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## What volume of #"0.200 M"# #"HCl"# should be mixed with #"0.300 M"# #"Na"_2"CO"_3# to achieve a #"pH"# of #9.45# in a #"1-L"# solution?

Michael
Featured 6 months ago

Mix 569.9 ml of 0.200M $\textsf{H C {l}_{\left(a q\right)}}$ with 430.1 ml of 0.300M $\textsf{N {a}_{2} C {O}_{3}}$ solution.

#### Explanation:

When the acid is added to the carbonate solution, the #sf(CO_3^(2-)# ions are first protonated:

$\textsf{{H}^{+} + C {O}_{3}^{2 -} \rightarrow H C {O}_{3}^{-}}$

This means we are interested in the 2nd dissociation of carbonic acid:

$\textsf{H C {O}_{3}^{-} r i g h t \le f t h a r p \infty n s {H}^{+} + C {O}_{3}^{2 -}}$

#sf(K_(a2)=([H^+][CO_3^(2-)])/([HCO_3^(-)])=4.69xx10^(-11)#

Rearranging:

$\textsf{\left[{H}^{+}\right] = {K}_{a 2} \times \frac{\left[H C {O}_{3}^{-}\right]}{\left[C {O}_{3}^{2 -}\right]}}$

$\textsf{p H = 9.45}$

$\therefore$$\textsf{- \log \left[{H}^{+}\right] = 9.45}$

$\textsf{\left[{H}^{+}\right] = 3.55 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

$\therefore$$\textsf{3.55 \times {10}^{- 10} = 4.69 \times {10}^{- 11} \times \frac{\left[H C {O}_{3}^{-}\right]}{\left[C {O}_{3}^{2 -}\right]}}$

$\textsf{\frac{\left[H C {O}_{3}^{-}\right]}{\left[C {O}_{3}^{2 -}\right]} = \frac{3.55 \times {10}^{- 10}}{4.69 \times {10}^{- 11}} = 7.569 \text{ } \textcolor{red}{\left(1\right)}}$

This is the ratio of acid : co-base we need to achieve to get the target pH of 9.45.

A certain volume of HCl which I will call V must be added such that the total volume of buffer created is 1 Litre.

From the original equation you can see that if n moles of HCl is added, then n moles of $\textsf{H C {O}_{3}^{-}}$ is formed.

Since $\textsf{n = c \times v}$ we can say that the no. moles $\textsf{{H}^{+}}$ added is given by:

$\textsf{{n}_{{H}^{+}} = 0.200 \times V}$

So we can say that the no. moles $\textsf{H C {O}_{3}^{-}}$ formed is given by:

$\textsf{{n}_{H C {O}_{3}^{-}} = 0.200 \times V}$

Since the total volume must be 1 Litre we can say that the initial moles of $\textsf{C {O}_{3}^{2 -}}$ must be $\textsf{0.3 \times \left(1 - V\right)}$.

So the no. moles of $\textsf{C {O}_{3}^{2 -}}$ remaining is given by:

$\textsf{{n}_{C {O}_{3}^{2 -}} = 0.300 \times \left(1 - V\right) - 0.200 \times V}$

Since the total volume is common, we can put these into $\textsf{\textcolor{red}{\left(1\right)} \Rightarrow}$

$\textsf{\frac{0.2 V}{0.3 \left(1 - V\right) - 0.2 V} = 7.569}$

$\textsf{\frac{0.2 V}{0.3 - 0.3 V - 0.2 V} = 7.569}$

$\textsf{\frac{0.2 V}{0.3 - 0.5 V} = 7.569}$

$\textsf{0.2 V = 7.569 \left(0.3 - 0.5 V\right)}$

$\textsf{0.2 V = 2.2707 - 3.7845 V}$

$\textsf{3.9845 V = 2.2707}$

$\textsf{V = \frac{2.2707}{3.9845} = 0.5699 \textcolor{w h i t e}{x} L}$

$\textsf{V = 569.9 \textcolor{w h i t e}{x} m l}$

This is the volume of $\textsf{0.200 M}$ HCl required.

The volume of $\textsf{0.300 M}$ $\textsf{N {a}_{2} C {O}_{3}}$ required will be $\textsf{1 - 0.5699 = 0.4301 \textcolor{w h i t e}{x} L}$

$\textsf{= 430.1 \textcolor{w h i t e}{x} m l}$

Iteration check:

$\textsf{\left[{H}^{+}\right] = {K}_{a 2} \times \frac{\left[H C {O}_{3}^{-}\right]}{\left[C {O}_{3}^{2 -}\right]}}$

$\textsf{\left[{H}^{+}\right] = 4.69 \times {10}^{- 11} \times \frac{0.2 V}{0.3 \left(1 - V\right) - 0.2 V}}$

#sf([H^+]=4.69xx10^(-11)xx(0.2xx0.5699)/((0.3xx0.4301)-0.2xx0.5699)#

$\textsf{\left[{H}^{+}\right] = 0.35517 \times {10}^{- 9} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left(0.35517 \times {10}^{- 9}\right) = 9.45}$

So that's all good.

## Ionic equilibrium problem...?proof needed....

Truong-Son N.
Featured 5 months ago

$\left[{\text{H}}^{+}\right] = \frac{1}{2} \left(\sqrt{{K}_{a 1} {C}_{1}} \pm \sqrt{{K}_{a 1} {C}_{1} + 4 {K}_{a 2} {C}_{2}}\right)$

Note that this works best if ${K}_{a 1}$ and ${K}_{a 2}$ are both on the order of ${10}^{- 5}$ or less, i.e. the acids are sufficiently weak.

DISCLAIMER: DERIVATION!

Well, concentration is a state function, so we can simply choose acid 1 to go first, and acid 2 can go second, suppressed by the equilibrium of the first.

By writing out an ICE table, you would construct the mass action expression for acid 1 (say, $\text{HA}$):

${\text{HA"(aq) rightleftharpoons "H"^(+)(aq) + "A}}^{-} \left(a q\right)$

${K}_{a 1} = \frac{{\left[{\text{H"^(+)]_1["A"^(-)])/(["HA"]) = (["H}}^{+}\right]}_{1} {\alpha}_{1} {C}_{1}}{\left(1 - {\alpha}_{1}\right) {C}_{1}}$

Although we know that $\left[{\text{H"^(+)]_1 = ["A}}^{-}\right]$ in this first process, we choose to write them distinct from each other.

Since we assume ${\alpha}_{1}$ $\text{<<}$ $1$, i.e. ${K}_{a 1} < {10}^{- 5}$ or so, then rewrite this as...

#K_(a1) ~~ (["H"^(+)]_1alpha_1C_1)/(C_1) = alpha_1["H"^(+)]_1#

So,

${\left[{\text{H}}^{+}\right]}_{1} \approx {K}_{a 1} / {\alpha}_{1}$

The second acid, say ${\text{BH}}^{+}$, now acts.

${\text{BH"^(+)(aq) rightleftharpoons "B"(aq) + "H}}^{+} \left(a q\right)$

And the ${\left[{\text{H}}^{+}\right]}_{1}$ from acid 1 now is the initial concentration for the ${\text{BH}}^{+}$ dissociation.

${K}_{a 2} = \frac{\left[{\text{B"]["H"^(+)])/(["BH"^(+)]) = (["H}}^{+}\right] {\alpha}_{2} {C}_{2}}{\left(1 - {\alpha}_{2}\right) {C}_{2}}$

And since we also have ${\alpha}_{2}$ $\text{<<}$ $1$,

${K}_{a 2} \approx {\alpha}_{2} \left[{\text{H}}^{+}\right]$

$\left[{\text{H}}^{+}\right] \approx {K}_{a 2} / {\alpha}_{2}$

with $\left[{\text{H}}^{+}\right]$ being the net $\left[{\text{H}}^{+}\right]$ concentration. By subtracting out the contribution from acid 1,

${\alpha}_{2} {C}_{2} = {\left[{\text{H}}^{+}\right]}_{2} = {K}_{a 2} / {\alpha}_{2} - {\alpha}_{1} {C}_{1}$.

So, one form of this is

${\left[{\text{H"^(+)] = ["H"^(+)]_1 + ["H}}^{+}\right]}_{2}$

$= {\alpha}_{1} {C}_{1} + {\alpha}_{2} {C}_{2}$

But we have built into ${\alpha}_{2}$ the suppressed equilibrium, and so,

$\textcolor{red}{\left[{\text{H}}^{+}\right] < \frac{{K}_{a 1}}{\alpha} _ 1 + {K}_{a 2} / {\alpha}_{2}}$

It would be convenient to determine ${\alpha}_{2}$ in terms of ${\alpha}_{1}$, but it won't look nice at first.

${K}_{a 2} / {\alpha}_{2} = {\alpha}_{1} {C}_{1} + {\alpha}_{2} {C}_{2}$

$0 = {C}_{2} {\alpha}_{2}^{2} + {\alpha}_{1} {C}_{1} {\alpha}_{2} - {K}_{a 2}$

This becomes a quadratic equation. If you wish to see it,

$\textcolor{g r e e n}{{\alpha}_{2}} = \frac{- \left({\alpha}_{1} {C}_{1}\right) \pm \sqrt{{\alpha}_{1}^{2} {C}_{1}^{2} - 4 {C}_{2} \left(- {K}_{a 2}\right)}}{2 {C}_{2}}$

$= \textcolor{g r e e n}{\frac{- \left({\alpha}_{1} {C}_{1}\right) \pm \sqrt{{\alpha}_{1}^{2} {C}_{1}^{2} + 4 {C}_{2} {K}_{a 2}}}{2 {C}_{2}}}$

And so,

$\left[{\text{H}}^{+}\right] = {\alpha}_{1} {C}_{1} + \left(- \frac{{\alpha}_{1} {C}_{1}}{2 {C}_{2}} \pm \frac{\sqrt{{\alpha}_{1}^{2} {C}_{1}^{2} + 4 {C}_{2} {K}_{a 2}}}{2 {C}_{2}}\right) {C}_{2}$

$= \frac{1}{2} {\alpha}_{1} {C}_{1} \pm \sqrt{{\left(\frac{{\alpha}_{1} {C}_{1}}{2}\right)}^{2} + {K}_{a 2} {C}_{2}}$

And lastly, it would be convenient to know this in terms of ${K}_{a 1}$ instead of ${\alpha}_{1}$, since that requires information we may not already have.

Since ${K}_{a 1} \approx {\alpha}_{1} {\left[{\text{H}}^{+}\right]}_{1} \approx {\alpha}_{1}^{2} {C}_{1}$, we can say that

${\alpha}_{1} = \sqrt{{K}_{a 1} / {C}_{1}}$

Therefore:

$\textcolor{b l u e}{\left[{\text{H}}^{+}\right]} = \frac{1}{2} \sqrt{{K}_{a 1} {C}_{1}} \pm \sqrt{{\left(\frac{{C}_{1}}{2}\right)}^{2} {K}_{a 1} / {C}_{1} + {K}_{a 2} {C}_{2}}$

$= \frac{1}{2} \sqrt{{K}_{a 1} {C}_{1}} \pm \sqrt{\frac{1}{4} {K}_{a 1} {C}_{1} + {K}_{a 2} {C}_{2}}$

$= \textcolor{b l u e}{\frac{1}{2} \left(\sqrt{{K}_{a 1} {C}_{1}} \pm \sqrt{{K}_{a 1} {C}_{1} + 4 {K}_{a 2} {C}_{2}}\right)}$

TESTING ON ACTUAL PROBLEM

And of course, we should try this on an actual problem.

Consider acetic acid (${K}_{a} = 1.8 \times {10}^{- 5}$) and formic acid (${K}_{a} = 1.8 \times {10}^{- 4}$), for ${C}_{1} = \text{0.50 M}$ and ${C}_{2} = \text{2.00 M}$. Suppose we let formic acid dissociate first.

${\text{HCHO"_2(aq) rightleftharpoons "H"^(+)(aq) + "CHO}}_{2}^{-} \left(a q\right)$

$1.8 \times {10}^{- 4} = {x}^{2} / \left(0.50 - x\right) \approx {x}^{2} / 0.50$

So,

#["H"^(+)]_1 ~~ sqrt(1.8 xx 10^(-4) cdot 0.50) = "0.00949 M"#

and

${\alpha}_{1} \approx \frac{x}{\left[{\text{HCHO}}_{2}\right]} = 0.019$

Let's see if we get the same ${K}_{a}$.

#1.8 xx 10^(-4) stackrel(?)(=) (alpha_1C_1)^2/((1 - alpha_1)C_1)#

#= (0.019 cdot 0.50)^2/((1 - 0.019)cdot0.50) ~~ 1.8 xx 10^(-4) color(blue)(sqrt"")#

${\text{HC"_2"H"_3"O"_2(aq) rightleftharpoons "H"^(+)(aq) + "C"_2"H"_3"O}}_{2}^{-} \left(a q\right)$

#1.8 xx 10^(-5) = ((["H"^(+)]_1 + x)(x))/(2.00 - x) ~~ (["H"^(+)]_1 + x)x/2#

$\approx {\left[{\text{H}}^{+}\right]}_{1} \frac{x}{2} + {x}^{2} / 2$

The quadratic equation solves to be

#x = ["C"_2"H"_3"O"_2^(-)] = "0.00291 M"#,

which is also the ${\text{H}}^{+}$ contributed by the second acid.

And so, the total ${\text{H}}^{+}$ is:

#color(green)(["H"^(+)]) = "0.00949 M" + "0.00291 M" = color(green)ul("0.0124 M")#

This acid's percent dissociation at this concentration is

#alpha_2 = x/(["HC"_2"H"_3"O"_2]) = "0.00291 M"/"2.00 M" = 0.00145#.

So from the first form of the derived equation,

$\textcolor{g r e e n}{\left[{\text{H}}^{+}\right] = {\alpha}_{1} {C}_{1} + {\alpha}_{2} {C}_{2}}$

$= 0.019 \cdot 0.50 + 0.00145 \cdot 2.00 = \textcolor{g r e e n}{\underline{\text{0.0124 M}}}$ #color(blue)(sqrt"")#

Now to check the general formula we derived above.

#color(green)(["H"^(+)]) stackrel(?)(=) 1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2))#

$= \frac{1}{2} \left(\sqrt{1.8 \times {10}^{- 4} \cdot 0.50} \pm \sqrt{1.8 \times {10}^{- 4} \cdot 0.50 + 4 \cdot 1.8 \times {10}^{- 5} \cdot 2.00}\right)$

$= \frac{1}{2} \left(\sqrt{9.00 \times {10}^{- 5}} \pm \sqrt{9.00 \times {10}^{- 5} + 1.44 \times {10}^{- 4}}\right)$

$=$ $\textcolor{g r e e n}{\underline{\text{0.0124 M}}}$ #color(blue)(sqrt"")#

## Determine the symmetry elements and point groups of #"NF"_3#, #"NClF"_2#, and #"NBrClF"#. What symmetry elements are lost as we descend from #"NF"_3# to #"NCl"_2"F"# and then to #"NBrClF"#?

Ernest Z.
Featured 5 months ago

Here's what I get.

#### Explanation:

Point group of ${\text{NF}}_{3}$

${\text{NF}}_{3}$ has a trigonal pyramidal geometry. It belongs to point group ${\text{C}}_{\textrm{3 v}}$, like ammonia.

It has a ${\text{C}}_{3}$ principal axis and three #Ïƒ_text(v)# vertical planes of symmetry.

The image below shows the three #Ïƒ_text(v)# planes.

If you look at the molecule from the top, the ${\text{C}}_{\textrm{3}}$ axis becomes more apparent. You can rotate ${120}^{\circ}$ about the $z$ axis to return an indistinguishable configuration of the molecule.

1. ${\text{NF}}_{3}$ to ${\text{NClF}}_{2}$

${\text{NClF}}_{2}$ is also a trigonal pyramidal molecule. It belongs to point group ${\text{C}}_{\textrm{s}}$, however, just like if you slightly tilted one $\text{F}$ inwards on ${\text{NF}}_{3}$.

Its structure is

Its only symmetry element is a single #Ïƒ_text(h)# plane that bisects the molecule into two mirror image halves.

Thus, it has lost two ${\sigma}_{\textrm{v}}$ mirror planes and its ${C}_{3}$ axis of rotation.

2. ${\text{NClF}}_{2}$ to $\text{NBrClF}$

$\text{NBrClF}$ is a trigonal pyramidal molecule, but it has no symmetry element except the trivial one of rotating 360Â° about a ${\text{C}}_{1}$ axis.

That now belongs to point group ${\text{C}}_{1}$ since it has only the identity element.

Its structure is

Thus, it has lost even the ${\sigma}_{\textrm{h}}$ plane of ${\text{NClF}}_{2}$ .

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

Stefan V.
Featured 3 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

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

## Sterling silver,an alloy of copper, contains 7.5% copper by mass and has a density of 10.3g/cm^3. How many copper atoms are in a sterling silver sphere whose radius is 2.0mm?

Nam D.
Featured 2 months ago

I get approximately $2.45 \cdot {10}^{20}$ copper atoms.

#### Explanation:

NOTE: The answer is kind of long!

We have a sphere of sterling silver with a radius of $2 \setminus \text{mm}$. First, we need to find the volume of the sphere.

The volume of a sphere is given by: $\frac{4}{3} \pi {r}^{3}$

where $r$ is the radius of the sphere.

Plugging that into the equation, we get

$V = \frac{4}{3} \cdot \pi \cdot {\left(2 \setminus \text{mm}\right)}^{3}$

$= \frac{4}{3} \cdot \pi \cdot 8 \setminus {\text{mm}}^{3}$

$\approx 33.5 \setminus {\text{mm}}^{3}$

So, the volume of this alloy in the shape of a sphere is $33.5 \setminus {\text{mm}}^{3}$. To find the mass, we need to use the density equation. But since the density is in ${\text{g/cm}}^{3}$, we might as well convert ${\text{mm}}^{3}$ into ${\text{cm}}^{3}$.

We know that $1 \setminus {\text{cm"=10 \ "mm", :.1 \ "cm"^3=1000 \ "mm}}^{3}$.

So, $33.5 \setminus {\text{mm"^3=0.0335 \ "cm}}^{3}$.

The density of this alloy is $10.3 \setminus {\text{g/cm}}^{3}$, so the mass of this alloy will be

#0.0335color(red)cancelcolor(black)("cm"^3)*(10.3 \ "g")/(color(red)cancelcolor(black)("cm"^3))=0.34505 \ "g"#

So, this alloy's mass is $0.34505 \setminus \text{g}$.

Since #7.5%# of its mass is copper, then the mass of copper in this alloy is

$0.34505 \setminus \text{g"*7.5%=0.02587875 \ "g}$

To find the amount of copper atoms in this alloy, we first need to find the number of moles of copper, and we can do that by dividing the total mass of copper atoms in here by copper's molar mass.

Copper $\left(C u\right)$ has a molar mass of $63.546 \setminus \text{g/mol}$.

Therefore, there are

#(0.02587875color(red)cancelcolor(black)"g")/(63.546color(red)cancelcolor(black)"g""/mol")=0.000407244358 \ "mol"#

To convert between moles to atoms, we multiply the number of moles by Avogadro's number $\left({N}_{A}\right)$, which is $6.02 \cdot {10}^{23}$.

So we have in total:

$0.000407244358 \cdot 6.02 \cdot {10}^{23} \approx \textcolor{red}{\underline{\textcolor{b l a c k}{2.45 \cdot {10}^{20}}}}$ copper atoms.

## 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:

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

(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,

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

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

#p = (3.840 color(red)(cancel(color(black)("mol"))) Ã— "0.082 06" color(red)(cancel(color(black)("L")))Â·"atm"Â·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) Ã— 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,

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

#P = (nRT)/(V-nb) â€“ (n^2a)/V^2#

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

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

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