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Given the following information and assuming the final solution will be diluted to 1.00L, how much more HCl should you add to achieve the desired pH?

Stefan V.
Featured 5 months ago

$\text{42.9 mL}$

Explanation:

The idea here is that adding sodium hydroxide, $\text{NaOH}$, to your hydrochloric acid solution will neutralize some, if not all, depending on how much you've added, of the acid.

$\text{100.0 mL "-> 6.00 * 10^(-2)"M HCl}$

$\text{100.0 mL " -> 5.00 * 10^(-2)"M NaOH}$

Now, you know that after you realize your error, you're left with $\text{81.0 mL}$ of hydrochloric acid and $\text{89.0 mL}$ of sodium hydroxide. This means that you've added to the resulting solution

$\text{100.0 mL " - " 81.0 mL" = "19.0 mL HCl}$

$\text{100.0 mL " - " 89.0 mL" = "11.0 mL NaOH}$

When you mix hydrochloric acid, a strong acid, and sodium hydroxide, a strong base, a neutralization reaction takes place

${\text{HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

Because hydrochloric acid and sodium hydroxide produce hydrogen cations, ${\text{H}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$, respectively, in a $1 : 1$ mole ratio, you will have

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

Notice that this reaction consumes hydrogen cations and hydroxide anions in a $1 : 1$ mole ratio, which means that for every mole of hydrochloric acid present it takes one mole of sodium hydroxide to neutralize it.

Use the molarities and volumes of the solutions you've mixed to calculate how many moles of each were added

#19.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace((6.00 * 10^(-2)"moles HCl")/(1color(red)(cancel(color(black)("L")))))^(color(blue)(=6.00 * 10^(-2)"M")) = "0.00114 moles H"^(+)#

#11.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace((5.00 * 10^(-2)"moles NaOH")/(1color(red)(cancel(color(black)("L")))))^(color(blue)(=6.00 * 10^(-2)"M")) = "0.000550 moles OH"^(-)#

So, you know that you've accidentally added $0.00114$ moles of hydrogen cations and $0.000550$ moles of hydroxide anions to your solution.

Since you have fewer moles of hydroxide anions present, you can say that the hydroxide anions will be completely consumed by the neutralization reaction, i.e. they will act as a limiting raegent.

Your resulting solution will thus contain

$0.000550 - 0.000550 = {\text{0 moles OH}}^{-} \to$ completely consumed

$0.00114 - 0.000550 = {\text{0.000590 moles H}}^{+}$

Now, the volume of this solution will be equal to the volume of hydrochloric acid and the volume of sodium hydroxide solutions you've mixed

${V}_{\text{sol" = "19.0 mL" + "11.0 mL" = "30.0 mL}}$

This means that the concentration of hydrogen cations in the resulting solution will be

#["H"^(+)] = "0.000590 moles"/(30.0 * 10^(-3)"L") = "0.01967 M"#

The pH of the solution is given by

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"^(+)])color(white)(a/a)|)))#

$\text{pH} = - \log \left(0.01967\right) = 1.71$

So, you know that your target solution must have a volume of $\text{1.00 L}$ and a pH of $2.50$. Use the above equation to find the concentration of hydrogen cations needed to have this target solution

#"pH" = - log(["H"^(+)]) implies ["H"^(+)] = 10^(-"pH")#

#["H"^(+)] = 10^(-2.50) = "0.003162 M"#

For a volume equal to $\text{1.00 L}$, this gives you $0.003162$ moles of hydrogen cations -- remember that when dealing with a liter of solution, molarity and number of moles of solute are interchangeable.

Your solution contains $0.000590$ moles of hydrogen cations and needs $0.003162$ moles, which means that you must add

#n_("H"^(+)"needed") = 0.003162 - 0.000590 = "0.002572 moles H"^(+)#

Use the molarity of the stock hydrochloric acid solution to see what volume would contain this many moles of acid

#0.002572 color(red)(cancel(color(black)("moles H"^(+)))) * "1.0 L"/(6.00 * 10^(-2)color(red)(cancel(color(black)("moles H"^(+))))) = "0.0429 L"#

This is equivalent to $\text{42.9 mL}$, which means that in order to prepare your target solution, you must add another $\text{42.9 mL}$ of stock hydrochloric acid solution to the existing $\text{30.0 mL}$ solution.

This will give you a volume of $\text{72.9 mL}$ of solution. To get your target solution, you must add enough distilled water to get the total volume to $\text{1000 mL}$.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{volume of HCl needed " = " 42.9 mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

Estimate the value of the equilibrium constant at #"610 K"# for each of the following reactions? #Delta_fH^@# and #S^@# for #"BrCl"(g)# are #"14.6 kJ/mol"# and #"240.0 J/mol"cdot"K"#, respectively.

Ernest Z.
Featured 5 months 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#

What are the methods in separating mixtures?

mrpauller
Featured 4 months ago

Several techniques can be used. They include - filtration, centrifugation, chromatography, crystallization...

Explanation:

Mixtures can be separated using various separation methods such filtration,separating funnel,sublimation,simple distillation and paper chromatography.

The methods stated above are all physical methods. There are also chemical methods, which are used by rearranging the particles so a certain substance no longer exists (chemical reaction). However here is my explanation on the four MAIN physical methods (the ones that show up on tests) of separation:

1. Distillation.
If two substances have different boiling points and are mixed together, you can boil them and the one with the lower boiling point will evaporate out.

Here is a video of an experiment which uses distillation to purify water from a solution of salt water.

Video from: Noel Pauller

2. Chromatography .
If you've ever done the experiments where you draw colored dots on a paper towel and dip it in water, this is that. Chromatography is when a substance is carried away (through the towel in this case) or spread around by the absorption of water. You can also think of a spill on a rug.

Here is a video which shows a paper chromatography experiment which was conducted to separate the pigments found in a black overhead marker.

Video from: Noel Pauller

3. Crystallization.
For a supersaturated solution, you can choose to let the solute crystallize out. Example: Rock candy (sugar solution)

4. Filtration/decanting.
Sort them out by particle size using a selective membrane such as filter paper.
The video below shows how filtration can be used to separate calcium carbonate (chalk) from water.

Video from: Noel Pauller

What is degenerate state and degeneracy? Are #p# and #d# orbitals in same shell of the same energy? Are #s# orbitals (for example) of every shell equal in energy? I want to know the energy of different orbital respective to same and different shell.

Truong-Son N.
Featured 3 months ago

This is a kind of complicated question.

• A degenerate state is a state in which the energy is the same as other states.
• A degeneracy is the number of states that have that same energy, and is described as $2 l + 1$.

So, the difference is that degeneracy describes how many states, and a degenerate state is specifically which ones count.

• Any $n s$ orbital is the same energy for the same $n$.
• No, $p$ and $d$ orbitals of the same $n$ don't necessarily have the same energy (not even in hydrogen atom).
• I cannot tell you the energy of every orbital, because their energies change throughout the periodic table, and sometimes the actual ordering is different.

DEGENERATE STATES

Atomic orbitals that share the same principal quantum number $n$ and angular momentum quantum number $l$, but not ${m}_{l}$ are necessarily degenerate. They are called degenerate states.

(if they share the same ${m}_{l}$ as well, they are the same orbital!)

On the other hand, if their $n$ and $l$ values differ, they might be degenerate by coincidence, but they are not necessarily degenerate.

For example, the $4 s$ and $4 p$ orbitals share the same $n$, but they do not share the same $l$. That means:

• They do not have the same shape (different $l$).
• They do not have the same total number of nodes (different $n - l - 1$).
• Possibly, they do not have the same number of each type of node (radial vs. planar).
• Maybe some combination of all three.

However, the $4 s$ and $3 d$ orbitals differ in both $n$ and $l$, and yet, they sometimes are very similar in energy, depending on what atom we are talking about.

One might call them degenerate once the energies get close enough. When comparing their $n$ and $l$, the net effects to the energies are counteracting (lower $n$ vs. higher $l$ is counteracting). That is why I said that these are "degenerate by coincidence".

DEGENERACY

Like I said, degeneracy is just the number of orbitals of the same energy. Typically we say, for example:

• The $n p$ atomic orbitals are triply degenerate, because there are three of them in a free atom, and they are all equivalent orbitals with equivalent energies.
• The $n p$ atomic orbitals have a degeneracy of $2 l + 1 = 3$, since their $l$ is equal to $1$.

For example, a $2 {p}^{4}$ configuration could be drawn as:

$\underline{\uparrow \downarrow} \text{ " ul(uarr color(white)(darr)) " } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underbrace{\text{ "" "" "" "" "" "" "" }}$
$2 {p}_{x} \text{ "" "2p_y" "" } 2 {p}_{z}$

If the uncertainty in the position of an electron is equal to its de Broglie wavelength, then its velocity in that direction becomes completely unstable. Explain?

Truong-Son N.
Featured 1 month ago

Well, referring to the Heisenberg Uncertainty Principle, there is one formulation of it that is fairly easy to use in calculations:

#bb(DeltaxDeltap_x >= ℏ)#,

or

$\boldsymbol{\Delta x \Delta {p}_{x} \ge \frac{h}{2 \pi}}$,

where #ℏ = h/(2pi)# is the reduced Planck's constant, and $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$. You may also see #ℏ/2#, or $\frac{h}{4 \pi}$, but that's beside the point. The point is, it's on the order of #ℏ#.

The de Broglie wavelength is:

$\lambda = \frac{h}{m v}$

If the uncertainty in the position becomes numerically equal to $\lambda$, then we can plug in $\lambda = \frac{h}{m {v}_{x}} = \Delta x$ to get:

$\frac{\cancel{h}}{m {v}_{x}} \Delta {p}_{x} \ge \frac{\cancel{h}}{2 \pi}$

Since $p = m v$, $\Delta p = m \Delta v$ for a particle experiencing little relativistic effects on its mass:

$\frac{1}{\cancel{m} {v}_{x}} \cancel{m} \Delta {v}_{x} \ge \frac{1}{2 \pi}$

$\frac{\Delta {v}_{x}}{v} _ x \ge \frac{1}{2 \pi}$

Flipping both sides, we get:

$\textcolor{b l u e}{{v}_{x} / \left(\Delta {v}_{x}\right) \le 2 \pi}$

Since $\lambda$ is very small (on the order of $n m$ for electrons), we expect that the uncertainty in the velocity is very large so that the inequality $\Delta x \Delta {p}_{x} \ge \frac{h}{2 \pi}$ is maintained.

That should make sense because if $\Delta {v}_{x}$ is very large, only then would ${v}_{x} / \left(\Delta {v}_{x}\right) \le 2 \pi$ hold true, since ${v}_{x}$ for an electron is generally quite large as well.

EXAMPLE

For instance, the $1 s$ electron in hydrogen atom travels at around $\frac{1}{137}$ times the speed of light, or ${v}_{x} = 2.998 \times {10}^{8} \times \frac{1}{137} = 2.188 \times {10}^{6} \text{m/s}$. That means:

$\frac{2.188 \times {10}^{6}}{\Delta {v}_{x}} \le 2 \pi$

$\textcolor{red}{\Delta {v}_{x} \ge 3.483 \times {10}^{5}}$ $\textcolor{red}{\text{m/s}}$

i.e. the uncertainty in the velocity of a $1 s$ electron is AT LEAST #~~15.92%# of the velocity of the electron (not at most... at least) when you are sure of the position of the electron to within $\text{nm}$ of horizontal distance.

It physically means that if we were to try to predict its velocity, we are extremely unsure of which way it's going and at what actual velocity.

In real life, if this were to be the case, then if you shined a laser through a slit of a few $\text{nm}$ of width, you would see a bunch of constructive and destructive interference on the far walls, indicating the high uncertainty in the velocity along the $x$ axis:

Of course, the de Broglie relation is for electrons, as photons have no mass, but both behave as waves, and so, the slit experiment applies to both.

What is bond order? How we can determine it? Please basics. Thank you so much.

Truong-Son N.
Featured 1 month ago

A conceptual approach is to simply count electrons in a bond and treat each bonding valence electron as half a bond order.

This works for many cases, except for when the highest-energy electron is in an antibonding molecular orbital.

SIMPLE CASE

For example, the bond order of $: \text{N"-="N} :$ fairly straightforward because it's a triple bond, and each bonding valence electron contributes half a bond order.

So:

#"BO"_"triple" = "BO"_sigma + 2"BO"_pi = 1/2 xx ("2 electrons") + 2(1/2 xx ("2 electrons")) = 3# for the bond order, as we should expect, since bond order tells you the "degree" of bonding.

MULTI-ATOM CASE

Or, in a more complicated example, like ${\text{NO}}_{3}^{-}$, a conceptual approach is to count the number of electrons in the bond and see how many bonds it is distributed across.

So, in ${\text{NO}}_{3}^{-}$, which has one double bond in its resonance structure, has $2$ electrons in its $\pi$ bond, distributed across three $\text{N"-"O}$ bonds.

That means its $\boldsymbol{\pi}$-bond order is simply $\frac{1}{2} \cdot \left(\text{2 pi electrons")/("3 N"-"O bonds") = color(blue)("0.333}\right)$, making the bond order for each $\text{N"-"O}$ bond overall be:

${\text{BO" = "BO"_sigma + "BO}}_{\pi} = 1 + 0.333 = 1.333$.

Therefore, ${\text{NO}}_{3}^{-}$ on average actually has three "$\boldsymbol{\text{1.333}}$" bonds overall (instead of one double bond and two single bonds), meaning it is one third of the way between a single bond and a double bond.

EXCEPTION EXAMPLE: O2

${\text{O}}_{2}$ actually has two singly-occupied ${\pi}^{\text{*}}$ antibonding orbitals.

If we were to calculate its bond order, we would get $2$ normally, corresponding to the #:stackrel(..)("O")=stackrel(..)"O":# Lewis structure.

But what if we wanted the bond order for ${\text{O}}_{2}^{+}$? From the discussion above, we may expect $1.5$, but it's NOT $1.5$, even though ${\text{O}}_{2}^{+}$ has one less valence electron. What is it actually?

You may realize that we would have removed one electron from an ${\pi}^{\text{*}}$ antibonding molecular orbital. That means we've removed half a bond order corresponding to antibonding character, which is the same as adding half a bond order corresponding to bonding character.

So, by removing an antibonding electron, we've done the equivalent of adding a bonding electron.

In other words, we've decreased a bond-weakening factor, thereby increasing the bonding ability of the molecule.

Therefore, the actual bond order of ${\text{O}}_{2}^{+}$ is $\boldsymbol{2.5}$, stronger than ${\text{O}}_{2}$!

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