclaiming certainty about linear momentum or position, which isnotbreaking the Uncertainty Principle.

Featured 6 days ago

Method

The blue solution that forms due to dissolving the dime in

#2("NO"_3^(-)(aq) + 4"H"^(+)(aq) + cancel(3e^(-)) -> "NO"(g) + 2"H"_2"O"(l))#

#3ul(("Cu"(s) -> "Cu"^(2+)(aq) + cancel(2e^(-)))" "" "" "" "" "" "" "" ")#

#2"NO"_3^(-)(aq) + 3"Cu"(s) + 8"H"^(+)(aq) -> 3"Cu"^(2+)(aq) + 2"NO"(g) + 4"H"_2"O"(l)#

A

*These both occur because nitrate reduction is more favorable than silver and copper reduction (its* *is more positive).*

If

#"50.00 mL"# of the original#"100.00 mL"# in the volumetric flask (which was made to contain#"100.00 mL"# !) was titrated with excess#"NaCl"# , then all of the#"Ag"^(+)# will precipitate with the#"Cl"^(-)# in the form of#"AgCl"(s)# .Upon washing, filtering, drying, weighing, drying, weighing, drying, etc... a mass of

#"1.375 g"# recovered of#"AgCl"# would suggest that:

#1.375 cancel"g AgCl" xx "107.8682 g Ag"/(143.32 cancel"g AgCl") = "1.035 g Ag"^+# And this mass of silver cation was what reacted with the

#"NaCl"# from within the#"50.00 mL"# aliquot.Therefore, since volume and mass are

, we really haveextensive#ul("2.070 g Ag"^(+))# dissolved in the original flask volume, and as a result, thepercent by massof#"Ag"# metal in the alloy is:

#"2.070 g Ag"/"2.490 g alloy" = color(blue)(83.12% "w/w Ag")#

Errors in this method include:

- loss of solid in filtration into the Erlenmeyer flask
- loss of solid due to not washing all of it into the filter paper
- dropping solid flakes of
#"AgCl"# on the groundAll of these lead to a less than

#100%# mass yield.

Using the given standard curve from the data previously collected:

#ul("Conc. (mg/mL)"" ""A""bsorbance")#

#5.00" "" "" "" "" "" "0.843#

#4.00" "" "" "" "" "" "0.672#

#3.00" "" "" "" "" "" "0.507#

#2.00" "" "" "" "" "" "0.336#

#1.00" "" "" "" "" "" "0.161# we would construct an Excel graph to obtain

#y = 0.1700x - 0.0062# for the Beer's law fit line equation:

And hence, with the absorbance of

#"Cu"^(2+)# being#0.420# for#lambda_(max) = "740 nm"# (absorbing red light, reflecting blue), we useBeer's Lawto get:

#A = epsilonbc + y"-int"#

#=> color(blue)(c) = (A - y"-int")/(epsilonb)#

#= (0.420 -(- 0.0062))/(0.1700) "mg"/"mL"#

#=# #"2.507 mg Cu"^(2+)"/mL"#

#=# #"250.7 mg Cu"^(2+)"/100 mL"#

#=# #color(blue)("0.2507 g Cu"^(2+)"/100 mL")# By subtraction,

#"2.490 g" - "0.2507 g Cu" = "2.239 g Ag"# Thus,

#"2.239 g Ag"/"2.490 g alloy" = color(blue)(89.93% "w/w Ag")#

This method is not void of errors. **They include:**

- Wrong choice of wavelength (it is a reasonable wavelength)
- Calibrating with the measured sample each time, making all the absorbance values too low (yes, students do this)
- Spilling sample into the spectrophotometer, saturating the absorbance readings (quite possible)

Of course, if all of these are done right, then this is quite accurate.

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