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5

## What is Ksp in chemistry?

anor277
Featured 2 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.

1

## When the atmospheric pressure is 6 times it’s average, how would you calculate the boiling and melting point of a molecule of water?

Ernest Z.
Featured 2 months ago

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

#### Explanation:

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

Δ_text(fus)H color(white)(ll)= the enthalpy of fusion of ice
$T \textcolor{w h i t e}{m m m} =$ the Kelvin temperature
Δv_text(fus) color(white)(ml)= the change in specific volume for the phase change

We can integrate the Clausius equation:

"int_(p₁)^(p₂)dp = (Δ_text(fus)H)/(Δv_text(fus))int_(T₁)^(T₂)(dT)/T

p_2 - p_1 = "6 bar- 1 bar = 5 bar" = (Δ_text(fus)H)/(Δv_text(fus))ln(T_2/T_1)

Δ_text(fus)H = "333.55 J·g"^"-1"

v_text(ice) = 1/("0.916 75 g·cm"^"-3") = "1.0908 cm"^3·"g"^"-1"
v_text(liq) = 1/("0.999 84 g·cm"^"-3") = "1.0001 cm"^3·"g"^"-1"

Δv_text(fus) = v_text(liq) - v_text(ice) = "(1.0001 - 1.0908) cm"^"-3""g"^"-1" = "-0.0907 cm"^3"g"^"-1"

"5 bar" = "-"("333.55 J"·color(red)(cancel(color(black)("g"^"-1"))))/("0.0907 cm"^3color(red)(cancel(color(black)("g"^"-1"))))ln(T_2/T_1)

$\ln \left({T}_{2} / {T}_{1}\right) = \text{-"(5 color(red)(cancel(color(black)("bar"))) × 0.0907 color(red)(cancel(color(black)("cm"^3))))/(333.55 color(red)(cancel(color(black)("J")))) × (1 color(red)(cancel(color(black)("J"))))/(1 color(red)(cancel(color(black)("Pa")))·color(red)(cancel(color(black)("m"^3)))) × ( 10^5 color(red)(cancel(color(black)("Pa"))))/(1 color(red)(cancel(color(black)("bar")))) × (1 color(red)(cancel(color(black)("m"^3))))/(10^6 color(red)(cancel(color(black)("cm"^3)))) = "-0.000 1360}$

${T}_{2} / {T}_{1} = {e}^{\text{-0.000 1360" = "0.999 86}}$

${T}_{2} = \text{273.15 K × 0.999 86 = 273.112 K = -0.037 °C}$

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

${p}_{1}$ and ${p}_{2}$ are the vapour pressures at temperatures
${T}_{1}$ and ${T}_{2}$

Δ_"vap"H = the enthalpy of vaporization of the liquid

$R \textcolor{w h i t e}{m m l l}$ = the Universal Gas Constant

For example. the boiling point of water is 99.6 °C at 1 bar and its Δ_text(vap)H = 40.66 "kJ·mol"^"-1".

We can calculate the boiling point at 6 bar.

${p}_{1} = \text{1 bar"; T_1 = "99.6 °C = 372.75 K}$

p_2 = "6 bar"; T_2 = ?

Δ_text(vap)H = "40.66 kJ·mol"^"-1"

$R = \text{8.314 J·K"^"-1""mol"^"-1}$

ln((6 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar"))))) = ("40 660" color(red)(cancel(color(black)("J·mol"))))/(8.314 color(red)(cancel(color(black)("J")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1"))))(1/"372.75 K" - 1/T_2)

$1.792 = \frac{4891}{372.75} - \text{4891 K"/T_2 = 13.12 - "4891 K"/T_2}$

$\text{4891 K"/T_2 = "13.12 - 1.792 = 11.33}$

${T}_{2} = \text{4891 K"/11.33 = "431.7 K" = "159 °C}$

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

3

## What can be the least energy of a system as per Heisenberg's uncertainty principle?

Truong-Son N.
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 $\to$ 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 calculate rotational energy for this system (hence, nothing is stated about the translational kinetic energy, which can be nonzero, so that momentum is nonzero):

${E}_{J} = h c B J \left(J + 1\right)$, $J = 0 , 1 , 2 , 3 , . . .$

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-state rotational energy is ${E}_{0} = h c B J \left(J + 1\right) = 0$.

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 translational part $\psi \left(r\right)$ (which just gives a phase factor to the total wave function psi(r,theta,phi)) and a solvable rotational part $\psi \left(\theta , \phi\right)$.

Hence, we are not claiming certainty about linear momentum or position, which is not breaking the Uncertainty Principle.

2

## Gibbs Free Energy. Why are equilibrium constants for a chemical reaction temperature dependent?

Truong-Son N.
Featured 1 month ago

Because rate constants are temperature-dependent, and by changing the temperature, you change the ratio of rate constants so that $K \equiv {k}_{f w d} / {k}_{r e v}$ changes.

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

$\Delta {G}^{\circ} = - R T \ln K$

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

$k = A {e}^{- {E}_{a} / R T}$

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 $A$ remains constant:

${k}_{2} / {k}_{1} = \text{exp} \left(- {E}_{a} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]\right)$

$\implies \textcolor{g r e e n}{\ln \left({k}_{2} / {k}_{1}\right) = - {E}_{a} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]}$

where $\text{exp} \left(x\right) = {e}^{x}$.

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 ${E}_{a , f w d}$ is the energy from reactant to transition state for the forward reaction, and ${E}_{a , r e v}$ is the energy from product to transition state for the reverse reaction.

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

$a A + b B r i g h t \le f t h a r p \infty n s c C + \mathrm{dD}$

The rate laws would be:

${r}_{f w d} \left(t\right) = {k}_{f w d} {\left[A\right]}^{a} {\left[B\right]}^{b}$
${r}_{r e v} \left(t\right) = {k}_{r e v} {\left[C\right]}^{c} {\left[D\right]}^{d}$

So we set them equal:

${k}_{f w d} {\left[A\right]}^{a} {\left[B\right]}^{b} = {k}_{r e v} {\left[C\right]}^{c} {\left[D\right]}^{d}$

By definition, the equilibrium constant is:

$\implies K \equiv \frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}} = \frac{{k}_{f w d}}{{k}_{r e v}}$

Now, take $K$ at different temperatures, so that:

${K}_{1} = \frac{{k}_{1 , f w d}}{{k}_{1 , r e v}}$$\text{ "" }$${K}_{2} = \frac{{k}_{2 , f w d}}{{k}_{2 , r e v}}$

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

$\ln \left({k}_{2 , f w d} / {k}_{1 , f w d}\right) = - {E}_{a , f w d} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]$

$\ln \left({k}_{2 , r e v} / {k}_{1 , r e v}\right) = - {E}_{a , r e v} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]$

Use the fact that ${K}_{i} = \frac{{k}_{i , f w d}}{{k}_{i , r e v}}$ to see that by subtracting these equations we proceed to get the left-hand side to be:

$\ln \left({k}_{2 , f w d} / {k}_{1 , f w d}\right) - \ln \left({k}_{2 , r e v} / {k}_{1 , r e v}\right)$

$= \ln \left({k}_{2 , f w d} / {k}_{1 , f w d} \cdot \frac{1}{{k}_{2 , r e v} / {k}_{1 , r e v}}\right)$

$= \ln \left(\frac{{k}_{2 , f w d} / {k}_{2 , r e v}}{{k}_{1 , r e v} / {k}_{1 , f w d}}\right)$

$= \ln \left({K}_{2} / {K}_{1}\right)$

The right-hand side by subtraction becomes:

$- {E}_{a , f w d} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right] - \left(- {E}_{a , r e v} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]\right)$

$= - \frac{{E}_{a , f w d} - {E}_{a , r e v}}{R} \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]$

But look up at the reaction coordinate diagram above. We can make the connection that $\textcolor{g r e e n}{{E}_{a , f w d} - {E}_{a , r e v} = \Delta {H}_{r x n}}$.

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

$\textcolor{b l u e}{\ln \left({K}_{2} / {K}_{1}\right) = - \frac{\Delta {H}_{r x n}}{R} \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]}$

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

Or put in another form,

$\textcolor{b l u e}{\ln K = - \frac{\Delta {H}_{r x n}}{R} \frac{1}{T} + \frac{\Delta {S}_{r x n}}{R}}$

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.

2

## What are the different kinds of f orbitals?

Truong-Son N.
Featured 1 month ago

They are quite complicated, and can often do combinations of $\sigma$, $\pi$, $\delta$, and even $\phi$ bonding.

For an introduction into these kinds of bonds:

$\sigma$ bonds are in every chemical bond. $\pi$ bonds start showing up in double and triple bonds (e.g. ${\text{O}}_{2}$, ${\text{N}}_{2}$, etc), $\delta$ bonds start showing up in quadruple bonds (see link), and $\phi$ bonds aren't seen until a sextuple bond is made (e.g. in ${\text{Mo}}_{2}$ or ${\text{W}}_{2}$).

The $4 f$ orbitals can be separated into three types (here, we use the convention that outer atoms point their $y$ axes inwards and $z$ axes upwards):

1) Two lobes - $\sigma$ bonding only (${m}_{l} = 0$)

• The ${f}_{{z}^{3}}$ (${m}_{l} = 0$) is the only one that only $\sigma$ bonds. It can bond head-on along the $z$ axis.

2) Six lobes - $\sigma$ and $\pi$ bonding, OR $\phi$ bonding only (${m}_{l} = - 3 , + 3 , - 1 , + 1$)

• The ${f}_{y \left(3 {x}^{2} - {y}^{2}\right)}$ (${m}_{l} = - 3$) 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 ${d}_{x y}$ orbital).

It can alternatively form a $\phi$ bond (a six-lobed side-on overlap) along the $x y$ plane (with another ${f}_{y \left(3 {x}^{2} - {y}^{2}\right)}$ orbital in a bimetallic complex).

• The ${f}_{x \left({x}^{2} - 3 {y}^{2}\right)}$ (${m}_{l} = + 3$) 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 ${d}_{x y}$ orbital).

It can alternatively form a $\phi$ bond (a six-lobed side-on overlap) along the $x y$ plane (with another ${f}_{x \left({x}^{2} - 3 {y}^{2}\right)}$ orbital in a bimetallic complex).

• The ${f}_{y {z}^{2}}$ (${m}_{l} = - 1$) 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 $y z$ plane (with another ${f}_{y {z}^{2}}$ orbital in a bimetallic complex).

• The ${f}_{x {z}^{2}}$ (${m}_{l} = + 1$) 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 $x z$ plane (with another ${f}_{x {z}^{2}}$ orbital in a bimetallic complex).

3) Eight lobes - $\pi$ bonding OR $\delta$ bonding (${m}_{l} = - 2 , + 2$)

• The ${f}_{z \left({x}^{2} - {y}^{2}\right)}$ (${m}_{l} = - 2$) 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 ${f}_{z \left({x}^{2} - {y}^{2}\right)}$ orbital in a bimetallic complex.

• The ${f}_{x y z}$ (${m}_{l} = + 2$) is for $\delta$ bonding along ANY of the planes ($x z , y z , x y$) (for example, with ${d}_{x y}$, ${d}_{x z}$, or ${d}_{y z}$ orbitals).

It can alternatively form a $\pi$ bond with another ${f}_{x y z}$ orbital in a bimetallic complex.

3

## Draw a diagram for this Galvanic cell, labeling the electron flow, the anode and cathode, and the positive and negative sides of the Galvanic cell?

Jacob T.
Featured 1 month ago

#### Explanation:

From the reaction formula:
Copper is reduced as its oxidation state decreases from $+ 2$ in $C {u}^{2 +} \left(a q\right)$ to $0$ in $C u \left(s\right)$.
Cobalt is oxidized; its oxidation state increases from $0$ in $C o \left(s\right)$ to $+ 2$ in $C {o}^{2 +} \left(a q\right)$

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 color(blue)("An")"ode" of a cell, voltaic or electrochemical, attracts anions (ions with negative charges) and thus must carries some positive charges. So it is losing electrons and thus undergoing oxidation. Similarly, the color(red)("Cat")"hode" attracts cations (positively-charged ions), possesses excessive negative charges and therefore undergoes reduction. It's apparently a rule of thumb but it works for me.

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

2

## A 1956 dime is made of a silver-copper alloy and has a mass of 2.490. The dime was dissolved in nitric acid and a blue solution remained. The solution quantitatively transferred to a 100mL volumetric flask and brought to volume with water...?

Truong-Son N.
Featured 1 month ago

Method $b$ is more accurate if the actual percent by mass is $\text{90 % w/w Ag}$, but is more prone to experimenter error than experimental/random error.

The blue solution that forms due to dissolving the dime in ${\text{HNO}}_{3}$ follows the nitric acid reduction and copper oxidation given here:

$2 \left(\text{NO"_3^(-)(aq) + 4"H"^(+)(aq) + cancel(3e^(-)) -> "NO"(g) + 2"H"_2"O} \left(l\right)\right)$
3ul(("Cu"(s) -> "Cu"^(2+)(aq) + cancel(2e^(-)))" "" "" "" "" "" "" "" ")
$2 \text{NO"_3^(-)(aq) + 3"Cu"(s) + 8"H"^(+)(aq) -> 3"Cu"^(2+)(aq) + 2"NO"(g) + 4"H"_2"O} \left(l\right)$

A $\textcolor{b l u e}{\text{blue}}$ copper solution is due to ${\text{Cu}}^{2 +}$ (a red solution would be due to ${\text{Cu}}^{+}$). $\text{Ag} \left(s\right)$ also gets dissolved in this way. I'll leave it to you to write the reaction this time.

These both occur because nitrate reduction is more favorable than silver and copper reduction (its ${E}_{red}^{\circ}$ is more positive).

a)

If $\text{50.00 mL}$ of the original $\text{100.00 mL}$ in the volumetric flask (which was made to contain $\text{100.00 mL}$!) was titrated with excess $\text{NaCl}$, then all of the ${\text{Ag}}^{+}$ will precipitate with the ${\text{Cl}}^{-}$ in the form of $\text{AgCl} \left(s\right)$.

Upon washing, filtering, drying, weighing, drying, weighing, drying, etc... a mass of $\text{1.375 g}$ recovered of $\text{AgCl}$ would suggest that:

$1.375 {\cancel{\text{g AgCl" xx "107.8682 g Ag"/(143.32 cancel"g AgCl") = "1.035 g Ag}}}^{+}$

And this mass of silver cation was what reacted with the $\text{NaCl}$ from within the $\text{50.00 mL}$ aliquot.

Therefore, since volume and mass are extensive, we really have $\underline{{\text{2.070 g Ag}}^{+}}$ dissolved in the original flask volume, and as a result, the percent by mass of $\text{Ag}$ 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 $\text{AgCl}$ on the ground

All of these lead to a less than 100% mass yield.

b)

Using the given standard curve from the data previously collected:

$\underline{\text{Conc. (mg/mL)"" ""A""bsorbance}}$
$5.00 \text{ "" "" "" "" "" } 0.843$
$4.00 \text{ "" "" "" "" "" } 0.672$
$3.00 \text{ "" "" "" "" "" } 0.507$
$2.00 \text{ "" "" "" "" "" } 0.336$
$1.00 \text{ "" "" "" "" "" } 0.161$

we would construct an Excel graph to obtain $y = 0.1700 x - 0.0062$ for the Beer's law fit line equation:

And hence, with the absorbance of ${\text{Cu}}^{2 +}$ being $0.420$ for ${\lambda}_{\max} = \text{740 nm}$ (absorbing red light, reflecting blue), we use Beer's Law to get:

$A = \epsilon b c + y \text{-int}$

$\implies \textcolor{b l u e}{c} = \frac{A - y \text{-int}}{\epsilon b}$

$= \frac{0.420 - \left(- 0.0062\right)}{0.1700} \text{mg"/"mL}$

$=$ $\text{2.507 mg Cu"^(2+)"/mL}$

$=$ $\text{250.7 mg Cu"^(2+)"/100 mL}$

$=$ $\textcolor{b l u e}{\text{0.2507 g Cu"^(2+)"/100 mL}}$

By subtraction,

$\text{2.490 g" - "0.2507 g Cu" = "2.239 g Ag}$

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.

2

## 100 g of acetylene ("C"_2"H"_2) gas is placed in a 20.0 L container at 298 K. (i) Use the ideal gas equation to calculate the pressure? (ii) Use van der Waals equation to calculate the pressure? (a="4.390 L"^2cdot"atm/mol"^2, b="0.0513 L/mol")

Ernest Z.
Featured 3 weeks ago

(i) ${p}_{\textrm{i \mathrm{de} a l}} = \text{4.70 atm}$: (ii) ${p}_{\textrm{v \mathrm{dW}}} = \text{4.58 atm}$

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.

#### Explanation:

${\text{Moles of C"_2"H"_2 = 100 color(red)(cancel(color(black)("g C"_2"H"_2))) × ("1 mol C"_2"H"_2)/(26.04 color(red)(cancel(color(black)("C"_2"H"_2)))) = "3.840 mol C"_2"H}}_{2}$

(a) Ideal gas

The equation for the Ideal Gas Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Then,

$p = \frac{n R T}{V}$

$n = \text{3.840 mol}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{298 K}$
$V = \text{20.0 L}$

p = (3.840 color(red)(cancel(color(black)("mol"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 298 color(red)(cancel(color(black)("K"))))/(20.0 color(red)(cancel(color(black)("L")))) = "4.70 atm"

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

(b) van der Waals Equation

The van der Waals equation is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \left(P + \frac{{n}^{2} a}{V} ^ 2\right) \left(V - n b\right) = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

$P + \frac{{n}^{2} a}{V} ^ 2 = \frac{n R T}{V - n b}$

$P = \frac{n R T}{V - n b} - \frac{{n}^{2} a}{V} ^ 2$

For this problem,

$n = \text{3.840 mol}$
$R = \text{0.082 06"color(white)(l)"L·atm·K"^"-1""mol"^"-1}$
$T = \text{298 K}$
$V = \text{20.0 L}$
$a = \text{4.390 atm·L"^2"mol"^"-2}$
$b = \text{0.0513 L·mol"^"-1}$

P = (nRT)/(V-nb) – (n^2a)/V^2

$= {\left(3.840 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"))) × "0.082 06 atm"color(red)(cancel(color(black)("L·""K"^"-1""mol"^"-1")))× 298 color(red)(cancel(color(black)("K"))))/(20.0 color(red)(cancel(color(black)("L"))) – 3.840 color(red)(cancel(color(black)("mol")))× 0.0513 color(red)(cancel(color(black)("L·mol"^(-1))))) - ((3.840 color(red)(cancel(color(black)("mol"))))^2 × "4.390 atm" color(red)(cancel(color(black)("L"^2"mol"^"-2"))))/(20.0 color(red)(cancel(color(black)("L}}}}\right)}^{2}$
$= \text{93.90 atm"/19.80 - "0.1618 atm "= "4.742 atm" - "0.1618 atm"= "4.58 atm}$

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

2

## Equilibrium constant problem help?

Ernest Z.
Featured 2 weeks ago

Warning! Long Answer. K_text(p) = 4.4 × 10^"-4"

#### Explanation:

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

We have three equations:

bb(1.) color(white)(m)"COCl"_2"(g)" + "H"_2"O(l)" ⇌ "CH"_2"Cl"_2"(g)" + "O"_2"(g)"; color(white)(mmmmmmm)K_text(p₁) = 0.015
bb(2.) color(white)(m)"4HCl(g)" + "O"_2"(g)" ⇌ 2"H"_2"O(l)" + "2Cl"_2"(g)";color(white)(mmmmmmmmll)K_text(p₂) = 6800
bb(3.) color(white)(m)"2CH"_2"Cl"_2"(g)" + "2H"_2"(g)" + "3O"_2"(g)" ⇌ 2"COCl"_2"(g)" + "4H"_2"O(l)"; K_text(p₃) = 5.8

From these, we must construct the target equation.

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

The target equation has $\text{2HCl(g)}$ oh the right, so we reverse Equation 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.

bb(4.) color(white)(m)"H"_2"O(l)" + "Cl"_2"(g)" ⇌ "2HCl(g)" + "½O"_2"(g)"; K_5 = 1/sqrtK_text(p₂)

The target equation has $\text{H"_2} \left(g\right)$ on the left.

We halve Equation 3.

bb(5.) color(white)(m)"CH"_2"Cl"_2"(g)" + "H"_2"(g)" + "³/₂O"_2"(g)" ⇌ "COCl"_2"(g)" + "2H"_2"O(l)" ;K_5 = sqrtK_text(p₃)

Equation 5. has $\text{CH"_2"Cl"_2"(g)}$ on the left, and that is not in the target equation.

We re-write Equation 1.

bb(6.) color(white)(m)"COCl"_2"(g)" + "H"_2"O(l)" ⇌ "CH"_2"Cl"_2"(g)" + "O"_2"(g)"; K_6 = K_text(p₁)

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.

bb(4.) color(white)(m)color(red)(cancel(color(black)("H"_2"O(l)"))) + "Cl"_2"(g)" ⇌ "2HCl(g)" + color(red)(cancel(color(black)("½O"_2"(g)"))); color(white)(mmmmmmmml)K_4 = 1/sqrtK_text(p₂)
bb(5.) color(white)(m)color(red)(cancel(color(black)("CH"_2"Cl"_2"(g)"))) + "H"_2"(g)" + color(red)(cancel(color(black)("³/₂O"_2"(g)"))) ⇌ color(red)(cancel(color(black)("COCl"_2"(g)"))) + color(red)(cancel(color(black)("2H"_2"O(l)"))) ;K_5"="sqrtK_text(p₃)

bb(6.) ul(color(white)(m)color(red)(cancel(color(black)("COCl"_2"(g)"))) + color(red)(cancel(color(black)("H"_2"O(l)"))) ⇌ color(red)(cancel(color(black)("CH"_2"Cl"_2"(g)"))) + color(red)(cancel(color(black)("O"_2"(g)"))); color(white)(mmmmmml)K_6 = K_text(p₁))
color(white)(mml)"H"_2"(g)" + "Cl"_2"(g)" ⇌ "2HCl(g)"; color(white)(mmmmmmmmmmmmmm)K_text(p) = K_4K_5K_6

K_text(p) = K_text(p₁)sqrt(K_text(p₃)/K_text(p₂)) = 0.015sqrt(5.8/6800) = 0.015sqrt(8.53 × 10^"-4") = 0.015 × 0.0292 = 4.4 × 10^"-4"

1

## Chemistry Acid and Base Help??

Ernest Z.
Featured 1 week ago

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

#### Explanation:

Step 1. Write the chemical equation

$\text{KOH + HI" → "KCl" + "H"_2"O}$

Step 2. Calculate the volume to reach the equivalence point

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {c}_{\textrm{A}} {V}_{\textrm{A}} = {c}_{\textrm{B}} {V}_{\textrm{B}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

V_text(B) = V_text(A) × c_text(C)/c_text(B) = "10.00 mL" × (0.125 color(red)(cancel(color(black)("mol/L"))))/(0.100 color(red)(cancel(color(black)("mol/L")))) = "12.5 mL"

The equivalence point is at 12.5 mL $\text{KOH}$.

Step 3. pH during titration

(i) At 0.000 mL

["H"_3"O"^"+"] = "0.125 mol/L"

$\text{pH" = "-log} \left(0.125\right) = 0.90$

(ii) pH at 3.125 mL

$\text{Initial moles of HI" = 10.0 color(red)(cancel(color(black)("mL HI"))) × ("0.125 mol I")/(1 color(red)(cancel(color(black)("mL HI")))) = "1.25 mmol HI}$

$\text{Moles of KOH added " = 3.125 color(red)(cancel(color(black)("mL KOHl"))) × "0.100 mmol KOH"/(1 color(red)(cancel(color(black)("mL KOH")))) = "0.3125 mmol KOH}$

$\text{Moles of HI remaining" = "(1.25 - 0.3125) mmol HI = 0.938 mmol HIH}$

$\text{Volume = (10.00 + 3.125) mL = 13.12 mL}$

["H"_3"O"^"+"] = "0.938 mmol"/"13.12 mL" = "0.0714 mol/L"

$\text{pH = -log(0.0714) = 1.15}$

(iii) pH at 6.25 mL

$\text{Moles of KOH added " = 6.25 color(red)(cancel(color(black)("mL KOHl"))) × "0.100 mmol KOH"/(1 color(red)(cancel(color(black)("mL KOH")))) = "0.625 mmol KOH}$

$\text{Moles of HI remaining" = "(1.25 - 0.625) mmol HI = 0.625 mmol HIH}$

$\text{Volume = (10.00 + 6.25) mL = 16.25 mL}$

["H"_3"O"^"+"] = "0.625 mmol"/"16.25 mL" = "0.0385 mol/L"

$\text{pH = -log(0.0385) = 1.41}$

(iv) pH at 11.50 mL

$\text{Moles of KOH added " = 11.50 color(red)(cancel(color(black)("mL KOHl"))) × "0.100 mmol KOH"/(1 color(red)(cancel(color(black)("mL KOH")))) = "1.15 mmol KOH}$

$\text{Moles of HI remaining" = "(1.25 - 1.15) mmol HI = 0.10 mmol HIH}$

$\text{Volume = (10.00 +11.50) mL = 21.50 mL}$

["H"_3"O"^"+"] = "0.10 mmol"/"21.5 mL" = "0.004 65 mmol/L"

$\text{pH = -log(0.004 65) = 2.33}$

(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 $\text{HI}$ and have a solution of excess $\text{KOH}$.

$\text{ Excess moles KOH " =1.00 color(red)(cancel(color(black)("mL KOH"))) × "0.100 mmol KOH"/(1 color(red)(cancel(color(black)("mL KOH")))) = "0.100 mmol KOH}$

$\text{Volume = (10.0 + 13.50) mL = 23.50 mL}$

["OH"^"-"] = "0.100 mmol"/"23.50 mL" = "0.004 25 mmol/L"

$\text{pOH = -log(0.004 25) = 2.37}$

$\text{pH = 14.00 -pOH = 14.00 - 2.37 = 11.62}$

(vii) pH at 18.75 mL

You are 6.25 mL past the equivalence point, so you have neutralized all the $\text{HI}$ and have a solution of excess $\text{KOH}$.

$\text{ Excess moles KOH " =6.25 color(red)(cancel(color(black)("mL KOH"))) × "0.100 mmol KOH"/(1 color(red)(cancel(color(black)("mL KOH")))) = "0.625 mmol KOH}$

$\text{Volume = (10.0 + 18.75) mL = 28.75 mL}$

["OH"^"-"] = "0.625 mmol"/"28.75 mL" = "0.0217 mmol/L"#

$\text{pOH = -log(0.0217) = 1.66}$

$\text{pH = 14.00 -pOH = 14.00 - 1.66 = 12.33}$

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.