See below:
When cobalt(II) chloride is dissolved in water the pink, hydrated
This is a complex ion of octahedral symmetry where the central cobalt is surrounded by 6 water ligands:
If concentrated hydrochloric acid is added we are adding a large XS of
The reason why only 4 ligands can be accomodated is due to the large size of the
This can be expressed as:
You can see that Le Chatelier would predict that adding XS
Similarly if an XS of water is added the position of equilibrium is driven to the left and the pink colour is restored.
The 2 solutions look like this:
You may have come across this reaction where cobalt chloride paper is used to test for the presence of water:
By using the ladder operators, we derived:
#barul|stackrel(" ")(" "l(l+1) >= m_l (m_l pm 1)" ")|#
and from this inequality we get that
INTRODUCTORY RELATIONS
Following this and this page, let us introduce the ladder operators for orbital angular momentum
#hatL_(pm) = hatL_x pm ihatL_y# where
#hatL_i# is the angular momentum operator for the#i# th direction in 3D space.
These satisfy the commutation relations:
#[hatL^2, hatL_(pm)] = hatL^2hatL_(pm) - hatL_(pm)hatL^2 = 0#
#[hatL_(pm), hatL_z] = hatL_(pm)hatL_z - hatL_zhatL_(pm) = ∓ℏhatL_(pm)# where
#hatL# is the orbital angular momentum operator and#hatL_z# is its#z# component.
Now, the eigenvalues we get when we operate on the angular wave function
#color(green)ul(hatL^2)Y_(l)^(m_l)(theta,phi) = color(green)ul(ℏ^2l(l+1))Y_(l)^(m_l)(theta,phi)#
#color(green)ul(hatL_z)Y_(l)^(m_l)(theta,phi) = color(green)ul(m_lℏ)Y_(l)^(m_l)(theta,phi)#
DO THESE LADDER OPERATORS CHANGE
Now we shall ask, what happens to the value of
#color(red)(hatL^2hatL_(pm)Y_(l)^(m_l)(theta,phi) = ???cdotY_(l)^(m_l)(theta,phi))#
Since
#color(green)(hatL^2hatL_(pm)Y_(l)^(m_l)(theta,phi)) = hatL_(pm)hatL^2Y_(l)^(m_l)(theta,phi)#
This eigenvalue is known, so that helps...
#hatL_(pm)hatL^2Y_(l)^(m_l)(theta,phi)#
#= hatL_(pm)[ℏ^2l(l+1)Y_(l)^(m_l)(theta,phi)]#
#= color(green)(ℏ^2l(l+1)hatL_(pm)Y_(l)^(m_l)(theta,phi))#
Nothing has happened to
DO THESE LADDER OPERATORS CHANGE
What about
#color(red)(hatL_zhatL_(pm)Y_(l)^(m_l)(theta,phi) = ???cdotY_(l)^(m_l)(theta,phi))#
These operators do not commute, so we use the commutation relation we put earlier to note that
#color(green)(hatL_zhatL_(pm)Y_(l)^(m_l)(theta,phi)_#
#= [hatL_(pm)hatL_z pm ℏhatL_(pm)]Y_(l)^(m_l)(theta,phi)#
#= hatL_(pm)hatL_zY_(l)^(m_l)(theta,phi) pm ℏhatL_(pm)Y_(l)^(m_l)(theta,phi)#
We know the
#= hatL_(pm)m_lℏY_(l)^(m_l)(theta,phi) pm ℏhatL_(pm)Y_(l)^(m_l)(theta,phi)#
#= color(green)((m_l pm 1)ℏhatL_(pm)Y_(l)^(m_l)(theta,phi))#
We now see that
WHAT ARE THE LIMITS OF
Now our final question is, when will
Now, the expectation value of the
#int_"allspace" Y_(l)^(m_l)(theta,phi)^"*"hatL_(∓)hatL_(pm)Y_(l)^(m_l)(theta,phi)d tau >= 0#
We can condense this notation down to:
#<< Y_(l)^(m_l) | hatL_(∓)hatL_(pm) | Y_(l)^(m_l) >> >= 0#
Now, it becomes physically useful to rewrite
#hatL_(∓)hatL_(pm) = hatL^2 - hatL_z^2 ∓ ℏhatL_z#
Finally, using this, we can derive limits on
#<< Y_(l)^(m_l) | hatL_(∓)hatL_(pm) | Y_(l)^(m_l) >>#
#= << Y_(l)^(m_l) | hatL^2 - hatL_z^2 ∓ ℏhatL_z | Y_(l)^(m_l) >>#
#= << Y_(l)^(m_l) | ℏ^2l(l+1) - m_l^2ℏ^2 ∓ m_lℏ^2 | Y_(l)^(m_l) >>#
#= << Y_(l)^(m_l) | (l(l+1) - m_l^2 ∓ m_l)ℏ^2 | Y_(l)^(m_l) >>#
The
#=> (l(l+1) - m_l^2 ∓ m_l)ℏ^2 cancel(<< Y_(l)^(m_l) | Y_(l)^(m_l) >>)^(1) >= 0#
We can divide out
#l(l+1) - m_l^2 ∓ m_l >= 0#
With further factoring and rearranging, we have the following inequality:
#color(blue)(barul|stackrel(" ")(" "l(l+1) >= m_l (m_l pm 1)" ")|)#
CHECKING THE LIMITS OF
Testing out values of
#0(0 + 1) >= 0(0 pm 1)# #" "" "" "" "color(blue)sqrt""#
#1(1 + 1) >= 1(1 pm 1)# #" "" "" "" "color(blue)sqrt""#
#1(1 + 1) >= 0(0 pm 1)# #" "" "" "" "color(blue)sqrt""#
#1(1 + 1) >= -1(-1 pm 1)# #" "" "color(blue)sqrt""#
#2(2 + 1) >= 2(2 pm 1)# #" "" "" "" "color(blue)sqrt""#
#2(2 + 1) >= 1(1 pm 1)# #" "" "" "" "color(blue)sqrt""#
#2(2 + 1) >= 0(0 pm 1)# #" "" "" "" "color(blue)sqrt""#
#2(2 + 1) >= -1(-1 pm 1)# #" "" "color(blue)sqrt""#
#2(2 + 1) >= -2(-2 pm 1)# #" "" "color(blue)sqrt""#
So, in order to satisfy this inequality,
#bb(|m_l| <= l)# .
#"Q.E.D."#
Lithium:
Oxygen:
Nitrogen:
Potassium:
Lithium:
From its position, we know that it has
We also know that its
Putting it all together, we get
Oxygen:
From its position in the periodic table, we know that it has
We also know that:
Putting it all together, we get
Nitrogen:
Nitrogen is directly to the left of oxygen in the periodic table. This tells us that it has one less electron than oxygen—therefore, its electron configuration is the exact same as oxygen's, except with one less electron in the valence energy level.
Oxygen's electron configuration is
After taking one electron from that, it becomes
Potassium:
From its position in the periodic table, we know that it has
We also know that:
Its
Its
Its
Its
Its
Putting it all together, we get
(This was originally a comment.)
@Brian M. wrote that it was
Method 1:
The expression that we can use to calculate the maximum number of electrons given energy level is
The fourth energy level, therefore, should have
Method 2:
The fourth energy level doesn't necessarily equate to the fourth period in the periodic table.
The maximum number of electrons to be gained in the fourth period is
However, this doesn't include other fourth energy level orbitals, like
The fourth energy level, however, is composed of the
That would give us:
Here's what's going on here.
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
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,
#4"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))#
Notice that the half-reaction is balanced in terms of charge because you have
#4 xx (1+) + (2-) + 2 xx (1-) = 0 + 0#
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
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
#(2-) = (1-) + (1-)#
So, you know that the balanced half-reactions look like this
#{(4"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"^(-)) :}#
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
#{(4"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"^(-)) :}#
Add the two half-reactions to find the balanced chemical equation that describes this disproportionation reaction.
#{(4"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"^(-)) :}#
#color(white)(a)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#
#4"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"_ ((l))#
You will end up with
#4"H"_ ((aq))^(+) + 3"MnO"_ (4(aq))^(2-) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + 2"H"_ 2"O"_ ((l))#
These solutions are well-known and you should get to know what they look like. Here are the wave function
DISCLAIMER: LONG ANSWER!
A hydrogen cation (presumably
The Hamiltonian for such a scenario is:
#hatH = hatK + cancel(hatV)^(0)#
#= -ℏ^2/(2m) (del^2)/(delx^2)#
The box looks like:
with boundary conditions
#V = {(0, x in (0,L)),(oo, x <= 0),(,x >= L):}#
The Schrodinger equation is then:
#hatHpsi = Epsi#
#=> -ℏ^2/(2m) (d^2psi)/(dx^2) = Epsi#
Rearrange to the standard form:
#(d^2psi)/(dx^2) + (2mE)/(ℏ^2)psi = 0#
Often we set the substitution
#(d^2psi)/(dx^2) + k^2psi = 0#
The general solution to this is assumed to be
#psi = e^(rx)#
and upon inserting it, we obtain the auxiliary equation:
#r^2 e^(rx) + k^2 e^(rx) = 0#
In the well,
#r = ik#
and we write a linear combination for
#psi = c_1e^(ikx) + c_2e^(-ikx)#
Using Euler's formula, we rewrite this in terms of real trig functions.
#psi = c_1(cos(kx) + isin(kx)) + c_2(cos(kx) - isin(kx))#
#= (c_1 + c_2)cos(kx) + (ic_1 - ic_2)sin(kx)#
Define
#psi = Acos(kx) + Bsin(kx)#
The boundary conditions state that since the potential goes to
#Acos(kcdot0) + Bsin(kcdot0) = Acos(kL) + Bsin(kL) = 0#
But since
#Bsin(kL) = 0#
And this is only when
Thus, since
#psi_n(x) = Bsin((npix)/L)#
And the probability distribution is then:
#psi_n^"*"(x)psi_n(x) = B^2 sin^2((npix)/L)#
A well-behaved wave function is normalized in its boundaries, so we say that
#int_(0)^(L) psi_n^"*"psi_ndx = 1#
From this we get the normalization constant.
#1 = B^2 int_(0)^(L) sin^2((npix)/L)dx#
Consider the following argument.
If
The identities
Therefore,
#color(blue)(psi_n^"*"(x)psi_n(x) = "Probability Distribution")#
#= color(blue)(2/L sin^2((npix)/L))#
Well, first I would sketch the titration curve of a diprotic acid (
Here we distinguish between the buffer region (wherein the half-equivalence point lies) and the equivalence point.
#K_(a1,2,3) = 7.1 xx 10^(-3), 6.3 xx 10^(-8), 4.5 xx 10^(-13)#
So we first find the equivalence points.
#"H"_3"PO"_4 + 3"NaOH"(aq) -> "Na"_3"PO"_4(aq) + 3"H"_2"O"(l)#
#"0.100 mol"/"L" xx "0.050 L" = "0.0050 mols H"_3"PO"_4#
#= "0.0150 mols OH"^(-)#
#=> "1000 mL"/"0.100 mol NaOH" xx "0.0150 mols OH"^(-)#
#=# #"150 mL"# for all three protons
#=># #"50 mL"# for each proton
Thus, the equivalence points are at
#"H"_3"PO"_4(aq) rightleftharpoons "H"_2"PO"_4^(-)(aq) + "H"^(+)(aq)#
#K_(a1) = 7.1 xx 10^(-3) = (["H"_2"PO"_4^(-)]["H"^(+)])/(["H"_3"PO"_4]) = x^2/(0.100 - x)# Although
#x# is not small enough, since#(K_(a1))/("0.100 M") < 1# , this can be done iteratively.
#x_1 ~~ sqrt(0.100K_(a1)) = "0.02665 M"#
#x_2 ~~ sqrt((0.100 - x_1)K_(a1)) = "0.02282 M"#
#x_3 ~~ sqrt((0.100 - x_2)K_(a1)) = "0.02341 M"#
#x_4 ~~ sqrt((0.100 - x_3)K_(a1)) = "0.02332 M"#
#x_5 = sqrt((0.100 - x_4)K_(a1)) = "0.02333 M"# And thus,
#x = "0.02333 M"# for#["H"^(+)]# . That gives
#color(blue)("pH"_1) = -log(0.02333) = color(blue)(1.63)# .
#"0.100 mol/L" xx "0.010 L" = "0.0010 mols OH"^(-)# This neutralizes that much
#"H"_3"PO"_4# to give
#"0.0050 mols H"_3"PO"_4 - "0.0010 mols OH"^(-)#
#= "0.0040 mols H"_3"PO"_4# leftover
#-> "0.0010 mols H"_2"PO"_4^(-)# producedThis is then a concentration of
#("0.0010 mols H"_2"PO"_4^(-))/("50.00 mL acid" + "10.00 mL base") cdot "1000 mL"/"1 L"#
#=# #"0.01667 M H"_2"PO"_4^(-)# and
#"0.06667 M H"_3"PO"_4# These again go into the full equilibrium expression. Remember to include the initial concentrations.
#K_(a1) = 7.1 xx 10^(-3) = (["H"_2"PO"_4^(-)]["H"^(+)])/(["H"_3"PO"_4])#
#= ((0.01667 + x)(x))/(0.06667 - x)# This must be solved in full.
#7.1 xx 10^(-3) cdot 0.06667 - 7.1 xx 10^(-3)x = 0.01667x + x^2#
#x^2 + (0.01667 + 7.1 xx 10^(-3))x - 7.1 xx 10^(-3) cdot 0.06667 = 0#
#x^2 + 0.02377x - 4.73 xx 10^(-4) = 0# and here we get
#["H"^(+)] = x = "0.01290 M"# , so
#color(blue)("pH"_2) = -log(0.01290) = color(blue)(1.89)#
I won't show much math here, but you can convince yourself that you are at the first half-equivalence point, so
#"pH"_3 = "pK"_(a1) = -log(7.1 xx 10^(-3)) ~~ 2.15# is expected.However, it's a bit larger, due to Le Chatelier's principle.
#["H"_2"PO"_4^(-)] = ["H"_3"PO"_4] = "0.0025 mols"/("50.00 mL acid" + "25.00 mL base") cdot "1000 mL"/"1 L"#
#=# #"0.03333 M"#
#K_(a1) = 7.1 xx 10^(-3) = (["H"_2"PO"_4^(-)]["H"^(+)])/(["H"_3"PO"_4])#
#= ((0.03333 + x)(x))/(0.03333 - x)# Solving this, we find
#0 = x^2 + (0.03333 + 7.1 xx 10^(-3))x - 7.1 xx 10^(-3) cdot 0.03333# for which
#x = ["H"^(+)] = "0.005188 M"# and
#color(blue)("pH"_3) = -log(0.005188) = color(blue)(2.29)#
Here we are halfway between the first and second half-equivalence points, which had
#"pH" = "pK"_(ai)# , so
#color(blue)("pH"_4 = 1/2("pK"_(a1) + "pK"_(a2)) = 4.67)#
#"0.0015 mols NaOH"# added to neutralize#"0.0015 mols H"_2"PO"_4^(-)# to give#"0.0015 mols HPO"_4^(2-)# and#"0.0035 mols H"_2"PO"_4^(-)# in#"50.00 + 65.00 mL"# solution.The Henderson-Hasselbalch equation here would apply, since
#K_(a2)# is small. The#K_(a2)# would be:
#6.3 xx 10^(-8) = ((0.01304 + x)(x))/(0.03043 - x) ~~ (0.01304x)/0.03043#
#=> x ~~ 1.470 xx 10^(-7) "M"# and
#color(blue)("pH"_5 ~~ 6.83)# .Or, we could have done...
#"pH"_5 = "pK"_(a2) + log((["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)]))#
#= -log(6.3 xx 10^(-8)) + log(0.01304/0.03043)#
#= 6.83#
And so, we get
#["H"_2"PO"_4^(-)] = ["HPO"_4^(2-)]# , which means
#color(blue)("pH"_6) ~~ "pK"_(a2) = -log(6.3 xx 10^(-8))#
#= color(blue)(7.20)#
Here we are halfway between the second and third half-equivalence points, which had
#"pH" = "pK"_(ai)# , so
#color(blue)("pH"_7 = 1/2("pK"_(a2) + "pK"_(a3)) = 9.77)#
Therefore, we are
#"0.100 mol/L" xx "0.010 L" = "0.0010 mols OH"^(-)# past the second equivalence point, which means we neutralized
#"0.0010 mols HPO"_4^(2-)# and made#"0.0010 mols PO"_4^(3-)# . This gives
#["HPO"_4^(2-)] = ("0.0040 mols HPO"_4^(2-))/("50.00 + 110.00 mL") cdot "1000 mL"/"1 L"#
#=# #"0.02500 M"#
#["PO"_4^(3-)] = ("0.0010 mols HPO"_4^(2-))/("50.00 + 110.00 mL") cdot "1000 mL"/"1 L"#
#=# #"0.00625 M"# As a result, and we can again make small
#x# approximations here,
#K_(a3) = 4.5 xx 10^(-13) = ((0.00625 + x)(x))/(0.02500 - x)#
#~~ 0.00625/0.02500 x#
#=> x ~~ 1.80 xx 10^(-12) "M"# And that gives
#color(blue)("pH"_8) = -log(1.80 xx 10^(-12)) = color(blue)(11.74)#
Here we "derive"
#"pH"_("1st-half-equiv") = "pK"_(a1) + log((["base"])/(["acid"]))#
#"pH"_("2nd-half-equiv") = "pK"_(a2) + log((["base"])/(["acid"]))#
At the half-equivalence point, the concentrations of weak base and weak acid are equal, so
A perfect titration would give a
That is,
the
the
Therefore, if we want the second equivalence point, we are equidistant from the second half-equivalence point and third half-equivalence point, i.e.
#"pH"_("equivpt",i) = 1/2("pK"_(ai) + "pK"_(a(i+1)))#
Can you label this graph with the half-equivalence and equivalence points? (There are two of each on this graph.)
Warning! Long Answer. Here's what I get.
(a)
(b)
(c)
(d)
(f)
(a) 0.10 mol/L
The equilibria are
The exact formula for calculating
The approximate formula for calculating
Rather than do the rest of the calculations manualy, I will let Microsoft Excel do them.
Here are my results for Part (a).
(b) 0.0010 mol/L
(c) 0.010 mol/L
The equations are
(d) 0.010 mol/L
The equations are
(e) No data
(f) 0.010 mol/L
The equations are
Starting from
#DeltaxDeltap_x >= ℏ"/"2#
isn't actually enlightening. In fact it adds hours of pain and suffering because you have to derive it from scratch to find the form
#DeltaxDeltap_x >= 1/2 |i << psi | [hatx"," hatp_x] | psi >> |# where
#[hatx, hatp_x] = hatxhatp_x - hatp_xhatx = iℏ# .
That would have shown that for normalized wave functions,
#DeltaxDeltap_x >= 1/2 |i cdot i ℏ |#
#>= ℏ/2 sqrt(i^"*"i)#
#>= ℏ/2#
Just know that the operators are:
#hatx = x#
#hatp_x = -iℏd/dx#
which satisfy
This is consistent with any Schrodinger equation in one Cartesian dimension, and is easy to prove. In fact, I have done it several times, and here is one of those times.
Here is an easy proof to go from here...
We know:
Thus,
#hatK = hatp^2/(2m) = (-iℏ)^2 /(2m) del^2/(delx^2) = -ℏ^2/(2m) (del^2)/(delx^2)#
So for any Schrodinger equation in one Cartesian dimension...
#hatH psi = Epsi#
#= (hatK + hatV)psi#
#= (-ℏ^2/(2m) (del^2)/(delx^2) + hatV)psi#
Thus,
#-ℏ^2/(2m) (del^2psi)/(delx^2) + hatVpsi = Epsi#
#(del^2psi)/(delx^2) - (2m)/(ℏ^2)[hatV-E]psi = 0#
#(del^2psi)/(delx^2) + (2m)/(ℏ^2)[E-hatV]psi = 0#
#color(blue)((del^2psi)/(delx^2) + (8pi^2m)/(h^2)[E-hatV]psi = 0)#
This of course does NOT tell us what the potential is. That's your responsibility to tell us.
Warning! Long Answer. They come from the integrated rate laws for zero-, first-, and second-order reactions.
Chemists like to plot graphs that are straight lines.
Zero-order reactions
The integrated rate law is
Compare this with the equation for a straight line.
If we let
Thus, if a plot of
First-order reactions
The integrated rate law is
Compare this with the equation for a straight line.
If we let
Thus, if a plot of
Second-order reactions
The integrated rate law is
Compare this with the equation for a straight line.
If we let
Thus, if a plot of
Thus, you plot [dye] vs time, ln[dye] vs time, and 1/[dye] vs time.
The graph that is a straight line gives you the order of the reaction.