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2

Answer:

See below:

Explanation:

When cobalt(II) chloride is dissolved in water the pink, hydrated #sf(Co_((aq))^(2+))# ion is formed:

www.docbrown.info

#sf([Co(H_2O)_6]^(2+))#

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 #sf(Cl^-)# ions. A ligand displacement reaction occurs forming the blue tetrahedral chloride complex:

www.il-eco.uft.uni-bremen.de

The reason why only 4 ligands can be accomodated is due to the large size of the #sf(Cl^-)# ion.

This can be expressed as:

#sf([Co(H_2O)_6]^(2+)+4Cl^(-)rightleftharpoons[CoCl_4]^(2-)+6H_2O)#

You can see that Le Chatelier would predict that adding XS #sf(Cl^-)# will shift the position of equilibrium to the right causing the solution to turn blue.

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:

images.fineartamerica.com

You may have come across this reaction where cobalt chloride paper is used to test for the presence of water:

images.fineartamerica.com

3

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 #|m_l| <= l#.


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 #Y_(l)^(m_l)(theta,phi)# are given by:

#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 #l#?

Now we shall ask, what happens to the value of #l# and #m_l# when these ladder operators are applied? That is, we want to know:

#color(red)(hatL^2hatL_(pm)Y_(l)^(m_l)(theta,phi) = ???cdotY_(l)^(m_l)(theta,phi))#

Since #hatL^2# and #hatL_(pm)# commute, it follows that

#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 #l# here, so these ladder operators do not touch #l#.

DO THESE LADDER OPERATORS CHANGE #m_l#?

What about #m_l#? We want to know:

#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 #hatL_zhatL_(pm) = hatL_(pm)hatL_z pm ℏhatL_(pm)#:

#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_z# eigenvalue, so we proceed with that:

#= 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 #m_l# was indeed affected. So the ladder operators raise #(+)# or lower #(-)# the value of #m_l#.

WHAT ARE THE LIMITS OF #m_l#?

Now our final question is, when will #m_l# stop decreasing, and when will it stop increasing?

Now, the expectation value of the #hatL_(pm)# and the #hatL_(∓)# is:

#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)# in terms of components of #hatL#. It turns out to be:

#hatL_(∓)hatL_(pm) = hatL^2 - hatL_z^2 ∓ ℏhatL_z#

Finally, using this, we can derive limits on #m_l#.

#<< 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 #Y_(l)^(m_l)# are normalized, so #<< Y_(l)^(m_l) | Y_(l)^(m_l) >> = 1#, and therefore:

#=> (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 #ℏ^2# to obtain the relation:

#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 #m_l#

Testing out values of #l# and #m_l#, we find:

#bbul(m_l = 0)#

#0(0 + 1) >= 0(0 pm 1)# #" "" "" "" "color(blue)sqrt""#

#bbul(m_l = -1,0,+1)#

#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""#

#bbul(m_l = -2,-1,0,+1,+2)#

#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."#

2

Answer:

Lithium: #1s^2 2s^1#
Oxygen: #1s^2 2s^2 2p^4#
Nitrogen: #1s^2 2s^2 2p^3#
Potassium: #1s^2 2s^2 2p^6 3s^2 3p^6 4s^1#

Explanation:

Lithium:

PTable

From its position, we know that it has #1# valence electron in the #2s# orbital series (because it's in the second period): #2s^1#.

We also know that its #1s# orbital is full, because to get to lithium in the periodic table, we have to pass #1s#. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #1s# orbital: #1s^2#.

Putting it all together, we get #1s^2 2s^1#.

Oxygen:

PTable

From its position in the periodic table, we know that it has #4# valence electrons in the #2p# orbital series (because it's in the second period): #2p^4#.
We also know that:

  • Its #1s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #1s# orbital: #1s^2#.
  • Its #2s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #2s# orbital: #2s^2#.

Putting it all together, we get #1s^2 2s^2 2p^4#.

Nitrogen:

PTable

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 #1s^2 2s^2 2p^4#.
After taking one electron from that, it becomes #1s^2 2s^2 2p^3#—nitrogen.

Potassium:

PTable

From its position in the periodic table, we know that it has #1# valence electron in the #4s# orbital series (because it's in the #s# block of the fourth period): #4s^1#.
We also know that:

  • Its #1s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #1s# orbital: #1s^2#.

  • Its #2s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #2s# orbital: #2s^2#.

  • Its #2p# orbital is full. There are #6# electrons in an #s# orbital; this means that it has #6# electrons in its #2p# orbital: #2p^6#.

  • Its #3s# orbital is full. There are #2# electrons in an #s# orbital; this means that it has #2# electrons in its #3s# orbital: #3s^2#.

  • Its #3p# orbital is full. There are #6# electrons in an #s# orbital; this means that it has #6# electrons in its #3p# orbital: #3p^6#.

Putting it all together, we get #1s^2 2s^2 2p^6 3s^2 3p^6 4s^1#.

3

Answer:

#32# electrons.

Explanation:

(This was originally a comment.)

@Brian M. wrote that it was #18# electrons, but here are my reasons for it being #32# electrons instead.

Method 1:
The expression that we can use to calculate the maximum number of electrons given energy level is #2n^2#, with #n# being the energy level.
The fourth energy level, therefore, should have #2*(4)^2 = 2*16 = 32# electrons.

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 #18#, because we have:

Alexia Allison on SlidePlayer

  • #2# electrons from the #4s# orbital series;
  • #10# electrons from #3d# orbital series; and
  • #6# electrons from #4p# orbital series.

#2 + 10 + 6 = 18# electrons.

However, this doesn't include other fourth energy level orbitals, like #4d# and #4f#. It also includes the #3d# orbitals, which is in the third energy level.

The fourth energy level, however, is composed of the #4s, 4d, 4p,# and #4f# orbital series—the #4# in front means that it's in the fourth energy level.

That would give us:

  • #2# electrons from the #4s# orbital series;
  • #6# electrons from #4p# orbital series;
  • #10# electrons from #4d# orbital series; and
  • #14# electrons from #4f# orbital series.

#2 + 6 + 10 + 14 = 32# electrons.

5

Answer:

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 #color(blue)(+6)# on the reactants' side to #color(blue)(+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, #"H"^(+)#, to the side that needs hydrogen.

#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#

#color(white)(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 #color(blue)(+6)# on the reactants' side to #color(blue)(+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

#(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 #2# to get

#{(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))#

2

#psi_n^"*"(x)psi_n(x) = 2/L sin^2((npix)/L)#

These solutions are well-known and you should get to know what they look like. Here are the wave function #psi# and the probability density #psi_n^"*"psi_n# plotted:

http://slideplayer.com/slide/4905211/16/images/21/


DISCLAIMER: LONG ANSWER!

A hydrogen cation (presumably #""^(1) "H"^(+)#) is a neutron and proton with no potential of interaction. Thus, we set this up as a particle in a box scenario where #V = 0#.

The Hamiltonian for such a scenario is:

#hatH = hatK + cancel(hatV)^(0)#

#= -ℏ^2/(2m) (del^2)/(delx^2)#

The box looks like:

http://www.nyu.edu/

with boundary conditions #psi(0) = psi(L) = 0#, and

#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 #k = sqrt(2mE//ℏ^2)#, so

#(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, #e^(rx) ne 0# so that the particle exists. Thus,

#r = ik#

and we write a linear combination for #psi#:

#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 #A = c_1 + c_2# and #B = ic_1 - ic_2#, because these are arbitrary constants. Therefore:

#psi = Acos(kx) + Bsin(kx)#

The boundary conditions state that since the potential goes to #oo# at #x = 0,L#, it follows that #psi(0) = psi(L) = 0#:

#Acos(kcdot0) + Bsin(kcdot0) = Acos(kL) + Bsin(kL) = 0#

But since #sin(0) = 0# and #cos(0) = 1#, it follows that #A = 0#. Furthermore, since #A = 0#, we have:

#Bsin(kL) = 0#

And this is only when #kL = npi#, where #n = 1, 2, 3, . . . #. (We could take negative #n#, but that would give a linearly dependent solution.)

Thus, since #k = (npi)/L#, the wave function becomes:

#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 #sin^2u + cos^2u = 1#, then integrating #sin^2u + cos^2u# gives #L# for the area under the curve, with height #1# and length #L#, regardless of what #u# is.

The identities #sin^2u = 1/2(1-cos(2u))# and #cos^2u = 1/2(1+cos(2u))# show that #sin^2u# contributes to half the area under the curve.

Therefore, #int_(0)^(L) sin^2udu = L/2#, and #B = sqrt(2/L)#. This means

#color(blue)(psi_n^"*"(x)psi_n(x) = "Probability Distribution")#

#= color(blue)(2/L sin^2((npix)/L))#

2

Well, first I would sketch the titration curve of a diprotic acid (#"HPO"_4^(2-)# is hard enough to form).

https://www3.nd.edu/~aseriann/CHAP2B.html/

Here we distinguish between the buffer region (wherein the half-equivalence point lies) and the equivalence point.

  • Buffer region --- Henderson-Hasselbalch equation does NOT apply UNLESS the current #K_a# is small. #K_(a1)# is not small enough, but #K_(a2)# and #K_(a3)# are.
  • Half-equivalence point --- #"pH" = "pK"_a#
  • Equivalence point --- Regular weak acid/base equilibrium!

#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)#

#"Total equivalence point"#:

#"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 #"50 mL"#, #"100 mL"#, and #"150 mL"#. Notice how we don't reach #"150 mL"#, so ignore that one.

#(ia)# #"0.00 mL"# base added

#"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)#.

#(ib)# #"10.00 mL"# of base added

#"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^(-)# produced

This 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)#

#(ic)# #"25.00 mL"# of base added, the first half-equivalence point

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)#

#(id)# #"50.00 mL"# of base added, the first equivalence point

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)#

#(ie)# #"65.00 mL"# of base added, or #"15.00 mL"# past first equivalence point

#"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#

#(i f)# #"75.00 mL"# is the second half-equivalence point

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)#

#(ig)# #"100.00 mL"# is the second equivalence point

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)#

#(ih)# #"110.00 mL"# of base added, #"10.00 mL"# past the second equivalence point

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"# formulas for the first and second equivalence points.

#"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 #"pH"_("1st-half-equiv") = "pK"_(a1)#, and #"pH"_("2nd-half-equiv") = "pK"_(a2)#.

A perfect titration would give a #V_(NaOH)# volume gap on the graph such that the graph has a certain odd function symmetry throughout.

That is,

  • the #"pH"# on either side of the equivalence point is equidistant from the equivalence point #"pH"#.

  • the #"pH"# on either side of the half-equivalence point is equidistant from the half-equivalence point #"pH"#.

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)))#

#(ii)# And the graph should be easy to graph with these data points. It won't look as fine since you don't have all the in-between data, but... here's what I got in Excel:

Can you label this graph with the half-equivalence and equivalence points? (There are two of each on this graph.)

And a tutorial is here.

2

Answer:

Warning! Long Answer. Here's what I get.

Explanation:

(a) #["H"_3"O"^"+"] = 9.95 × 10^"-6"color(white)(l) "mol/L"; "pH = 5.00"#
#color(white)(ml)["H"_3"O"^"+"] = 1.00 × 10^"-6"color(white)(l) "mol/L"; "pH = 5.00"#

(b) #["H"_3"O"^"+"] = 7.07 × 10^"-6"color(white)(l) "mol/L"; "pH = 5.15"#
#color(white)(ml)["H"_3"O"^"+"] = 1.00 × 10^"-5"color(white)(l) "mol/L"; "pH = 5.00"#

(c) #["H"_3"O"^"+"] = 1.64 × 10^"-5"color(white)(l) "mol/L"; "pH = 4.79"#
#color(white)(ml)["H"_3"O"^"+"] = 2.16 × 10^"-5"color(white)(l) "mol/L"; "pH = 4.66"#

(d) #["H"_3"O"^"+"] = 4.59 × 10^"-9"color(white)(l) "mol/L"; "pH = 8.54"#
#color(white)(ml)["H"_3"O"^"+"] = 4.54 × 10^"-9"color(white)(l) "mol/L"; "pH = 8.54"#

(f) #["H"_3"O"^"+"] = 8.26 × 10^"-8"color(white)(l) "mol/L"; "pH = 7.08"#
#color(white)(ml)["H"_3"O"^"+"] = 8.26 × 10^"-8"color(white)(l) "mol/L"; "pH = 7.08"#

(a) 0.10 mol/L #"NaHB"#

The equilibria are

#"H"_2"B" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HB"^"-"; "p"K_text(a1) = 3.0#
#"HB"^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "B"^"2-"; color(white)(ll)"p"K_text(a2) = 7.0#

#K_text(a1) = 10^"-3.0" = 1.00 × 10^"-3"#
#K_text(a2) = 10^"-7.0" = 1.00 × 10^"-7"#

The exact formula for calculating #["H"_3"O"^"+"]# is

#["H"_3"O"^"+"] = sqrt((K_text(a1)K_text(a2)["HB"^"-"] + K_text(a1)K_text(w))/(["HB"^"-"] + K_text(a1)#

#["H"_3"O"^"+"] = sqrt((1.00 × 10^"-3" × 1.00 ×10^"-7" × 0.10 + 1.00 × 10^"-3"× 1.00 × 10^"-14")/(0.10 + 1.00 × 10^"-3")) = sqrt((1.00 ×10^"-11" + 1.00 × 10^"-17")/0.101) = sqrt((1.00× 10^"-11")/0.101) = sqrt(9.90 × 10^"-11")#
#= 9.95 ×10^"-6"#

#"pH = -log"["H"_3"O"^"+"] = "-log"(9.95 × 10^"-6") = 5.00#

The approximate formula for calculating #["H"_3"O"^"+"]# is

#["H"_3"O"^"+"] = sqrt(K_text(a1)K_text(a2))#

#["H"_3"O"^"+"] = sqrt(1.00 × 10^"-3" × 1.00 × 10^"-7") = sqrt(1.00 ×10^"-10") = 1.00 ×10^"-5"#

#"pH = -log"(1.00 × 10^"-5") = 5.00#

Rather than do the rest of the calculations manualy, I will let Microsoft Excel do them.

Here are my results for Part (a).

Part (a )

(b) 0.0010 mol/L #["HB"^"-"]#

Part (b)

(c) 0.010 mol/L #"NaH"_2"PO"_4#

The equations are

#"H"_3"PO"_4 + "H"_2"O" ⇌ "H"_3"O"^"+" + "H"_2"PO"_4"-"; K_text(a1) = 7.52 ×10^"-3"#
#"H"_2"PO"_4"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; color(white)(l)K_text(a2) = 6.23 × 10^"-8"#

Part (c)

(d) 0.010 mol/L #"NaHCO"_3#

The equations are

#"H"_2"CO"_3 + "H"_2"O" ⇌ "H"_3"O"^"+" + "HCO"_3^"-"; K_text(a1) =4.3 × 10^"-7" #
#"HCO"_3^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "CO"_3^"2-"; color(white)(m)K_text(a2) = 4.8 × 10^"-11" #

Part (d)

(e) No data

(f) 0.010 mol/L #"NaSH"#

The equations are

#"H"_2"S" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HS"^"-"; K_text(a1) = 1.1 × 10^"-7"#
#"HS"^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "S"^"2-"; color(white)(ll)K_text(a2) = 6.2 × 10^"-8"#

Part (f)

2

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 #[hatx, hatp_x] = hatxhatp_x - hatp_xhatx = iℏ#.

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.

https://socratic.org/questions/in-heisenberg-s-uncertainty-principle-is-the-uncertainty-in-the-position-a-2-dim?source=search

Here is an easy proof to go from here...

We know:

  • #p = mv#
  • #K = 1/2mv^2#

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.

3

Answer:

Warning! Long Answer. They come from the integrated rate laws for zero-, first-, and second-order reactions.

Explanation:

Chemists like to plot graphs that are straight lines.

Zero-order reactions

The integrated rate law is

#["A"] = "-"kt + ["A"]_0#

Compare this with the equation for a straight line.

#["A"] = "-"kt + ["A"]_0#
#color(white)(l)ycolor(white)(ll) = mx +color(white)(l) b#

If we let #y = ["A"]# and #x = t#, a plot of # ["A"]# vs #t# is a straight line with slope #m = "-"k# and #x#-intercept #["A"]_0#

Thus, if a plot of # ["A"]# vs #t# is a straight line with a negative slope, the reaction is zero-order.

www.chem.purdue.edu

First-order reactions

The integrated rate law is

#ln["A"] = -kt + ln["A"]_0#

Compare this with the equation for a straight line.

#ln["A"] = "-"kt + ["A"]_0#
#color(white)(m)ycolor(white)(ll) = color(white)(l)mx +b#

If we let #y = ln["A"]# and #x = t#, a plot of #ln["A"]# vs #t# is a straight line with slope #m = "-"k# and #x#-intercept #ln["A"]_0]#

Thus, if a plot of #ln["A"]# vs #t# is a straight line with a negative slope, the reaction is first-order.

www.chem.purdue.edu

Second-order reactions

The integrated rate law is

#1/(["A"]) = kt + 1/(["A"]_0)#

Compare this with the equation for a straight line.

#1/(["A"]) = color(white)(l)kt + 1/(["A"]_0)#
#color(white)(ll)ycolor(white)(ll) = color(white)()mx + color(white)(ll)b#

If we let #y = 1/(["A"])# and #x = t#, a plot of #1/(["A"])# vs #t# is a straight line with slope #m = k# and #x#-intercept #1/(["A"]_0#.

Thus, if a plot of #ln1/(["A"])# vs #t# is a straight line with a positive slope, the reaction is second-order.

www.chem.purdue.edu

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.