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## What is Ksp in chemistry?

anor277
Featured 3 months ago

${K}_{s p}$ is the so-called solubility product, that quantifies the solubility of a salt in water.

#### Explanation:

Consider a sparingly soluble salt, $M X$, in water.

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

$M X \left(s\right) r i g h t \le f t h a r p \infty n s {M}^{+} + {X}^{-}$

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

$\frac{\left[{M}^{+}\right] \left[{X}^{-}\right]}{\left[M X \left(s\right)\right]}$ $=$ ${K}_{s p}$

But $\left[M X \left(s\right)\right]$ is meaningless, as you cannot have the concentration of a solid, so we are left the solubility expression:

${K}_{s p} = \left[{M}^{+}\right] \left[{X}^{-}\right]$

${K}_{s p}$ have been measured for a great variety of insoluble and sparingly soluble salts . Why? Because suppose you were trying to isolate precious metal salts, i.e. those of gold, or rhodium, or iridium. You don't want to throw precious metals away. Likewise, if you had lead, or cadmium, or mercury salts, you don't want to throw these metals away, for the reason that you might poison the waterways.

${K}_{s p} , \text{lead chloride } = 1.62 \times {10}^{-} 5$ at $25$ #""^@C#. A temperature is specified because a hot solution can normally hold more solute than a cold one.

$P b C {l}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 C {l}^{-}$

And, ${K}_{s p} = \left[P {b}^{2 +}\right] {\left[C {l}^{-}\right]}^{2} = 1.62 \times {10}^{-} 5$.

If we say $\left[P {b}^{2 +}\right] = S$, then ${K}_{s p} = \left(S\right) {\left(2 S\right)}^{2}$.

i.e. ${K}_{s p} = 4 {S}^{3}$.

And thus $S$ $=$ #""^3sqrt{{(1.62xx10^-5)/(4)}# $=$ #??*mol*L^-1#.

I leave it to you to solve for the solubility of lead chloride in water in $g \cdot {L}^{-} 1$ under standard conditions.

## How are there #pi# bonds in #B_2# molecule without #sigma# bonds?

Truong-Son N.
Featured 6 months ago

There shouldn't be, and that's why ${\text{B}}_{2}$ is very unstable; it has two orthogonally-localized half-$\pi$ bonds, which is quite a weak bond.

Imagine that each $\text{B}$ takes the place of a $\text{CH}$ on $\text{HC"-="CH}$.

Then, take away the $\sigma$ bond, the two $\text{H}$ atoms, and one electron from each of the two full $\pi$ bonds, and you have a structure similar to ${\text{B}}_{2}$.

What about ${\text{Li}}_{2}$ "molecule"? Is that stable? (Its bond length is $\text{267.3 pm}$, over twice the length of an average bond.)

The first chemical bond made in a molecule is preferentially a $\boldsymbol{\sigma}$ bond.

$\sigma$ bonds are formed from a direct atomic orbital overlap. In comparison, $\pi$ bonds are sidelong overlaps and thus, $\sigma$ overlaps are made preferentially because they form the stronger bond.

${\text{B}}_{2}$ contains two boron atoms, which each use a basis of a $1 s$, a $2 s$, and three $2 p$ atomic orbitals.

For ${\text{Li}}_{2}$ through ${\text{N}}_{2}$, there exists an orbital mixing effect that makes the $\sigma$ molecular orbital for a $2 {p}_{z} - 2 {p}_{z}$ overlap (${\sigma}_{g \left(2 p\right)}$) higher in energy than the $\pi$ molecular orbitals for a $2 {p}_{x / y} - 2 {p}_{x / y}$ overlaps (${\pi}_{u \left(2 p\right)}$).

So, this particular orbital energy ordering takes place in the molecular orbital diagram, similar to the ${\text{C}}_{2}$ molecule:

where $| {E}_{{\sigma}_{2 {p}_{z}}} | > \left\{| {E}_{{\pi}_{2 {p}_{x}}} | = | {E}_{{\pi}_{2 {p}_{y}}} |\right\}$.

Since $\text{C}$ has one more electron, ${\text{B}}_{2}$ would have a similar MO diagram, EXCEPT for two fewer electrons. This, at first glance, seems to suggest that ${\text{B}}_{2}$ has two half-$\pi$ bonds, with a molecular electron configuration of:

$\textcolor{g r e e n}{{\left({\sigma}_{1 s}\right)}^{2} {\left({\sigma}_{1 {s}^{\text{*"))^2(sigma_(2s))^2(sigma_(2s^"*}}}\right)}^{2} {\left({\pi}_{2 {p}_{x}}\right)}^{1} {\left({\pi}_{2 {p}_{y}}\right)}^{1}}$

Since ${\text{B}}_{2}$ is paramagnetic with two electrons, in order to make that bond, there are indeed two half-$\boldsymbol{\pi}$ bonds, which form what we represent improperly in line notation as a $\sigma$ bond... and it isn't actually a $\sigma$ bond.

$: \text{B"-"B} :$

That isn't a $\sigma$ bond, but two half-$\pi$ bonds. We would then expect ${\text{B}}_{2}$ to be very unstable, which it is. (The lone pairs are from the ${\sigma}_{2 s}$ and ${\sigma}_{2 s}^{\text{*}}$, since bonding + antibonding filled = nonbonding.)

## Find (a) the mols required to break the buffer (i.e. when #Delta"pH" = 1.00#), and (b) change in pH due to adding #"0.005 mols OH"^(-)# in a #"1-L"# acetic acid buffer containing #"0.010 M"# of each component? Then find the buffer capacity in (b).

Ernest Z.
Featured 6 months ago

(a) mols to break the buffer = $\text{0.008 mol}$; (b) $\Delta \text{pH} = 0.35$
Buffer capacity = $\text{0.010 mol/L"cdot"pH}$

#### Explanation:

(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 $\text{NaOH}$ to the buffer.

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

Then we have

$\textcolor{w h i t e}{m m m m m l} \text{HA + H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$
$\text{I/mol} : \textcolor{w h i t e}{m l l} 0.01 \textcolor{w h i t e}{m m m m m m m m m l l} 0.01$
$\text{C/mol":color(white)(mll)"-"xcolor(white)(mmmmmmmmmmll)"+} x$
$\text{E/mol":color(white)(m)"0.01-"xcolor(white)(mmmmmmmml)"0.01+x}$

$\text{p} {K}_{\textrm{a}} = 4.76$

The pH will increase to 5.76.

#"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))#

$5.76 = 4.76 + \log \left(\frac{0.01 + x}{0.01 - x}\right)$

$1.00 = \log \left(\frac{0.01 + x}{0.01 - x}\right)$

$\frac{0.01 + x}{0.01 - x} = {10}^{1.00} = 10.0$

$0.01 + x = 10.0 \left(0.01 - x\right) = 0.10 - 10.0 x$

$11.0 x = 0.09$

$x = \frac{0.09}{11.0} = 0.008$

The mols required is 0.008 mol.

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

$\textcolor{w h i t e}{m m m m m m l l} \text{HA + H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m} 0.010 \textcolor{w h i t e}{m m m m m m m m m l} 0.010$
$\text{C/mol·L"^"-1":color(white)(ll)"-0.005"color(white)(mmmmmmmll)"+0.005}$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m} 0.005 \textcolor{w h i t e}{m m m m m m m m l l} 0.015$

For the original buffer,

#"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"])) = 4.76 + log(0.01/0.01) = 4.76 + log1 = 4.76 + 0 = 4.76#

#"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))#

$\text{pH} = 4.76 + \log \left(\frac{0.015}{0.005}\right) = 4.76 + \log 3 = 4.76 + 0.48 = 5.24$

#Δ"pH" = 5.24 - 4.76 = 0.48#

So, the buffer capacity at this point is:

#beta = (("0.005 mols OH"^(-))/("1 L buffer"))/("0.48 pH")#

$= \text{0.010 mol/L"cdot"pH}$

## How do we calculate ionic strength on the molal scale?

Truong-Son N.
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):

${I}_{m} = \frac{1}{2} {\sum}_{i} {z}_{i}^{2} {m}_{i}$

where for a strong electrolyte represented by

${M}_{{\nu}_{+}} {X}_{{\nu}_{-}} \left(s\right) \stackrel{{H}_{2} O \left(l\right) \text{ }}{\to} {\nu}_{+} {M}^{{z}_{+}} \left(a q\right) + {\nu}_{-} {X}^{{z}_{-}} \left(a q\right)$,

we have:

• ${m}_{i}$ is the molality of the entire $i$th strong electrolyte.

For instance, a cation within $\text{HCl}$ would have a molality of ${m}_{+} = {\nu}_{+} {m}_{i}$, and an anion within $\text{HCl}$ would have a molality of ${m}_{-} = {\nu}_{-} {m}_{i}$. The ${m}_{i}$ would then correspond to $\text{HCl}$.

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 $\text{HCl}$ of concentration $\text{0.01 m}$.

Its ionic strength is given by:

${I}_{m} = \frac{1}{2} \left({z}_{{H}^{+}}^{2} {m}_{{H}^{+}} + {z}_{C {l}^{-}}^{2} {m}_{C {l}^{-}}\right)$

$= \frac{1}{2} \left({z}_{{H}^{+}}^{2} {\nu}_{{H}^{+}} {m}_{H C l} + {z}_{C {l}^{-}}^{2} {\nu}_{C {l}^{-}} {m}_{H C l}\right)$

But since they are the same charge magnitudes, $| {z}_{+} {z}_{-} | = {z}_{\pm}^{2}$.

$\implies \frac{1}{2} \left(| {z}_{{H}^{+}} {z}_{C {l}^{-}} | {\nu}_{{H}^{+}} {m}_{H C l} + | {z}_{{H}^{+}} {z}_{C {l}^{-}} | {\nu}_{C {l}^{-}} {m}_{H C l}\right)$

$= \frac{1}{2} | {z}_{{H}^{+}} {z}_{C {l}^{-}} | \left({\nu}_{{H}^{+}} {m}_{H C l} + {\nu}_{C {l}^{-}} {m}_{H C l}\right)$

$= \frac{1}{2} | {z}_{{H}^{+}} {z}_{C {l}^{-}} | \left({\nu}_{{H}^{+}} + {\nu}_{C {l}^{-}}\right) {m}_{H C l}$

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:

• $\text{0.01 m}$ $\text{NaCl}$
• $\text{0.02 m}$ $\text{HCl}$
• $\text{0.03 m}$ ${\text{BaCl}}_{2}$

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

${I}_{m} = \frac{1}{2} {\sum}_{i} {z}_{i}^{2} {m}_{i}$

$= \frac{1}{2} \left[{z}_{N {a}^{+}}^{2} {m}_{N {a}^{+}} + {z}_{{H}^{+}}^{2} {m}_{{H}^{+}} + {z}_{B {a}^{2 +}}^{2} {m}_{B {a}^{2 +}} + {z}_{C {l}^{-}}^{2} {m}_{C {l}^{-}}\right]$

We should note that the chloride comes from all three electrolytes, so ${\nu}_{C {l}^{-}} = 4$ in total:

${\text{NaCl"(aq) -> "Na"^(+)(aq) + "Cl}}^{-}$
${\text{HCl"(aq) -> "H"^(+)(aq) + "Cl}}^{-} \left(a q\right)$
${\text{BaCl"_2(aq) -> "Ba"^(2+)(aq) + 2"Cl}}^{-} \left(a q\right)$

This means:

${I}_{m} = \frac{1}{2} \left[| {z}_{+} {z}_{-} | {m}_{N {a}^{+}} + | {z}_{+} {z}_{-} | {m}_{{H}^{+}} + {z}_{B {a}^{2 +}}^{2} {m}_{B {a}^{2 +}} + | {z}_{+} {z}_{-} | {m}_{C {l}^{-}}\right]$

$= \frac{1}{2} \left[| {z}_{+} {z}_{-} | {\nu}_{N {a}^{+}} {m}_{N a C l} + | {z}_{+} {z}_{-} | {\nu}_{{H}^{+}} {m}_{H C l} + {z}_{B {a}^{2 +}}^{2} {\nu}_{B {a}^{2 +}} {m}_{B a C {l}_{2}} + | {z}_{+} {z}_{-} | \left({\nu}_{C {l}^{-}} {m}_{N a C l} + {\nu}_{C {l}^{-}} {m}_{H C l} + {\nu}_{C {l}^{-}} {m}_{B a C {l}_{2}}\right)\right]$

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)]#

$=$ $\textcolor{b l u e}{\text{0.12 mol solutes/kg solvent}}$

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.

$\textcolor{b l u e}{{I}_{m}} = \frac{1}{2} \left[{1}^{2} \cdot 1 \cdot {\text{0.01 m Na"^(+) + {1^2 cdot 1 cdot "0.01 m Cl"^(-) + 1^2 cdot 1 cdot "0.02 m Cl"^(-) + 1^2 cdot 2 cdot "0.03 m Cl"^(-)} + 1^2 cdot 1 cdot "0.02 m H"^(+) + 2^2 cdot 1 cdot "0.03 m Ba}}^{2 +}\right]$

$=$ $\textcolor{b l u e}{\text{0.12 mol solutes/kg solvent}}$

## Consider Hydrogen ion in a box having one dimension, then find electron probability distribution?

Truong-Son N.
Featured 4 months ago

${\psi}_{n}^{\text{*}} \left(x\right) {\psi}_{n} \left(x\right) = \frac{2}{L} {\sin}^{2} \left(\frac{n \pi x}{L}\right)$

These solutions are well-known and you should get to know what they look like. Here are the wave function $\psi$ and the probability density ${\psi}_{n}^{\text{*}} {\psi}_{n}$ plotted:

A hydrogen cation (presumably ${\text{^(1) "H}}^{+}$) is a neutron and proton with no potential of interaction. Thus, we set this up as a particle in a box scenario where $V = 0$.

The Hamiltonian for such a scenario is:

$\hat{H} = \hat{K} + {\cancel{\hat{V}}}^{0}$

#= -ℏ^2/(2m) (del^2)/(delx^2)#

The box looks like:

with boundary conditions $\psi \left(0\right) = \psi \left(L\right) = 0$, and

$V = \left\{\begin{matrix}0 & x \in \left(0 L\right) \\ \infty & x \le 0 \\ \null & x \ge L\end{matrix}\right.$

The Schrodinger equation is then:

$\hat{H} \psi = E \psi$

#=> -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 #k = sqrt(2mE//ℏ^2)#, so

$\frac{{d}^{2} \psi}{{\mathrm{dx}}^{2}} + {k}^{2} \psi = 0$

The general solution to this is assumed to be

$\psi = {e}^{r x}$

and upon inserting it, we obtain the auxiliary equation:

${r}^{2} {e}^{r x} + {k}^{2} {e}^{r x} = 0$

In the well, ${e}^{r x} \ne 0$ so that the particle exists. Thus,

$r = i k$

and we write a linear combination for $\psi$:

$\psi = {c}_{1} {e}^{i k x} + {c}_{2} {e}^{- i k x}$

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

$\psi = {c}_{1} \left(\cos \left(k x\right) + i \sin \left(k x\right)\right) + {c}_{2} \left(\cos \left(k x\right) - i \sin \left(k x\right)\right)$

$= \left({c}_{1} + {c}_{2}\right) \cos \left(k x\right) + \left(i {c}_{1} - i {c}_{2}\right) \sin \left(k x\right)$

Define $A = {c}_{1} + {c}_{2}$ and $B = i {c}_{1} - i {c}_{2}$, because these are arbitrary constants. Therefore:

$\psi = A \cos \left(k x\right) + B \sin \left(k x\right)$

The boundary conditions state that since the potential goes to $\infty$ at $x = 0 , L$, it follows that $\psi \left(0\right) = \psi \left(L\right) = 0$:

$A \cos \left(k \cdot 0\right) + B \sin \left(k \cdot 0\right) = A \cos \left(k L\right) + B \sin \left(k L\right) = 0$

But since $\sin \left(0\right) = 0$ and $\cos \left(0\right) = 1$, it follows that $A = 0$. Furthermore, since $A = 0$, we have:

$B \sin \left(k L\right) = 0$

And this is only when $k L = n \pi$, where $n = 1 , 2 , 3 , . . .$. (We could take negative $n$, but that would give a linearly dependent solution.)

Thus, since $k = \frac{n \pi}{L}$, the wave function becomes:

${\psi}_{n} \left(x\right) = B \sin \left(\frac{n \pi x}{L}\right)$

And the probability distribution is then:

${\psi}_{n}^{\text{*}} \left(x\right) {\psi}_{n} \left(x\right) = {B}^{2} {\sin}^{2} \left(\frac{n \pi x}{L}\right)$

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

${\int}_{0}^{L} {\psi}_{n}^{\text{*}} {\psi}_{n} \mathrm{dx} = 1$

From this we get the normalization constant.

$1 = {B}^{2} {\int}_{0}^{L} {\sin}^{2} \left(\frac{n \pi x}{L}\right) \mathrm{dx}$

Consider the following argument.

If ${\sin}^{2} u + {\cos}^{2} u = 1$, then integrating ${\sin}^{2} u + {\cos}^{2} u$ gives $L$ for the area under the curve, with height $1$ and length $L$, regardless of what $u$ is.

The identities ${\sin}^{2} u = \frac{1}{2} \left(1 - \cos \left(2 u\right)\right)$ and ${\cos}^{2} u = \frac{1}{2} \left(1 + \cos \left(2 u\right)\right)$ show that ${\sin}^{2} u$ contributes to half the area under the curve.

Therefore, ${\int}_{0}^{L} {\sin}^{2} u \mathrm{du} = \frac{L}{2}$, and $B = \sqrt{\frac{2}{L}}$. This means

$\textcolor{b l u e}{{\psi}_{n}^{\text{*"(x)psi_n(x) = "Probability Distribution}}}$

$= \textcolor{b l u e}{\frac{2}{L} {\sin}^{2} \left(\frac{n \pi x}{L}\right)}$

## Number of structural isomers possible in C3H6O ???

Ernest Z.
Featured 4 days ago

I count nine structural isomers.

#### Explanation:

A saturated compound would have the formula $\text{C"_3"H"_8"O}$.

This compound is missing two $\text{H}$ atoms, so it must have a double bond or a ring.

A. Alkenes with an $\text{OH}$ 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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