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## Can someone check my work please and please tell me if I made a mistake where I error?

Stefan V.
Featured 3 months ago

Here's my take on this.

#### Explanation:

The problem wants you to predict the precipitate produced by mixing two solutions, one that contains aluminium nitrate, #"Al"("NO"_3)_3#, a soluble ionic compound, and one that contains sodium hydroxide, $\text{NaOH}$, another soluble ionic compound.

The reaction takes place in aqueous solution, so right from the start you should be aware that you're dealing with ions.

Soluble ionic compounds dissociate completely in aqueous solution to form cations, which are positively charged ions, and anions, which are negatively charged ions.

The two solutions can thus be written as

${\text{Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO}}_{3 \left(a q\right)}^{-}$

It's absolutely crucial to make sure that you add the charges of the ions to the chemical equation.

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

Now, you should be familiar with the solubility rules for aqueous solutions.

The aluminium cations, ${\text{Al}}^{3 +}$, will combine with the hydroxide anions, ${\text{OH}}^{-}$, to form the insoluble aluminium hydroxide, which precipitates out of solution.

Notice that the aluminium cations have a $3 +$ charge and the hydroxide anions have a $1 -$ charge. This means that you're going to need $3$ hydroxide anions in order to balance the positive charge of the cation.

Therefore, the chemical formula for aluminium hydroxide is #"Al"("OH")_3#, .

The other product of the reaction will be aqueous sodium nitrate. The sodium cations have a $1 +$ charge and the nitrate anions have a $1 -$ charge, so the chemical formula for sodium nitrate will be ${\text{NaNO}}_{3}$.

The balanced chemical equation will thus looks like this

${\text{Al"("NO"_ 3)_ (3(aq)) + 3"NaOH"_ ((aq)) -> "Al"("OH")_ (3(s)) darr + 3"NaNO}}_{3 \left(a q\right)}$

The complete ionic equation for this reaction looks like this

${\text{Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-) + 3"Na"_ ((aq))^(+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + 3"Na"_ ((aq))^(+) + 3"NO}}_{3 \left(a q\right)}^{-}$

To get the net ionic equation, eliminate spectator ions, which are those ions that are found on both sides of the equation

#"Al"_ ((aq))^(3+) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-))))#

This will get you

#color(green)(|bar(ul(color(white)(a/a)color(black)("Al"_ ((aq))^(3+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr)color(white)(a/a)|)))#

Now, I'm not really sure if the math formatting is to blame for some of the equations you wrote, but you're missing subscripts and charges in all of them.

Take the first equation, for example. #"Al"("NO"_3)# is actually #"Al"("NO"_3)_3#.

I'm assuming that $\text{Al"^3"OH}$ is actually ${\text{Al"^(3+)"OH}}^{-}$. If that is the case, then the correct version would be

${\text{Al"^(3+)("OH"^(-))_3 implies "Al"^(3+) + 3"OH}}^{-}$

There's no such thing as ${\text{Na}}_{3}$. Sodium cannot form molecules, it can exist either as a solid, $\text{Na}$, or a cation, ${\text{Na}}^{+}$.

In your case, the sodium cation was present in solution, so ${\text{Na}}_{3}$ is actually $3 {\text{Na}}^{+}$.

Once again, remember to always add charges when dealing with ions. An ion without an added charge is not actually an ion.

For example, there's no such thing as ${\text{NO}}_{3}$. Instead, add the $1 -$ charge that belongs to the ions to get ${\text{NO}}_{3}^{-}$ anion.

Also, keep in mind that a coefficient add to a soluble ionic compound gets distributed to all the ions that are produced by said compound in solution.

For example,

$\textcolor{red}{3} {\text{NaOH" = color(red)(3)("Na"^(+) + "OH"^(-)) = color(red)(3)"Na"^(+) + color(red)(3)"OH}}^{-}$

$\textcolor{b l u e}{2} {\text{Al"("NO"_3)_3 = color(blue)(2)("Al"^(3+) + 3"NO"_3^(-)) = color(blue)(2)"Al"^(3+) + 6"NO}}_{3}^{-}$

Notice that the charges must remain balanced at all times.

All in all, you should review ionic compounds before diving into complete and net ionic equations.

## #"A"#, #"B"#, and #"C"# are 3 metals. Under standard conditions, when #"B"# is placed in a solution of either #"A"_((aq))^(2+)# or #"C"_((aq))^(2+)#, it is oxidized. When #"C"# is placed in a solution of #"A"_((aq))^(2+)#, #"C"# is not oxidized ?

Stefan V.
Featured 2 months ago

$\text{Pb}$, $\text{Zn}$, and $\text{Cu}$.

#### Explanation:

The idea here is that you need to use the standard reduction potentials, ${E}^{\circ}$, given to you to determine the identity of the three metals.

You know that you have

$\text{Zn"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Zn"_ ((s))" "E^@ = -"0.76 V}$

$\text{Pb"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Pb"_ ((s))" "E^@ = -"0.13 V}$

$\text{Cu"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Cu"_ ((s))" "E^@ = +"0.34 V}$

Now, the standard reduction potentials measure the tendency of a chemical species to release electrons and form cations when compared to that of hydrogen.

A negative ${E}^{\circ}$ means that the chemical species loses electrons more readily than hydrogen, and that its reduction equilibrium lies to the left.

A positive ${E}^{\circ}$ means that the chemical species loses electrons less readily than hydrogen, and that its reduction equilibrium lies to the right.

Now, chemical species that lose electrons more readily are stronger reducing agents than chemical species that tend to lose electrons less readily.

When you compare the ${E}^{\circ}$ values for two reduction equilibria, you can say that

• the equilibrium with the less positive / more negative ${E}^{\circ}$ value will lie further to the left
• the equilibrium with the less negative / more positive ${E}^{\circ}$ value will lie further to the right

Let's take the first two reduction equilibria. You have

#E^@("Zn"^(2+), "Zn") = -"0.76 V"#

#E^@("Pb"^(2+), "Pb") = -"0.13 V"#

Here ${E}^{\circ} = - \text{0.76 V}$ is more negative than ${E}^{\circ} = - \text{0.13 V}$, which means that the first equilibrium lies further to the left than the second one.

You can thus say that zinc metal, $\text{Zn}$, will reduce ${\text{Pb}}^{2 +}$ because zinc loses electrons more readily than lead metal does.

Consequently, zinc will also reduce ${\text{Cu}}^{2 +}$, since ${E}^{\circ} = - \text{0.76 V}$ is more negative than ${E}^{\circ} = + \text{0.34 V}$.

In other words, when zinc metal is placed in a solution that contains ${\text{Pb}}^{2 +}$ and ${\text{Cu}}^{2 +}$ cations, the cations will be reduced to lead metal and copper metal, respectively, and zinc will be oxidized to ${\text{Zn}}^{2 +}$.

This implies that zinc is metal $\text{B}$.

Finally, notice that metal $\text{C}$ cannot reduce ${\text{A}}^{2 +}$, since it will not be oxidized when placed in a solution that contains ${\text{A}}^{2 +}$. This means that the ${E}^{\circ}$ of metal $\text{C}$ is more positive than ${E}^{\circ}$ for metal $\text{A}$. As a result, metal $\text{C}$ will be copper and metal $\text{A}$ will be lead.

$\text{Metal A "-> " Lead}$

$\text{Metal B " -> " Zinc}$

$\text{Metal C " -> " Copper}$

So, to answer such problems quickly, list the reduction half-reactions in order of increasing ${E}^{\circ}$ values

$\text{Zn"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons color(blue)("Zn"_ ((s)))" "E^@ = -"0.76 V}$

#color(red)("Pb"_ ((aq))^(2+)) + 2"e"^(-) rightleftharpoons "Pb"_ ((s))" "E^@ = -"0.13 V"#

#color(red)("Cu"_ ((aq))^(2+)) + 2"e"^(-) rightleftharpoons "Cu"_ ((s))" "E^@ = +"0.34 V"#

and keep in mind that the chemical species listed in the top right can reduce the chemical species listed in the bottom left.

For example, zinc metal is $\textcolor{b l u e}{\text{top right}}$, so it can reduce ${\text{Pb}}^{2 +}$ and ${\text{Cu}}^{2 +}$ because they are located in the $\textcolor{red}{\text{bottom left}}$.

Similarly, you can say that the chemical species located in the bottom left can oxidize the chemical species located to the top right.

Here ${\text{Cu}}^{2 +}$ and ${\text{Pb}}^{2 +}$ are $\textcolor{red}{\text{bottom left}}$, so they can oxidize zinc metal, which is located in the $\textcolor{b l u e}{\text{top right}}$.

$\left\{\left(\textcolor{red}{\text{Bottom left")color(white)(a)"oxidizes" color(white)(a)color(blue)("top right ")), (color(blue)("Top right")color(white)(a)"reduces" color(white)(a)color(red)("bottom left}}\right)\right.$

## Consider the following reaction: #Xe(g) + 2F_2(g) -> XeF_4(g)#. A reaction mixture initially contains 2.24 atm #Xe# and 4.27 atm #F_2#. lf the equilibrium pressure of #Xe# is 0.34 atm, how do you find the equilibrium constant (#K_p#) for the reaction?

Stefan V.
Featured 2 months ago

${K}_{p} = 25.3$

#### Explanation:

The equilibrium reaction given to you looks like this

${\text{Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF}}_{4 \left(g\right)}$

You know that the mixture initially contains $\text{2.24 atm}$ of xenon, $\text{Xe}$, and $\text{4.27 atm}$ of fluorine gas, ${\text{F}}_{2}$.

It's worth mentioning that the problem provides you with pressures instead of moles because when the volume of the container remains unchanged, and the temperature of the reaction is kept constant, the pressure of a gas is directly proportional to the number of moles of gas present.

Now, notice that the pressure of xenon is decreasing to an equilibrium value of $\text{0.34 atm}$. This is a significant decrease, which can only mean that the equilibrium constant, ${K}_{p}$, is greater than $1$.

In other words, the forward reaction is favored at the temperature at which the reaction takes place.

Consequently, you can expect the equilibrium pressure of fluorine gas to be significantly lower than its initial value.

So, use an ICE table to find the equilibrium pressures of fluorine gas and xenon tetrafluoride

${\text{ ""Xe"_ ((g)) " "+" " color(red)(2)"F"_ (2(g)) " "rightleftharpoons" " "XeF}}_{4 \left(g\right)}$

#color(purple)("I")color(white)(aaaaacolor(black)(2.24)aaaaaaaacolor(black)(4.27)aaaaaaaaaaacolor(black)(0)#
#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaacolor(black)((-color(red)(2)x))aaaaaaaaacolor(black)((+x))#
#color(purple)("E")color(white)(aaacolor(black)(2.24-x)aaaacolor(black)(4.27-color(red)(2)x)aaaaaaaaacolor(black)(x)#

Now, you know that the equilibrium concentration of xenon is $\text{0.34 atm}$. This means that you have

$2.24 - x = 0.34 \implies x = 1.90$

The equilibrium pressures of fluorine gas and xenon tetrafluoride will thus be

#("F"_2) = 4.27 - color(red)(2) * 1.90 = "0.47 atm"#

#("XeF"_4) = "1.90 atm"#

By definition, ${K}_{p}$ will be equal to

${K}_{p} = \left({\left({\text{XeF"_4))/(("Xe") * ("F}}_{2}\right)}^{\textcolor{red}{2}}\right)$

Plug in your values to find

#K_p = (1.90 color(red)(cancel(color(black)("atm"))))/(0.34color(red)(cancel(color(black)("atm"))) * ("0.47 atm")^color(red)(2)) = "25.3 atm"^(-2)#

${K}_{p} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{25.3} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ rounded to three sig figs

As predicted, the equilibrium constant turned out to be grater than $1$.

## What is the percent dissociation if 1 mole of x2y is introduced into 1l vessel at 1000k?kc for the dissociation of x2y=8*10^-6. 2x2y→←2x2+y2.

dk_ch
Featured 2 months ago

The given reversible gaseous reaction :

$2 {X}_{2} Y \left(g\right) \text{ "rightleftharpoons" "2X_2(g)" "+" } {Y}_{2} \left(g\right)$

$\textcolor{red}{I} \text{ "1" "mol" " 0" "mol" " 0" " } m o l$

$\textcolor{red}{C} \text{ "-alpha" "mol" " alpha" "mol" " alpha/2" " } m o l$

$\textcolor{red}{E} \text{ "1-alpha" "mol" " alpha" "mol" " alpha/2" " } m o l$

Where $\alpha$ is the degree of dissociation of ${X}_{2} Y \left(g\right)$

Volume of the vessel being 1 L
the concentrations of the reactants and products at equilibrium will be as follows

$\left[{X}_{2} Y \left(g\right)\right] = \left(1 - \alpha\right) M$

$\left[{X}_{2} \left(g\right)\right] = \alpha M$

$\left[{Y}_{2} \left(g\right)\right] = \frac{\alpha}{2} M$

Now concentration equilibrium constant

${K}_{c} = \frac{{\left[{X}_{2} \left(g\right)\right]}^{2} \left[{Y}_{2} \left(g\right)\right]}{{X}_{2} Y \left(g\right)} ^ 2$

$\implies 8 \cdot {10}^{-} 6 = \frac{{\left(\alpha\right)}^{2} \cdot \frac{\alpha}{2}}{1 - \alpha} ^ 2$

Neglecting $\textcolor{red}{\alpha}$ comparing with 1

$\implies {\alpha}^{3} / 2 = 8 \cdot {10}^{-} 6$

$\implies \alpha = {16}^{\frac{1}{3}} \cdot {10}^{-} 2 \approx 2.5 \cdot {10}^{-} 2 \to$ No. of moles dissociated per mole of reactant

So the percent dissociation #=2.5*10^-2*100=2.5%#

## How do you calculate the change in pH when 3.00 mL of 0.100 M #"HCl"(aq)# is added to 100.0 mL of a buffer solution that is 0.100 M in #"NH"_3(aq)# and 0.100 M in #"NH"_4"Cl"(aq)# ?

Stefan V.
Featured 1 month ago

${\Delta}_{\text{pH}} = - 0.026$

#### Explanation:

You're adding hydrochloric acid, $\text{HCl}$, a strong acid, to your buffer, so right from the start you should expect it pH to decrease.

This implies that the change in pH will be negative

${\Delta}_{\text{pH" = "pH"_ "final" - "pH"_ "initial}} < 0$

However, the fact that you're dealing with a buffer solution lets you know that this change will not be significant since the role of a buffer is to resist significant changes in pH that result from the addition of strong acid or strong bases.

So, hydrochloric acid will react with ammonia, ${\text{NH}}_{3}$, a weak base, to produce the ammonium cation, ${\text{NH}}_{4}^{+}$, the ammonia's conjugate acid, and water.

${\text{HCl"_ ((aq)) + "NH"_ (3(aq)) -> "NH"_ (4(aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

The reaction consumes hydrochloric acid and ammonia in a $1 : 1$ mole ratio. Also, notice that for very mole of hydrochloric acid or ammonia consumed by the reaction, one mole of ammonium cations is produced. Keep this in mind.

Use the molarity of the ammonia solution and the volume of the buffer to calculate how many moles it contains

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will have

#n_("NH"_3) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#

$= {\text{0.0100 moles NH}}_{3}$

Do the same for the ammonium cations, ${\text{NH}}_{4}^{+}$

#n_("NH"_4^(+)) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#

$= {\text{0.0100 moles NH}}_{4}^{+}$

Calculate how many moles of hydrochloric acid are being added to the buffer

#n_("HCl") = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 3.00 * 10^(-3)color(red)(cancel(color(black)("L")))#

$= \text{0.000300 moles HCl}$

You know that hydrochloric acid and ammonia react in a $1 : 1$ mole ratio, which means that the resulting solution will contain

#n_("HCl") = "0 moles HCl" -># completely consumed

#n_("NH"_3) = "0.0100 moles" - "0.000300 moles"#

$= {\text{0.0097 moles NH}}_{3}$

#n_("NH"_4^(+)) = "0.0100 moles" + "0.000300 moles"#

$= {\text{0.0103 moles NH}}_{4}^{+}$

The total volume of the buffer will be

${V}_{\text{total" = "100.0 mL" + "3.00 mL" = "103.0 mL}}$

The new concentrations of ammonia and ammonium cations will be

#["NH"_3] = "0.0097 moles"/(103.0 * 10^(-3)"L") = "0.094175 M"#

#["NH"_4^(+)] = "0.0103 moles"/(103.0 * 10^(-3)"L") = "0.100 M"#

Notice that the concentration of ammonium cations remained virtually unchanged because the increase in the number of moles was counteracted by the increase in volume.

Now, you can find the change in pH by using the Henderson - Hasselbalch equation, which for a buffer that contains a weak base and its conjugate acid looks like this

#color(blue)(|bar(ul(color(white)(a/a)"pOH" = "p"K_b + log((["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#

Notice that the the initial buffer solution had equal concentrations of weak base and conjugate acid. This means that its pOH was equal to

#"pOH" = "p"K_b + log( color(red)(cancel(color(black)("0.100 M")))/(color(red)(cancel(color(black)("0.100 M")))))#

$\text{pOH" = "p} {K}_{b}$

After the strong acid is added to the buffer, the pOH of the buffer will be

#"pOH" = "p"K_b + log( (0.100 color(red)(cancel(color(black)("M"))))/(0.095175color(red)(cancel(color(black)("M")))))#

$\text{pOH" = "p} {K}_{b} + 0.026$

You know that an aqueous solution at room temperature has

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " pOH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

For the initial solution, you have

$\text{pH"_ "initial" = 14 - "p} {K}_{b}$

For the final solution, you have

#"pH"_ "final" = 14 - ("p"K_b + 0.026)#

$\text{pH"_ "final" = 14 - "p} {K}_{b} - 0.026$

Therefore, you can say that the change in pH is equal to

#Delta_ "pH" = color(red)(cancel(color(black)(14))) - color(red)(cancel(color(black)("p"K_b))) - 0.026 - color(red)(cancel(color(black)(14))) + color(red)(cancel(color(black)("p"K_b)))#

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\Delta}_{\text{pH}} = - 0.026} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As predicted, the change in pH is negative because the pH of the buffer decreased as a result of the addition of a strong acid.

## Estimate the value of the equilibrium constant at 610 K for each of the following reactions. ΔH∘f and S∘ for BrCl(g) is 14.6 kJ/mol and 240.0 Jmol⋅K, respectively. 2NO2(g)⇌N2O4(g)?

Ernest Z.
Featured 1 month ago

I get #K = 1.93 × 10^11# for $\text{BrCl}$ and #2.03 × 10^4# for ${\text{N"_2"O}}_{4}$.

#### Explanation:

For $\text{BrCl}$

$\text{Br"_2"(g)" + "Cl"_2"(g)" ⇌ "2BrCl(g)}$
#Δ_fH^° = "14.6 kJ·mol"^"-1"#
#ΔS^° = "240.0 J·K"^"-1""mol"^"-1"#

#Δ_fG^° = Δ_fH^° - TΔS^° = "14.6 kJ·mol"^"-1" - 610 color(red)(cancel(color(black)("K"))) × "0.2400 kJ"·color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1"= "14.6 kJ·mol"^"-1" - "146.4 kJ·mol"^"-1" = "-131.8 kJ·mol"^"-1"#

#ΔG = "-"RTlnK#

#lnK = ("-"ΔG)/(RT) = ("131 800" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 610 color(red)(cancel(color(black)("K")))) = 25.99#

#K = e^25.99 = 1.93 × 10^11#

For $\text{N"_2"O"_4}$

$\textcolor{w h i t e}{m m m m m m m m m} 2 {\text{NO"_2 ⇌ "N"_2"O}}_{4}$
#Δ_fH^°"/kJ·mol"^"-1":color(white)(ll)33.10color(white)(mmm)9.08#
#S^°"/J·K"^"-1""mol"^"-1": color(white)(ll)240.04color(white)(mm)304.38#

The formula for enthalpy of reaction is

#color(blue)(|bar(ul(color(white)(a/a) Δ_rH = sumΔ_fH_text(products) - sumΔ_fH_text(reactants)color(white)(a/a)|)))" "#

# Δ_rH = "[1(9.08) - 2(33.01)] kJ" = "-56.94 kJ"#

The formula for the entropy of reaction is

#color(blue)(|bar(ul(color(white)(a/a)Δ_rS = sumS_text(products) - sumS_text(reactants)color(white)(a/a)|)))" "#

#Δ_rS = "(1×304.38 - 2×240.04) J/K" = "-175.80 J/K"#

#color(blue)(|bar(ul(color(white)(a/a)ΔG = ΔH - TΔScolor(white)(a/a)|)))" "#

#Δ_rG = "-56.94 kJ" - 610 color(red)(cancel(color(black)("K"))) × ("-0.175 80 kJ"·color(red)(cancel(color(black)("K"^"-1")))) = "-56.94 kJ" + "107.238 kJ" = "-50.30 kJ"#

The formula for $K$ is

#color(blue)(|bar(ul(color(white)(a/a)ΔG = -RTlnKcolor(white)(a/a)|)))" "#

#lnK = ("-"ΔG)/(RT) = ("50 300" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 610 color(red)(cancel(color(black)("K")))) = 9.918#

#K = e^"9.918" = 2.03 × 10^4#

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