Featured 2 months ago

Consider a sparingly soluble salt,

We can represent its solubility in water in the following way:

As for any equilibrium, we can write (and quantify) this equilibrium:

But

And,

If we say

i.e.

And thus

I leave it to you to solve for the solubility of lead chloride in water in

Featured 2 months ago

**Warning! Long Answer.** Here's how I would do it.

An individual molecule does not have a melting point or boiling point.

It is the bulk substance, which consists of many molecules, that has these physical properties.

**Pressure and melting point**

We can use the **Clausius equation** to calculate the effect of pressure on the melting point.

One form of the Clausius equation is

#color(blue)(bar(ul(|color(white)(a/a)(dp)/(dT) = (Δ_(fus)text(H))/(TΔv_text(fus))color(white)(a/a)|)))" "#

where

We can integrate the Clausius equation:

**Pressure and boiling point**

Chemists often use the **Clausius-Clapeyron equation** to calculate the vapour pressure of a liquid at different temperatures:

#color(blue)(bar(ul(|color(white)(a/a) ln(p_2/p_1) = (Δ_"vap"H)/R(1/T_1- 1/T_2)color(white)(a/a)|)))" "#

where

For example. the boiling point of water is 99.6 °C at 1 bar and its

We can calculate the boiling point at 6 bar.

At 6 bar, the boiling point of water is 159 °C.

Featured 1 month 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 1 month ago

Because rate constants are temperature-dependent, and by changing the temperature, you change the ratio of rate constants so that

Although you may be familiar with this in the context of this equation:

#DeltaG^@ = -RTlnK#

That isn't actually where the temperature dependence is readily seen. Consider the **Arrhenius equation**:

#k = Ae^(-E_a//RT)# where

#k# is the rate constant at temperature#T# ,#A# is the frequency factor of the reaction (i.e. the limiting rate constant at high temperature), and#E_a# is the activation energy.

Now, take two temperatures and two rate constants, assuming

#k_2/k_1 = "exp"(-E_a/R[1/T_2 - 1/T_1])#

#=> color(green)(ln(k_2/k_1) = -E_a/R[1/T_2 - 1/T_1])# where

#"exp"(x) = e^x# .

We currently have the relationship for two rate constants each at different temperatures. Now, think about where activation energy is on a reaction coordinate diagram:

We see that **reactant** to transition state for the ** forward** reaction, and

So now, consider that the rate of the forward reaction equals the rate of the reverse reaction at **equilibrium**. Take the reaction as:

#aA + bB rightleftharpoons cC + dD#

The **rate laws** would be:

#r_(fwd)(t) = k_(fwd)[A]^a[B]^b#

#r_(rev)(t) = k_(rev)[C]^c[D]^d#

So we set them equal:

#k_(fwd)[A]^a[B]^b = k_(rev)[C]^c[D]^d#

By definition, the **equilibrium constant** is:

#=> K -= ([C]^c[D]^d)/([A]^a[B]^b) = (k_(fwd))/(k_(rev))#

Now, take

#K_1 = (k_(1,fwd))/(k_(1,rev))# #" "" "# #K_2 = (k_(2,fwd))/(k_(2,rev))#

Substitute back into the Arrhenius equation for the forward and reverse reactions.

#ln(k_(2,fwd)/k_(1,fwd)) = -E_(a,fwd)/R[1/T_2 - 1/T_1]#

#ln(k_(2,rev)/k_(1,rev)) = -E_(a,rev)/R[1/T_2 - 1/T_1]#

Use the fact that

#ln(k_(2,fwd)/k_(1,fwd)) - ln(k_(2,rev)/k_(1,rev))#

#= ln(k_(2,fwd)/k_(1,fwd) cdot 1/(k_(2,rev)//k_(1,rev)))#

#= ln((k_(2,fwd)//k_(2,rev)) / (k_(1,rev)//k_(1,fwd)))#

#= ln(K_2/K_1)#

The right-hand side by subtraction becomes:

#-E_(a,fwd)/R[1/T_2 - 1/T_1] - (-E_(a,rev)/R[1/T_2 - 1/T_1])#

#= -(E_(a,fwd) - E_(a,rev))/R[1/T_2 - 1/T_1]#

But look up at the reaction coordinate diagram above. We can make the connection that

Therefore, we arrive at the **van't Hoff equation** for *equilibrium constants* vs. *temperature*:

#color(blue)(ln(K_2/K_1) = -(DeltaH_(rxn))/R[1/T_2 - 1/T_1])#

which looks strikingly like the Arrhenius equation(not a coincidence!).

Or put in another form,

#color(blue)(ln K = -(DeltaH_(rxn))/R 1/T + (DeltaS_(rxn))/R)#

Hence, we have shown that equilibrium constants certainly are temperature-dependent.

**Endothermic reactions have increased #K# at higher temperatures, and exothermic reactions have decreased #K# at lower temperatures.**

Featured 1 month ago

They are quite complicated, and can often do combinations of

For an introduction into these kinds of bonds:

The **three types** (here, we use the convention that outer atoms point their

**Two lobes** -

- The
#f_(z^3)# (#m_l = 0# ) is the only one that only#sigma# bonds. It can bond head-on along the#z# axis.

**Six lobes** -

- The
#f_(y(3x^2 - y^2))# (#m_l = -3# ) can#sigma# bond along the#x# axes (for example, with a#p_y# orbital) AND#pi# bond along the#y# axes (for example, with a#p_x# orbital, or a#d_(xy)# orbital).

It can alternatively form a

#phi# bond (a six-lobed side-on overlap) along the#xy# plane (with another#f_(y(3x^2 - y^2))# orbital in a bimetallic complex).

- The
#f_(x(x^2 - 3y^2))# (#m_l = +3# ) can#sigma# bond along the#y# axes (for example, with a#p_y# orbital) AND#pi# bond along the#x# axes (for example, with a#p_x# orbital, or a#d_(xy)# orbital).

It can alternatively form a

#phi# bond (a six-lobed side-on overlap) along the#xy# plane (with another#f_(x(x^2 - 3y^2))# orbital in a bimetallic complex).

- The
#f_(yz^2)# (#m_l = -1# ) can form decent#sigma# bonds along the#y# axes, AND/OR#pi# bonds along the#y# AND#z# axes.

It can alternatively form a

#phi# bond (a six-lobed side-on overlap) along the#yz# plane (with another#f_(yz^2)# orbital in a bimetallic complex).

- The
#f_(xz^2)# (#m_l = +1# ) can form decent#sigma# bonds along the#x# axes, AND/OR#pi# bonds along the#x# AND#z# axes.

It can alternatively form a

#phi# bond (a six-lobed side-on overlap) along the#xz# plane (with another#f_(xz^2)# orbital in a bimetallic complex).

**Eight lobes** -

- The
#f_(z(x^2 - y^2))# (#m_l = -2# ) is for#pi# bonding along ANY of the axes,#x,y# , or#z# . The lobes lie above and below each of the axes, but also along them.

It can alternatively form a

#delta# bond with another#f_(z(x^2 - y^2))# orbital in a bimetallic complex.

- The
#f_(xyz)# (#m_l = +2# ) is for#delta# bonding along ANY of the planes (#xz, yz, xy# ) (for example, with#d_(xy)# ,#d_(xz)# , or#d_(yz)# orbitals).

It can alternatively form a

#pi# bond with another#f_(xyz)# orbital in a bimetallic complex.

Featured 1 month ago

Image adapted from https://www.ck12.org/section/Galvanic-Cells/ [1]

From the reaction formula:

Copper is reduced as its oxidation state decreases from

Cobalt is oxidized; its oxidation state increases from

**Direction of electron flow**

An element gains electrons as it undergoes reduction and loses electron when it undergoes oxidation. Therefore there's going to be a flow of electron *from cobalt to copper* through the external circuit.

**Anode or cathode**

"The *cathode* is where the *reduction* take place and *oxidation* takes place at the *anode*". (Chemistry Libretexts) [2]

Cobalt is being oxidized to form cobalt (II) ions so the cobalt electrode would be the anode. Copper (II) ions are reduced to elementary copper at the copper electrode, so that would be the cathode.

The way I memorize this is by considering where the two names for the voltaic electrodes came from. The

**Positive or negative terminal**

Electrons always flow from the negative terminal to the positive terminal- exactly the opposite as the direction of the current flow. Therefore the cathode would be the positive terminal and the anode the negative terminal.

References

[1] "Galvanic Cells", CK-12 Editor, https://www.ck12.org/section/Galvanic-Cells/

[2] "Voltaic Cells", LibreText Chemistry, https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Voltaic_Cells

Featured 1 month 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.

Featured 3 weeks ago

(i)

This tells you that the van der Waals equation of state picks up on the attractive interactions in acetylene that the ideal gas law assumes is not present.

**(a) Ideal gas**

The equation for the **Ideal Gas Law** is

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

Then,

#p = (nRT)/V#

The pressure predicted by the Ideal Gas Law is **4.70 atm**.

**(b) van der Waals Equation**

The van der Waals equation is

#color(blue)(bar(ul(|color(white)(a/a) (P + (n^2a)/V^2)(V - nb) = nRTcolor(white)(a/a)|)))" "#

#P + (n^2a)/V^2 = (nRT)/(V - nb)#

#P = (nRT)/(V - nb)- (n^2a)/V^2#

For this problem,

The pressure predicted by the van der Waals equation is **4.58 atm**, lower than for the ideal gas law.

Featured 2 weeks ago

**Warning! Long Answer.**

We have here a disguised **Hess' Law problem** using equilibrium constants.

We have three equations:

From these, we must construct the target equation.

#"H"_2"(g)" + "O"_2"(g)" ⇌ "2HCl(g)"; K_text(p) = ?#

The target equation has **2.** and halve it.

When we reverse an equation, we **take the reciprocal** of its equilibrium constant.

When we halve an equation, we take **the square root** of its equilibrium constant.

The target equation has

We halve Equation **3.**

Equation **5.** has

We re-write Equation **1.**

Now, we add Equations **4.**, **5.**, and **6.**, cancelling terms that appear on opposite sides of the equation.

When you add equations, you **multiply** their equilibrium constants.

Featured 1 week ago

**Warning! Long Answer.** Here's what I get.

**Step 1. Write the chemical equation**

#"KOH + HI" → "KCl" + "H"_2"O"#

**Step 2. Calculate the volume to reach the equivalence point**

#color(blue)(bar(ul(|color(white)(a/a)c_text(A)V_text(A)= c_text(B)V_text(B)color(white)(a/a)|)))" "#

The equivalence point is at 12.5 mL

**Step 3. pH during titration**

**(i)** At 0.000 mL

**(ii)** pH at 3.125 mL

**(iii)** pH at 6.25 mL

**(iv)** pH at 11.50 mL

**(v)** pH at 12.50 mL

You are at the equivalence point.

#"pH = 7.00**

**(vi)** pH at 13.50 mL

You are 1.00 mL past the equivalence point, so you have neutralized all the

**(vii)** pH at 18.75 mL

You are 6.25 mL past the equivalence point, so you have neutralized all the

**Step 4. Construct the titration curve**

Plot the points on a graph and draw a smooth curve between them.

Your graph should look something like this.

I marked your points with the red dots.