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Featured 3 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 6 months ago

There shouldn't be, and that's why

Imagine that each

Then, take away the

What about

**The first chemical bond made in a molecule is preferentially a #bbsigma# bond.**

For *orbital mixing effect* that makes the ** higher** in energy than the

So, this particular orbital energy ordering takes place in the **molecular orbital diagram**, similar to the

where

#|E_(sigma_(2p_z))| > {|E_(pi_(2p_x))| = |E_(pi_(2p_y))|}# .

Since *fewer* electrons. This, at first glance, seems to suggest that

#color(green)((sigma_(1s))^2(sigma_(1s^"*"))^2(sigma_(2s))^2(sigma_(2s^"*"))^2(pi_(2p_x))^1(pi_(2p_y))^1)#

Since *paramagnetic* with two electrons, in order to make that bond, there are indeed **two half-****bonds**, which form what we represent improperly in line notation as a

#:"B"-"B":#

That isn't a

Featured 6 months ago

(a) mols to break the buffer =

Buffer capacity =

**(a) Calculate the mols required**

**Buffer capacity** is the amount of a monoprotic strong acid or base that must be added to 1 L of a buffer to change its pH by one unit.

Let's assume that we are adding

The base will decrease the amount of acetic acid and increase the amount of acetate.

Then we have

The pH will increase to 5.76.

The mols required is 0.008 mol.

**(b) Calculate the change in pH on adding 0.005 mols of base**

For the original buffer,

After adding the base,

So, the **buffer capacity** at this point is:

Featured 6 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 4 months ago

These solutions are well-known and you should get to know what they look like. Here are the wave function

**DISCLAIMER:** *LONG ANSWER!*

A **hydrogen cation** (presumably *particle in a box* scenario where

The **Hamiltonian** for such a scenario is:

#hatH = hatK + cancel(hatV)^(0)#

#= -ℏ^2/(2m) (del^2)/(delx^2)#

The box looks like:

with **boundary conditions**

#V = {(0, x in (0,L)),(oo, x <= 0),(,x >= L):}#

The **Schrodinger equation** is then:

#hatHpsi = Epsi#

#=> -ℏ^2/(2m) (d^2psi)/(dx^2) = Epsi#

Rearrange to the standard form:

#(d^2psi)/(dx^2) + (2mE)/(ℏ^2)psi = 0#

Often we set the substitution

#(d^2psi)/(dx^2) + k^2psi = 0#

The general solution to this is assumed to be

#psi = e^(rx)#

and upon inserting it, we obtain the auxiliary equation:

#r^2 e^(rx) + k^2 e^(rx) = 0#

In the well,

#r = ik#

and we write a **linear combination** for

#psi = c_1e^(ikx) + c_2e^(-ikx)#

Using Euler's formula, we rewrite this in terms of real trig functions.

#psi = c_1(cos(kx) + isin(kx)) + c_2(cos(kx) - isin(kx))#

#= (c_1 + c_2)cos(kx) + (ic_1 - ic_2)sin(kx)#

Define

#psi = Acos(kx) + Bsin(kx)#

The **boundary conditions** state that since the potential goes to

#Acos(kcdot0) + Bsin(kcdot0) = Acos(kL) + Bsin(kL) = 0#

But since

#Bsin(kL) = 0#

And this is only when

Thus, since

#psi_n(x) = Bsin((npix)/L)#

And the probability distribution is then:

#psi_n^"*"(x)psi_n(x) = B^2 sin^2((npix)/L)#

A well-behaved wave function is ** normalized** in its boundaries, so we say that

#int_(0)^(L) psi_n^"*"psi_ndx = 1#

From this we get the **normalization constant**.

#1 = B^2 int_(0)^(L) sin^2((npix)/L)dx#

Consider the following argument.

If

The identities

Therefore,

#color(blue)(psi_n^"*"(x)psi_n(x) = "Probability Distribution")#

#= color(blue)(2/L sin^2((npix)/L))#

Featured 4 days ago

I count nine structural isomers.

A saturated compound would have the formula

This compound is missing two

**A. Alkenes with an #"OH"# group**

**1.** Prop-1-en-1-ol

**2.** Prop-1-en-2-ol

**3.** Prop-2-en-1-ol

**B. Alkenes with an ether group**

**4.** Methoxyethene

**C. Carbonyl compounds**

**5.** Propanal

**6.** Propanone

**D. Cyclic alcohols**

**7.** Cyclopropanol

**E. Cyclic ethers**

**8.** Oxetane

**9.** Methyloxirane

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