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## Which of these liquids is miscible with water? Ethanol, methanol, hexane... Why? I thought since ethanol has a nonpolar tail, it isn't soluble in water.

anor277
Featured 4 months ago

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

#### Explanation:

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 $78.4$ #""^@C#; but ethane has a normal boiling point of $- 89.0$ #""^@C#; and propane has a normal boiling point of $- 42.0$ #""^@C#.

Throw in some polarity (but not hydrogen bonding), and we find that $\text{dimethyl ether}$ has a normal boiling point of $- 21.0$ #""^@C#, and $\text{diethyl ether}$ has a normal boiling point of $34.6$ #""^@C#.

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 $\text{methylated spirit}$). And if you drink this stuff you go blind. But why should anyone want to drink ethanol?

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

Truong-Son N.
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):

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

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.

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

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

## What is the formula for calculating a nodal point?

Truong-Son N.
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 $n = 3$.

As a simple but not undercomplicated example, consider one of the $2 p$ orbitals:

The hydrogenic atom wave function for the $2 {p}_{z}$ is:

${\psi}_{2 p z} = {R}_{21} \left(r\right) {Y}_{1}^{0} \left(\theta , \phi\right)$

$= \frac{1}{4 \sqrt{2 \pi}} {\left(\frac{Z}{a} _ 0\right)}^{3 / 2} \sigma {e}^{- \sigma / 2} \cos \theta$,

where:

• $\sigma = Z r / {a}_{0}$.
• $Z$ is the atomic number.
• ${a}_{0} = \text{0.0529177 nm}$ is the Bohr radius.
• $r$ is the radial distance away from the nucleus.
• ${R}_{n l} \left(r\right)$ is the radial component of the wave function.
• ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$ is the angular component of the wave function.

The radial nodes are where ${R}_{n l} \left(r\right) = 0$, and the angular nodes are where ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right) = 0$.

Since nonzero constants are clearly never zero, we just have to pick out the functions that have $r$ or $\theta$. That is, we just need the fact that:

${R}_{21} \left(r\right) \propto \sigma {e}^{- \sigma / 2}$

${Y}_{1}^{0} \left(\theta , \phi\right) \propto \cos \theta$

So, solving for the radial nodes, with $\sigma = Z r / {a}_{0}$:

$\frac{Z r}{{a}_{0}} {e}^{- Z r / 2 {a}_{0}} = 0$

This exponential never goes to zero except at $\infty$, at which the orbital doesn't exist, so we can eliminate it.

$\implies \frac{Z r}{{a}_{0}} = 0$

From here we can realize that there are no radial nodes in the $2 {p}_{z}$ orbital.

The only solution to this is $r = 0$, and at $r = 0$, no electrons exist because there is no electron that can fit in a radius of $\text{0 meters}$.

ANGULAR NODES?

Solving for the angular nodes,

$\cos \theta = 0$

And this only nonredundantly applies for $\theta \in \left[0 , \pi\right]$.

So, in this domain, $\cos \theta = 0$ at $\theta = \frac{\pi}{2}$, or ${90}^{\circ}$, relative to the axis of the orbital (the $z$ axis).

Furthermore, there is no $\phi$ dependence, so integration over the $\phi$ variable to generate this orbital probability density contributes only a constant of $2 \pi$:

${\text{Probability Density of 2p}}_{z}$

$= {\int}_{0}^{2 \pi} {\int}_{0}^{\pi} {\int}_{0}^{\infty} {\psi}_{2 p z}^{\text{*}} {\psi}_{2 p z} {r}^{2} \mathrm{dr} \sin \theta d \theta d \phi$

$= \textcolor{red}{\frac{1}{4 \sqrt{2 \pi}} {\left(\frac{Z}{a} _ 0\right)}^{5} {\int}_{0}^{2 \pi} d \phi} {\int}_{0}^{\pi} {\cos}^{2} \theta \sin \theta d \theta {\int}_{0}^{\infty} {r}^{2} {e}^{- Z r / {a}_{0}} \mathrm{dr}$

$= \textcolor{red}{\frac{2 \pi}{4 \sqrt{2 \pi}} {\left(\frac{Z}{a} _ 0\right)}^{5}} {\int}_{0}^{\pi} {\cos}^{2} \theta \sin \theta d \theta {\int}_{0}^{\infty} {r}^{2} {e}^{- Z r / {a}_{0}} \mathrm{dr}$

Having no dependence on the value of $\phi$ for a wave function that goes to zero at a particular value of $\theta$ means that a nodal plane is generated.

Therefore, the angular node is a plane perpendicular to the $z$ axis of the $2 {p}_{z}$ orbital, and is the $x y$ plane, just like it shows in the above diagram.

## Balance this reaction? #"H"_ ((aq))^(+) + "MnO"_ (4(aq))^(2-) -> "MnO"_ ((aq))^(-) + "MnO"_ (2(s)) + "H"_ 2"O"_ ((l))#

Stefan V.
Featured 2 months ago

Here's what's going on here.

#### Explanation:

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 $2$ electrons, which is why the oxidation number of manganese goes from $\textcolor{b l u e}{+ 6}$ on the reactants' side to $\textcolor{b l u e}{+ 4}$ on the products' side.

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, ${\text{H}}^{+}$, to the side that needs hydrogen.

$4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O}}_{\left(l\right)}$

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

$4 \times \left(1 +\right) + \left(2 -\right) + 2 \times \left(1 -\right) = 0 + 0$

$\textcolor{w h i t e}{a}$
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 $1$ electron, which is why the oxidation number of manganese goes from $\textcolor{b l u e}{+ 6}$ on the reactants' side to $\textcolor{b l u e}{+ 7}$ on the products' side.

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

$\left(2 -\right) = \left(1 -\right) + \left(1 -\right)$

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

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaaa)stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + "e}}^{-}\end{matrix}\right.$

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 $2$ to get

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + 2"e}}^{-}\end{matrix}\right.$

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

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + 2"e}}^{-}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}$
$4 {\text{H"_ ((aq))^(+) + ["MnO"_ (4(aq))^(2-) + 2"MnO"_ (4(aq))^(2-)] + color(red)(cancel(color(black)(2"e"^(-)))) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + color(red)(cancel(color(black)(2"e"^(-)))) + 2"H"_ 2"O}}_{\left(l\right)}$

You will end up with

$4 {\text{H"_ ((aq))^(+) + 3"MnO"_ (4(aq))^(2-) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + 2"H"_ 2"O}}_{\left(l\right)}$

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

Truong-Son N.
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 $\to$ 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.

## 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 6 days 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.

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