Exam 3 - Chem 327
Explain the difference between a gradient and isocratic separation.
- change in solvent composition to increase eluant strength - no change in solvent composition
the greater the resistance
...the smaller the current that flows
S.H.E. E =
0
A 100.0 mL solution of 0.0500 M Mx+ buffered to pH 9.00 was titrated with 0.0500 M EDTA. Calculate the concentration of Mx+ at V = 0.5 Veq.
0.0167
A solution 0.0500 M EDTA is used as a titrant for a metal solution buffered to pH 9.00. What fraction of free EDTA is in the form Y4- at pH 9.00? Note: for EDTA, pK1 = 0.0, pK2 = 1.5, pK3 = 2.00, pK4 = 2.69, pK5 = 6.13, pK6 = 10.37
0.0409
under standard conditions the activity of Ag+ =
1 the right side is E=+0.799V
1 A =
1 C/s
1 V =
1 J/C
What are the 2 purposes of buffers in electrophoresis?
1)Control the pH 2)Need buffer to soak up the H+ and OH- generated during the separation!
Find the conditional formation constant for Mg(EDTA)2− at pH 9.00.
2.5x10^7
The formation constant for some metal-EDTA complex, Kf, is 1012.00. Calculate the conditional formation constant (Kf') at pH 9.00. for EDTA pK1 = 0.0, pK2 = 1.5, pK3 = 2.00, pK4 = 2.69, pK5 = 6.13, pK6 = 10.37
4.1 x 10 ^10
Find the concentration of free Mg2+ in 0.050 M Na2[Mg(EDTA)] at pH 9.00.
4.5x10^7
Faraday constant
96,485 C/mol e-
Please explain to me why peak C is wider than peak A. HINT: There are 3 things you need to consider!
A = "Path Length" Term ~ How many paths can one molecule take when it travels through the column? B = "Longitudinal Diffusion" Term ~How long does it take for the molecules to diffuse against or with the flow? C = "Resistance to Mass Transfer" Term ~How long does it take to change between phases (i.e. stationary vs. mobile phase)?
redox reaction
A chemical reaction involving the transfer of one or more electrons from one reactant to another; also called oxidation-reduction reaction.
Observe the chromatogram below. Which compound interacts with the mobile phase more? Stationary phase more?
A likes the mobile phase the most E likes the stationary phase the most
Explain why a silver electrode can be an indicator electrode for Ag+ and for halides
A silver electrode serves as an indicator for Ag+ by virtue of the equilibrium Ag+ + e- -><- Ag(s) that occurs at its surface. If the solution is saturated with silver halide, then [Ag+] is affected by changes in halide concentration. Therefore, the electrode is also an indicator for halide.
salt bridge
A tube that allows the slow transfer of ions and maintains the neutrality of the electrolyte solutions.
Please name all the following reference electrodes.
A. Ag/AgCl electrode B. Saturated Calomel Electrode (S.C.E.) C. Standard Hydrogen Electrode (S.H.E.)
The following cell, as written, has a potential of -0.812 V. Ag(s) | AgCl(s) | KCl(aq) (saturated) || Ag+(aq) | Ag(s) What species is being oxidized?
Ag (s) Since we have a negative voltage, the electrons are flowing from right to left (when normally they flow left to right in a galvanic cell). With that in mind, we can see the Ag(s) will be oxidized to Ag+(aq), producing electrons.
Write the line diagram for the "Silver Electrode Study of Equilibrium" electrochemical cell.
Ag(s) | AgCl(s) | NH4Cl(aq)|| AgNO3(aq) | Ag(s)
EDTA (ethylenediaminetetraacetic acid)
Anticoagulant that preserves cell shape and structure and inhibits platelet clumping
Let's calculate the shape of the titration curve for the reaction of 50.0 mL of 0.040 0 M Ca2+ (buffered to pH 10.00) with 0.080 0 M EDTA
Ca2+ + EDTA→CaY2− Kf′ = αY4− Kf = (0.30)*(10^10.65) = 1.34×1010 The value of αY4− comes from Table 12-1.
electric charge (q)
Coulomb, C 1.602x10^-19 C
An electrochemical cell has an Eanode = 0.999 V and Ecathode = 0.333 V. Will this cell transfer electrons spontaneously? NOTES: Number of electrons transferred = 2 Faraday's Constant = 96,500 C/mol
Ecell = Ecathode-Eanode = (0.333 V) - (0.999 V) = -0.666 V ΔG = -nFE = (-2)(96,500 C/mol)(-0.666 V) = +1.29*105 J/mol Since ΔG is positive, the reaction is NOT spontaneous!
The below reaction has a reduction potential of Eo=-0.799 V. Ag(s) <-> Ag+ + e- What is the Eo of "3Ag(s) <-> 3Ag+ + 3e-"?
Eo= -0.799 V When you multiply the reaction, it does not change the standard potential! NOTE: It is important to remember this when you are working with electrochemical cells.
conditional formation constant
Equilibrium constant for formation of a complex under a particular stated set of conditions, such as pH, ionic strength, and concentration of auxiliary complexing species. Also called effective formation constant.
standard reduction potential
E˚ a higher E˚ means a greater tendency for reduction to occur a lower E˚ means a greater tendency for oxidation to occur
Multidentate
Forms two or more co-ordinate bonds
Please describe a HETP and how it relates to resolution and separation.
HETP = Height Equivalent of a Theoretical Plate = σ2/L = column length (L) / number theoretical plates (N) <- = down arrow -> = up arrow <-HETP = ->N (HETP = N/L) <-HETP = ->Resolution <-HETP = ->Separation Think about HETP as stairs...more stairs to hit falling down = more separation!
Calcium ion was titrated with EDTA at pH 11, using Calmagite as indicator (Table 12-3). Which is the principal species of Calmagite at pH 11? What color was observed before the equivalence point? After the equivalence point?
HIn2- is the principal species of calmagite at pH 11. The color before equivalence point is burgundy red. After the equivalence point the color is blue
The pKa's of EDTA are as follows: 0.0, 1.5, 2.0, 2.69, 6.13, 10.37 What species has the second highest alpha fraction at pH=5.5?
HY3-
Ohm's Law
I(A) = E(orV)/R(Ω)
EDTA right before equivalence point
In this region, there is excess Mn+ left in solution after the EDTA has been consumed. The concentration of free metal ion is equal to the concentration of excess, unreacted Mn+. The dissociation of MYn−4 is negligible.
How can the following electrode help us to determine what the junction? Ag(s) | AgCl(s) | HCl(aq) || KCl(aq) | Ag(s) | AgCl(s)
Junction Potential: occurs when two solutions of different concentrations come in contact with each other; due to different diffusion properties of other components, a small potential may exist even with equal concentration solutions. Since both sides of the electrodes are the same, the only thing that is causing a potential is the different excess ions (i.e. HCl and KCl), which are causing a junction potential.
Ag2S is added to water, and only some of it dissolves. Using the same electrode as in the "Silver Electrode Lab," the ΑAg+ was found to be 3.00*10-17 M. What is the Ksp of Ag2S including activities? NOTE: The activity coefficient of S2- is 0.66, and the activity coefficient of Ag+ is 0.90. HINT: Look at the stoichiometry between silver and sulfur.
Ksp = [Ag+]2[S2-] = AAg+2 AS2- NOTE: Sulfur concentration should be half the Ag+ concentration. AAg+ = ϒAg+[Ag+] = 3.00*10-17 M Thus: [Ag+] = AAg+/ϒAg+ = (3.00*10-17)/(0.90) = 3.33*10-17 M AS2- = ϒS2-[S2-] = ϒS2- ([Ag+]/2) = ϒS2- ((3.33*10-17)/2) = (0.66) (1.67*10-17) AS2- = 1.1*10-17 M Ksp = AAg+2 AS2- = (3.00*10-17)2(1.11*10-17) Ksp = 9.9*10-51
End point detection methods
Metal ion indicators Ion-selective electrode Mercury electrode Glass (pH) electrode
EDTA after equivalence point
Now there is excess EDTA, and virtually all the metal ion is in the form MYn−4. The concentration of free EDTA can be equated to the concentration of excess EDTA added after the equivalence point.
The stoichiometry is 1:1 regardless of the charge on the ion.
One mole of EDTA reacts with one mole of metal ion.
What is the difference between the anode and cathode in an electrochemical cell?
Oxidation occurs at the Anode where Reduction occurs at the Cathode
Power equation
P=W/t = E * q/s
Describe what isoelectric focusing is.
Separates proteins by pI (isoelectric point) in a pH gradient gel by applying an electric field.
electric potential difference (E)
The change in potential energy per unit charge in an electric field.
electric current
The continuous flow of electric charges through a material
oxidizing agent
The electron acceptor in a redox reaction.
Stability constant
The equilibrium constant for an equilibrium existing between a transition metal ion surrounded by water ligands and the complex formed when the same ion has undergone a ligand substitution
electric potential
The work done per unit charge in bringing a positive test charge from infinity to that point in the field.
EDTA at equivalence point
There is exactly as much EDTA as metal in the solution. We treat the solution as if it had been made by dissolving pure MYn−4. Some free Mn+ is generated by the slight dissociation of MYn−4: MYn−4⇌Mn++EDTA In this reaction, EDTA represents free EDTA in all its forms. At the equivalence point, [Mn+]=[EDTA].
Bidentate
When a ligand can form two co-ordinate bonds in a complex ion.
galvanic cell (voltaic cell)
a cell in which energy released from a spontaneous chemical reaction is used to generate electricity
metal ion indicator
a compound whose color changes when it binds to a metal ion
chelating ligand
a ligand that binds to a metal through more than one atom
Nernst equation definition
a mathematical relationship used to calculate an ionic equilibrium potential. net driving force for a redox reaction
monodentate ligand
a species that forms only a single coordinate bond to a metal ion in a complex
complexometric titration
a titration involving the formation of a complex between metal ions and a reagent such as edta. in this type of titration the end point is marked by a sharp decrease in the concentration of free metal ions
For the cell represented by the line diagram below, write the equilibrium reaction where the spontaneous reaction is the forward reaction. (Hint: Yes, this does involve the Nernst equation.) Pt(s) | Br2(l) |HBr(aq, 0.10 M) || Al(NO3)3(aq, 0.10 M) | Al(s) Standard potentials: EAl^(3+) = -1.677 V EBr2 (l) = 1.078 V
see image
q = +
system gains heat
Voltage of Galvanic cell
tells us how much work can be done by electrons flowing from one side to the other
Chelate Effect
the ability of multidentate ligands to form more stable metal complexes than those formed by similar monodentate ligands
Anode
the electrode at which oxidation occurs
Cathode
the electrode at which reduction occurs
formation constant (Kf)
the equilibrium constant associated with reactions for the formation of complex ions
If ΔG for a battery is negative
the free energy of the battery decreases and the battery can do work on its surroundings, so w is negative
Standard Hydrogen Electrode (SHE)
the half-cell consisting of an inert platinum electrode immersed in 1 M HCl with hydrogen gas at 1 atm bubbling through the solution; used as the standard of a cell potential of zero (left side = 0V)
E-
the potential of the electrode attached to the negative terminal
E+
the potential of the electrode attached to the positive terminal of the potentiometer
Ampere
unit of electric current
w = +
work done on the system
Work equation
work(J) = E(V) * q(C)
free energy difference (ΔG) equation
ΔG(j/mol) = -n * F(C/mol) * E(V) E = maximum voltage
For a spontaneous electrochemical reaction, what are the signs (+/-) of ΔG and E?
ΔG: negative (-) E: positive (+) galvanic cell vice versa for a non-spontaneous reaction (electrolytic cell)
If a voltmeter measured 0.023 V when an indicator electrode measured in conjunction with an Ag|AgCl electrode saturated with KCl, what would the indicator electrode measure against a saturated calomel electrode saturated with KCl? Note: •standard hydrogen electrode, E = 0.0 V •Ag/AgCl, saturated KCl, E = 0.197 V •Hg/Hg2Cl2, saturated KCl, E = 0.241 V)
• Emeasured= Eind -Eref • Eind = E(meas w/AgCl) + E(Ag/AgCl) = 0.023 V + 0.197 V = 0.220 V • E(meas, SCE) = Eind -ESCE =0.220V - 0.241V = -0.021 V
•Supposing the activities of OH- and H+ are 0.0100 and the fugacity of H2Se is 0.0100, calculate the cell potential for: Ni(s)|Ni(OH)2(s)|OH- (aq)||H+ (aq)|H2Se(g)|Se(s)|Au(s) •The standard reduction potentials for the left and right half-cells are -0.714 V and -0.082 V. • NOTE: Fugacity is the activity value for a gas.
•Left half-cell: Ni(OH)2(s) + 2 e- <-> Ni(s) + 2 OH-(aq) E˚ = -0.714 E-= E˚ - (0.05916/n) log(AOH-2) = -0.714 - (0.05916/2) * -4 = -0.596 V •Right half-cell: H2Se(g) + 2 e- <-> Se(s) + 2 H+(aq) E˚ = -0.082 V E+= E˚ - (0.05916/n) log(AH+2/fH2Se) = -0.082 - (0.05916/2) * -2 = -0.0228 V •Net cell Ecell = E+ - E- = 0.573 V
Consider the titration of 25.0 mL of 0.020 0 M MnSO4 with 0.010 0 M EDTA in a solution buffered to pH 8.00. Calculate pMn2+ at the following volumes of added EDTA and sketch the titration curve: 0 mL 20.0 mL 40.0 mL 49.0 mL 49.9 mL
a) at 0 mL [Mn+2] = 0.02 pMn+2 = -log[Mn+2] = 1.698 = 1.7 (approx) b) at 20mL Mn+2 added = 25 X 0.02 = 0.5 millimoles EDTA = 0.01 X 20 = 0.2 millimoles Mn+2 remained = 0.5-0.2 = 0.3 millimoles Concentration of Mn+2 = 0.3 / 25 + 20 = 0.0067 pMn+2 = -log 0.0067 = 2.17 c) at 40mL of EDTA Mn+2 added = 25 X 0.02 = 0.5 millimoles EDTA = 0.01 X 40 = 0.4 millimoles Mn+2 remained = 0.5-0.4 = 0.1 millimoles Mn+2 = 0.1 / 25 + 40 = 0.0015 pMn+2 = -log 0.0015 = 2.82 d) at 49 mL Mn+2 added = 25 X 0.02 = 0.5 millimoles EDTA = 0.01 X 49 = 0.49 millimoles Mn+2 remained = 0.5-0.49 = 0.01 millimoles Mn+2 = 0.01 / 25 + 49 = 1.35 X 10^-4 pMn+2 = -log 1.35 X 10^-4 = 3.86 e) at 49.9 mL moles of Mn+2 added = 25 X 0.02 = 0.5 millimoles Moles of EDTA = 0.01 X 49.9 = 0.499 millimoles so moles of Mn+2 remained unreacted = 0.5-0.49.9 = 0.001 millimoles Concentration of Mn+2 = Millimoles / total volume of solution in mL = 0.001 / 25 + 49.9 = 1.34 X 10^-5 So pMn+2 = -log 1.34 X 10^-5 =4.87 d) at 50mL Already calculated , this is the equivalence point so moles of MnSO4= moles of EDTA So all ions are in the form of metal-EDTA complex [MnY^-2] = moles / volume = 0.5 millimoles / 25 + 50 = 0.0067 Now here we need the formation constant of Mn+2-EDTA complex Mn+2 + Y-4 --> MnY-2 Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4] Let us apply ICE Mn+2 + Y-4 --> MnY-2 Initial 0 0 0.0067 Change x x -x Equilibrium x x 0.0067-x Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4] = 0.0067 -x / x^2 We may ignore x in numerator as the Kf is very high so 4.2 X 10^11 = 0.0067 / x^2 x^2 = 0.0067 / 4.2 X 10^11 = 0.0016 X 10^-11 x = 0.126 X 10^-6 pMn+2 = -logx = 6. 9
indicator electrode
an electrode placed in a sample in order to measure the concentration of an ion in the sample
Pt electrode
conducts electrons into or out of a chemical species in the redox reaction. Platinum is a common inert electrode. It does not participate in the redox chemistry except as a conductor of electrons.
Consider the titration of 25.0 mL of 0.020 0 M MnSO4 with 0.010 0 M EDTA in a solution buffered to pH 8.00. Calculate pMn2+ at the following volumes of added EDTA and sketch the titration curve:50.0 mL 50.1 mL 55.0 mL 60.0 mL
e) at 50.1 mL Excess of EDTA = 50.1 X 0.01 - 25 X 0.02 = 0.001 millimoles EDTA = 0.001 millimoles / 75.1 = 1.33 X 10^-5 metal ion complex = 0.5 millimoles / 75.1 mL = 0.0066 Molar Again we will apply the ICE table here Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4] 4.2 X 10^11 = 0.0066-x / x (1.33X 10^-5+x) we may ignore x 4.2 X 10^11 = 0.0066 / x (1.33X 10^-5) x = 0.0066 / (1.33X 10^-5)(4.2 X 10^11) x = 0.00118 X 10^-6 pMn+2 = 8.92 f) at 55mL We can calcualte similarly Here the moles of EDTA will exceed the moles of Mn+2 ions Excess of EDTA = Moles of EDTA added - Moles of Mn+2 present Excess of EDTA = 55 X 0.01 - 25 X 0.02 = 0.05 millimoles concentration of excess EDTA = 0.05 millimoles / 80 = 0.000625 Concentration of metal ion complex = 0.5 millimoles / 80 mL = 0.00625 Molar Again we will apply the ICE table here Mn+2 + Y-4 --> MnY-2 Initial 0 0.000625 0.00625 Change x x -x Equilibrium x 0.000625 +x 0.00625-x Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4] 4.2 X 10^11 = 0.00625-x / x (0.000625+x) we may ignore x 4.2 X 10^11 = 0.00625 / x (0.000625) x = 10 / 4.2 X 10^11 pMn+2 = 10.62 g) at 66mL Here the moles of EDTA will exceed the moles of Mn+2 ions Excess of EDTA = Moles of EDTA added - Moles of Mn+2 present Excess of EDTA = 66 X 0.01 - 25 X 0.02 = 0.16 millimoles concentration of excess EDTA = 0.16 millimoles / 91 = 0.0017 Concentration of metal ion complex = 0.5 millimoles / 91 mL = 0.0055 Molar Again we will apply the ICE table here Mn+2 + Y-4 --> MnY-2 Initial 0 0.0017 0.0055 Change x x -x Equilibrium x 0.0017+x 0.0055-x Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4] 4.2 X 10^11 = 0.0055-x / x (0.0017+x) we may ignore x 4.2 X 10^11 = 0.0055 / x (0.0017) x = 0.77 X 10^-11 pMn+2 = 11.11
reducing agent
electron donor
Lewis acid
electron pair acceptor
Lewis base
electron pair donor
Kf'
formation constant at the fixed pH of the solution
cell voltage is proportional to
free energy change for the electrochemical reaction
reduced
gains electrons
the potentiometer indicates a positive voltage when
hen electrons flow into the negative terminal of the meter vice versa
Write out the Nernst equation for the Ag/AgCl electrode.
look at image
You pipet 10.00 mL 10.0 mM ammonium chloride into 100.0 mL of 0.1000 mM silver nitrate with 10.00 mM ammonium nitrate, then you measure a cell potential of 72.7 mV. Calculate the formation constant of silver chloride, AgCl. Use a Nernst constant of 59.16 mV, cell constant of 455.0 mV, and activity coefficients of 0.90.
look at image
oxidized
loses electrons
A mass spectrometer separates chemicals based on what property?
mass-to-charge ratio (m/z)
open circuit potential
measured potential difference
Volts
measures potential difference
Potentiometer
measures the difference in electric potential (voltage) between the two metal electrodes. Emeas = E+ - E-
the greater the voltage =
more current flow
Please determine the oxidized species, reduced species, oxidizing agent, and the reducing agent of the following reaction. NiO2 + 2 H2O + Fe <-> Ni(OH)2 + Fe(OH)2
oxidized species: Fe solid (0 -> +2) reduced species: Ni4+ in NiO2 (+4 -> +2) oxidizing agent: Ni4+ in NiO2 reducing agent: Fe solid
P = E*I = (IR)*I = I^2 *R
power (watts) = work per second
relation between charge and moles
q = n * (mol) * F
electric current is proportional to
rate of reaction
line notation
reactant in anode | product in anode || reactant in cathode | product in cathode |
write half-reactions as
reductions
Draw the Bermuda Triangle of Separations
see image
