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Answer:

#K_"eq" = 6.5 * 10^9#

Explanation:

The first thing to do here is look up the values of the solubility product constant, #K_(sp)#, for lead(II) chloride, #"PbCl"_2#, and the formation constant, #K_f#, for the trihydroxoplumbate(II) complex ion, #"Pb"("OH")_3^(-)#

#K_(sp) = 1.70 * 10^(-5)#

#K_f = 3.8 * 10^(14)#

http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf

http://bilbo.chm.uri.edu/CHM112/tables/Kftable.htm

Now, the idea here is that lead(II) chloride is considered Insoluble in aqueous solution, which means that a dissociation equilibrium is established when you place this salt in water

#"PbCl"_ (2(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#

As shown by the value of #K_(sp)#, this equilibrium lies to the left, meaning that you will get very small concentrations of lead(II) cations and chloride anions in solution.

Now, if you add a strong base, which can be symbolized by #"OH"^(-)#, to this solution, a complexation reaction will take place

#"Pb"_ ((aq))^(2+) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-)#

Since #K_f# has such a large value, this equilibrium will lie to the right, meaning that the reaction will proceed in the forward direction

#"Pb"_ ((aq))^(2+) + 3"OH"_ ((aq))^(-) -> "Pb"("OH")_ (3(aq))^(-)#

The thing to notice here is that adding hydroxide anions, to the aqueous solution of lead(II) chloride will consume the lead(II) cations dissolved in solution.

As a result, the dissociation equilibrium for lead(II) chloride will shift to the right to compensate for the fact that the concentration of lead(II) cations is decreasing #-># think Le Chatelier's Principle here.

This means that you'll have

#"PbCl"_ (2(s)) -> "Pb"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#

If you were to add these two reactions, you would get

#{(color(white)(aaaaaaa)"PbCl"_ (2(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)), ("Pb"_ ((aq))^(2+) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-)) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#"PbCl"_ (2(s)) + color(red)(cancel(color(black)("Pb"_ ((aq))^(2+)))) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-) + color(red)(cancel(color(black)("Pb"_ ((aq))^(2+)))) + 2"Cl"_ ((aq))^(-)#

This is equivalent to

#"PbCl"_ (2(s)) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-) + 2"Cl"_ ((aq))^(-)#

Now, because you're added two equilibrium reactions to get your overall equilibrium, you must multiply the value of their equilibrium constants to get the equilibrium constant, let's say #K_"eq"#, for this overall equilibrium

#K_"eq" = K_(sp) * K_f#

Plug in your values to find

#K_"eq" = 1.70 * 10^(-5) * 3.8 * 10^(14) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.5 * 10^9)color(white)(a/a)|)))#

Once again, the fact the #K_"eq " ">> " 1# tells you that this equilibrium will lie mostly to the right, which essentially tells you that you can increase the solubility of lead(II) chloride by increasing the pH of the solution.

Answer:

The surface tension is #"73 mJ·m"^"-2"#.

Explanation:

One method to measure the surface tension of a liquid is to measure the height the liquid rises in a capillary tube.

www.chem1.com

The formula is

#color(blue)(|bar(ul(color(white)(a/a) γ = "rhρg"/"2cosθ" color(white)(a/a)|)))" "#

where

#γ# = the surface tension
#r# = the radius of the capillary
#h# = the height the liquid rises in the capillary
#ρ# = the density of the liquid
#g# = the acceleration due to gravity
#θ# = the angle of contact with the surface

For pure water and clean glass, the contact angle is nearly zero.

If #θ ≈ 0#, then #cosθ ≈ 1#, and the equation reduces to

#color(blue)(|bar(ul(color(white)(a/a) γ = "rhρg"/2 color(white)(a/a)|)))" "#

In your problem,

#r = "0.2 mm" = 2 × 10^"-4"color(white)(l) "m"#
#h = "6.6 cm" = "0.066 m"#
#ρ = "1.008 g/mL" = "1.008 kg/L" = "1008 kg·m"^"-3"#
#g = "9.81 m·s"^"-2"#

#γ = (2 × 10^"-4" color(red)(cancel(color(black)("m"))) × 0.066 color(red)(cancel(color(black)("m"))) × 1008 color(red)(cancel(color(black)("kg·m"^"-3"))) × 9.81 color(red)(cancel(color(black)("m·s"^"-2"))))/2 × ("1 J")/(1 color(red)(cancel(color(black)("kg")))·"m"^2color(red)(cancel(color(black)("s"^"-2"))))= "0.065 J·m"^"-2" = "65 mJ·m"^"-2" #

For comparison, the surface tension of pure water at 20 °C is #"73 mJ·m"^"-2"#.

Answer:

WARNING! Long answer! The minimum energy needed is #2.31 × 10^5color(white)(l) "J·mol"^"-1"# and the maximum wavelength is 518 nm.

Explanation:

The photoelectric effect

When a photon of sufficient energy hits the surface of a metal, some of the energy is used to eject an electron from the surface of the metal (the work function) and the rest goes into the kinetic energy of the electron.

www.daviddarling.info

#"Energy of photon" = "work function + kinetic energy"#

#color(blue)(bar(ul(|color(white)(a/a)E = φ + KEcolor(white)(a/a)|)))" "#

Minimum energy

The energy #E# of a photon is given by the formula

#color(blue)(bar(ul(|color(white)(a/a) E = hf color(white)(a/a)|)))" "#

where

#h# = Planck's constant
#f# = the frequency of the light.

Another important formula is

#color(blue)(bar(ul(|color(white)(a/a) fλ = c color(white)(a/a)|)))" "#

where

#λ# = the wavelength of the light
#c# = the speed of light

We can combine these two expressions and get the formula

#color(blue)(bar(ul(|color(white)(a/a) E = (hc)/λ color(white)(a/a)|)))" "#

For a mole of photons,

#E = N_"A"(hc)/λ#

where

#N_"A"# = Avogadro's number

The minimum energy will be that for which the photon has just enough energy to remove the electron with nothing left over (i.e. the work function)

#E = φ + KE#

#φ = E - KE#

#E = (N_"A"hc)/λ = (6.022 × 10^23color(white)(l) "mol"^"-1" × 6.626 × 10^"-34" color(white)(l)"J"·color(red)(cancel(color(black)("s"))) × 2.998 × 10^8 color(red)(cancel(color(black)("m·s"^"-1"))))/(300 × 10^"-9" color(red)(cancel(color(black)("m")))) = 3.99 × 10^5color(white)(l) "J·mol"^"-1"#

#KE = 1.68 × 10^5color(white)(l) "J·mol"^"-1"#

#φ = 3.99 × 10^5color(white)(l) "J·mol"^"-1" - 1.68 × 10^5color(white)(l) "J·mol"^"-1" = 2.31 × 10^5color(white)(l) "J·mol"^"-1"#

Maximum wavelength

A photon with just enough energy to equal the work function will have the maximum wavelength.

#E = N_"A"(hc)/λ = φ#

#λ = (N_"A"hc)/φ = (6.022 × 10^23 color(red)(cancel(color(black)("mol"^"-1"))) × 6.626 × 10^"-34" color(red)(cancel(color(black)("J·s"))) × 2.998 × 10^8 "m"·color(red)(cancel(color(black)("s"^"-1"))))/(2.31 × 10^5 color(red)(cancel(color(black)("J·mol"^"-1")))) = 5.18 × 10^"-7"color(white)(l) "m" = 518 × 10^"-9"color(white)(l) "m" = "518 nm"#

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

#bb(DeltaxDeltap_x >= h/(2pi))#,

where #ℏ = h/(2pi)# is the reduced Planck's constant, and #h = 6.626xx10^(-34) "J"cdot"s"#. You may also see #ℏ/2#, or #h/(4pi)#, but that's beside the point. The point is, it's on the order of #ℏ#.

The de Broglie wavelength is:

#lambda = h/(mv)#

If the uncertainty in the position becomes numerically equal to #lambda#, then we can plug in #lambda = h/(mv_x) = Deltax# to get:

#cancel(h)/(mv_x)Deltap_x >= cancel(h)/(2pi)#

Since #p = mv#, #Deltap = mDeltav# for a particle experiencing little relativistic effects on its mass:

#1/(cancel(m)v_x)cancel(m)Deltav_x >= 1/(2pi)#

#(Deltav_x)/v_x >= 1/(2pi)#

Flipping both sides, we get:

#color(blue)(v_x/(Deltav_x) <= 2pi)#

Since #lambda# is very small (on the order of #nm# for electrons), we expect that the uncertainty in the velocity is very large so that the inequality #DeltaxDeltap_x >= h/(2pi)# is maintained.

That should make sense because if #Deltav_x# is very large, only then would #v_x/(Deltav_x) <= 2pi# hold true, since #v_x# for an electron is generally quite large as well.

EXAMPLE

For instance, the #1s# electron in hydrogen atom travels at around #1/137# times the speed of light, or #v_x = 2.998xx10^8 xx 1/137 = 2.188xx10^6 "m/s"#. That means:

#(2.188xx10^6)/(Deltav_x) <= 2pi#

#color(red)(Deltav_x >= 3.483 xx 10^5)# #color(red)("m/s")#

i.e. the uncertainty in the velocity of a #1s# 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 #"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 #"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.

Answer:

The arrows give the orientation of a characteristic of electrons (and many other particles for that matter) known as spin .

Explanation:

Every electron in an atom must be unique in terms of four characteristics. This is known as the Pauli Exclusion Principle and is fundamental to quantum mechanics.

These characteristics are given by the four quantum numbers assigned to each electron during the solution of Schroedinger's equation - #n, l, m_l and s#.

The first three of these determine the shell, subshell and orbital (respectively) in which the electron is located. The fourth, the spin quantum number can only have two values #+1/2 and -1/2# and so, the limit to the number of electrons in any orbital is two. A third electron in an orbital would have to be identical to one of the other two, and this is what the Pauli Exclusion Principle prohibits.

Unfortunately there is no property in the "big" world we live in that corresponds to spin, and so it is a difficult quantity to grasp.

The unfortunate(?) choice for the name of this quantity has led many people to think that electrons really spin in some way. They do not. Spin does however result in a small magnetic character for an electron that is similar to what would be caused by a circulating charge so the name is not all bad!

Answer:

WARNING! Long answer!

Explanation:

The solubility of #"Ni(OH)"_2# is #1.52 × 10^"-5" color(white)(l)"mol/L"# or #"1.41 mg/L"#.

In #"0.100 mol/L NiSO"_4#, the solubility is #1.87 × 10^"-7" "mol/L" = "17.3 µg/L"#.

Solubility in water

I think your value of #K_"sp"# is in error. A better value would be #1.40 × 10^"-14"#.

The solubility equilibrium is

#color(white)(mmmmmm)"Ni(OH)"_2"(s)" ⇌ "Ni"^"2+"(aq) + "2OH"^"-"(aq)#; #K_text(sp) = 1.40 × 10^"-14"#
#"E/mol·L"^"-1":color(white)(mmmmmmmmm)xcolor(white)(mmmmm)2x#

Since #x color(white)(l)"mol of Ni(OH)"_2"(s)"# gives #xcolor(white)(l) "mol of Ni"^"2+"#, the solubility of #"Ni(OH)"_2# is #xcolor(white)(l) "mol/L"#.

The solubility constant expression is

#K_"sp" = ["Ni"^"2+"]["OH"^"-"]^2#

#["Ni"^"2+"]["OH"^"-"]^2 = x × (2x)^2 = 4x^3 = 1.40 × 10^"-14"#

#x^3 = (1.40 × 10^"-14")/4 = 3.50 × 10^"-15"#

#x = root(3)( 3.50 × 10^"-15") = 1.52 × 10^"-5"#

∴ The solubility of #"Ni(OH)"_2# is #1.52 × 10^"-5" color(white)(l)"mol/L"#

#M_r = 92.71#, so

#"Solubility" = (1.52 × 10^"-5" color(red)(cancel(color(black)("mol"))))/"1 L" × "92.71 g"/(1 color(red)(cancel(color(black)("mol")))) = 1.41 × 10^"-3" color(white)(l)"g/L" = "1.41 mg/L"#

The solubility listed here is 0.001 27 g/L (close enough!).

Solubility in 0.100 mol/L #"NiSO"_4#

The solubility of the #"Ni(OH)"_2# will be much less because of the common ion effect of the #"Ni"^"2+"# ion.

The solubility equilibrium is now

#color(white)(mmmmmm)"Ni(OH)"_2"(s)" ⇌ "Ni"^"2+"(aq) + "2OH"^"-"(aq)#; #K_text(sp) = 1.40 × 10^"-14"#
#"E/mol·L"^"-1":color(white)(mmmmmmmll)0.100 + xcolor(white)(mmmll)2x#

#K_"sp" =["Ni"^"2+"]["OH"^"-"]^2 = (0.100 + x)(2x)^2 = 1.40 × 10^"-14"#

We know that #x < 1.52 × 10^"-5"#, so #x ≪ 0.100#.

#0.100(2x)^2 = 0.400x^2 = 1.40 × 10^"-14"#

#x^2 = (1.40 × 10^"-14")/0.400 = 3.50 × 10^"-14"#

#x =sqrt(3.50 × 10^"-14") = 1.87 × 10^"-7"#

∴ The solubility of #"Ni(OH)"_2# is #1.87 × 10^"-7"color(white)(l) "mol/L"#

Also,

#"Solubility" = (1.87 × 10^"-7" color(red)(cancel(color(black)("mol"))))/"1 L" × "92.71 g"/(1 color(red)(cancel(color(black)("mol")))) = 1.73 × 10^"-5" "g/L" = "17.3 µg/L"#

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