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Day 1 > Analytical Chemistry > Flashcards

Analytical Chemistry Flashcards

1
Q

Dilution Formula

A

C1V1=C2V2

V2=V1+V(H2O)

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2
Q

f for Redox Reaction

A

Number of lost or gained electrons

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3
Q

f for Acid-Base Reactions

A

Number of participating OH- or H+

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4
Q

Cr2 (SO4)3 • 18 H2O is precipitated as hydroxide. What is the value of f?

A

f=3 for Cr (OH)3
f=6 for Cr2 (SO4)3

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5
Q

Gravimetric Factor Formula

A

(MW unknown/ MW given residue) • (mol unknown/ mol given residue)

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6
Q

Percent element in a sample for gravimetry

A

%X = (mass of residue × GF × 100%) / mass of sample

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7
Q

Aliquot is a portion of the sample. How to relate mass of aliquot to mass of sample?

A

Use ratio and proportion. Aliquot has the same properties as the sample.

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8
Q

If given total mass of two compounds in a mixture, how to attain their individual masses?

A

Set up two equations.
Eq 1: X+Y=total mass
Eq 2: set up reactions for X and Y to get the products

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9
Q

Gravimetry where mass of precipitate is the signal

A

Precipitation Gravimetry

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10
Q

Arrhenius Acid and Base

A

Acid - produces hydronium ion in aqueous solution

Base - produces hydroxide ion in aqueous solution

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11
Q

Bronsted-Lowry Acid and Base

A

Acid - proton donor
Base - proton acceptor

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12
Q

Lewis Acid and Base

A

Acid - electron pair acceptor
Base - electron pair donor

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13
Q

Product of neutralization reaction aside from water

A

Salt

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14
Q

When benzoic acid is dissolved in water, it partially ionizes forming benzoate and hydronium ion.

Name the parts of the reaction

A

Benzoic acid - Weak acid
Water - Base
Benzoate - Conjugate base
Hydronium Ion - Conjugate acid

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15
Q

Water at room temperature is more basic. True or False?

A

False, it is more acidic due to formation of carbonic acid

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16
Q

Differentiate strong acids and weak acids in terms of ionization

A

Strong acids - 100℅ ionization
Weak acids - partial ionization

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17
Q

Acids that yield more than 1 hydronium ion

A

Polyprotic Acids

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18
Q

For polyprotic compounds that yields two hydronium ions, K1 is _____ than K2?

smaller? larger? or equal?

A

smaller

Kn&raquo_space;> … K2 > K1

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19
Q

A mixture of ammonia and potassium hydroxide is considered a ____ base.

Strong or Weak?

A

Strong Base
pOH = -log (concentration of solution)

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20
Q

pH computation for strong acids and bases

A

pH = -log [H+]
pOH = -log [OH-]
pH + pOH = 14

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21
Q

pH means…

A

potential of hydrogen

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22
Q

computation of pH for weak acids

A

Ka = x² / (Cao - x)

Cao = [(Va)(Ca) / Va+Vt]

a - analyte
t - titrant added

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23
Q

Kw at 25°C

A

1x10^-14 = [H+][OH-]

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24
Q

Kw at 50 °C

A

5.5x10^-14

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25
Q

Formula of Kw in terms of equilibrium constants

A

Kw = Ka•Kb

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26
Q

How to solve equivalence volume

A

Veq = [(Va)(Ca) / Ct]

At end point = Veq = V(titrant)

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27
Q

Formula for pka, pkb, and pkw

A

pka = -log (Ka)
pkb = -log (Kb)
pka + pkb = pKw

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28
Q

Combination of weak acid and base and their respective conjugates.

A

Buffer system

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29
Q

Explain buffer

A

Weak analyte + Strong titrant = Conjugate + water

Buffer is the mixture of the weak analyte and its conjugate

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30
Q

Equation used when dealing with buffer solutions

A

Henderson-Hasselbalch Equation

ph = pKa + log (conjugate base/ weak acid) (both at equilibrium)

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31
Q

Increased in buffer capacity will result to…

A

ratio of base and acid components approaching 1

32
Q

pH = pKa happens when…

A

Vt = Veq/2
A-/HA = 1
at maximum buffer capacity

33
Q

Which of the following acids would be most suitable to prepare a buffer with pH of 4?

a. Alex acid pKa= 2.565
b. Xander acid pKa = 4.150
c. Busto acid pKa = 7.004
c. Harvard acid pKa = 10.000

A

B. Choose an acid with a pH close to the desired pH of the buffer solution

34
Q

Kjeldahl Method

A

Determination of Nitrogen from the conversion of organic Nitrogen to Ammonia, which is separated by distillation and quantified using titration

NH3 + HCl = NH4+ + Cl-

35
Q

How to get %N in the sample using Kjeldahl method?

A

NH3 reacted with HCl = HCl used - HCl reacted with NaOH

1 mol of N/ 1 mol of NH3

%N (used gravimetric procedure)

36
Q

In saponification, esters react with hydroxide to form what products?

A

Carboxylate ions (RCOO-)
Alcohol (R’OH)

37
Q

Why does hydroxide and bicarbonate can’t coexist in a solution?

A

They react with each other forming carbonate ion and water

38
Q

pH of Phenolphthalein and alkaline ions it detect

A

8.3
hydroxide and carbonate

39
Q

pH of bromocresol green and alkalines it detect

A

4.5
Bicarbonate

40
Q

Volume of Bromocresol green = 0

A

Hydroxide

41
Q

Volume of Phenolphthalein = 0

A

Bicarbonate

42
Q

Volume of BCG = Volume of Ph

A

Carbonate

43
Q

Volume of BCG > Volume of Ph

A

Carbonate (Volume of ph)
Bicarbonate (Volume at the end point of BCG - 2 Volume of ph)

44
Q

Volume of BCG < Volume of Ph

A

Hydroxide
Carbonate

45
Q

Uses silver ion as titrant or analyte

A

Argentometric Titration (Precipitation Titration)

46
Q

Mohr Method’s Indicator

A

Alkali Metal Chromate (e.g., K2CrO4)

47
Q

Titration reaction of Argentometric process

A

Ag+ + Cl- = AgCl (white precipitate)

Ag+ is supplied in excess

48
Q

Final indicator of Mohr Method

A

Ag2•CrO4
brick red precipitate

49
Q

Favorable pH for Mohr Method

A

6.5 < pH < 10.3

lower than 6.5 - soluble complex
higher than 10.3 - insoluble hydroxide

50
Q

Fajan Method’s Indicator

A

Dichlorofluorescein (DCF-) (green color)
Eosin

51
Q

Final indicator product of Fajans Method

A

Ag+• AgCl • DCF- (Pink precipitate)

52
Q

Indicator of Volhard’s Method

A

FeSCN^2+ from ferric alum
bloody red precipitate

53
Q

Favorable condition for Volhard method

A

Acidic (pH 4-7) to avoid Fe^3+ precipitation

54
Q

Nickel + Cyanide will form _____ and in what ratio?

A

Tetracyanonickelate
1 Ni^2+ : 4 CN^-

55
Q

EDTA

A

Ethylenediaminetetraacetic acid

C10•H16•N2•O8

56
Q

Who develop the metal ligand titration methods for determining CN- and Cl-?

Titrants used for the aforementioned ligands?

A

Julius von Liebig, 1850

Ag+ for CN-
Hg2+ for Cl-

57
Q

EDTA in neutral form is …

A

H4Y (tetraprotic, meaning it can exist in variety of protonated protons)

58
Q

Explain chelating effect

A

EDTA has a cleating effect, which is the ability to form more stable metal complexes than those formed by similar mono-dentate ligand

59
Q

Direct titration method for determination of carbonates and hydroxides

A

Warder Method

60
Q

Indirect titration method for determination of carbonates and hydroxide

A

Winkler Method

61
Q

Titration method used to determine protein content of food products

A

Kjeldahl Analysis

62
Q

Milk of magnesia, which is used for antacid and laxative

A

Mg(OH)2

63
Q

Slaked lime used in mortar

A

Ca(OH)2

64
Q

Titrants for standardization of potassium permanganate (KMnO4)

A

Sodium oxalate (Na2C2O4)
Pure iron wire (Fe)
Arsenic oxide

65
Q

Reaction for standardization of KMnO4 using Na2C2O4

A

2 KMnO4 + 5 H2C2O4 + 6 H+ —
2 Mn2+ + 10 CO2 + 8 H2O

66
Q

Permanganimetry Redox Couple (pH < 2)

A

MnO4^- (violet) — Mn^2+ (colorless)

67
Q

Permanganimetry Redox Couple (5 < pH < 9)

A

KMnO4^- (violet) — MnO2 (brownish precipitate)

68
Q

Redox Couple for Cerimetry

A

Ce^4+ (bright yellow) — Ce^3+ (colorless)

69
Q

Titrant for cerimetry

A

Cerium (IV) Sulphate

70
Q

Titrant for Dichromate Titrations

A

K2Cr2O7/Potassium Permanganate

71
Q

Redox couple for dichromate titrations

A

Cr2O7^-2 — Cr^3+

72
Q

Advantages of dichromate titration. What are the conditions maintained?

A

stable, low cost, available in primary standard grade purity

strongly acidic conditions (maintained by adding dilute H2SO4)

73
Q

Iodimetry vs Iodometry

A

Iodimetry - use of I^2 (iodine) as oxidant
Iodometry - use of I^- (iodide ion) as reductant

74
Q

Advantages and disadvantages of iodimetry

A

+++
1. Availability of excellent indicator, starch

  1. Iodine is a weaker oxidant
  2. Weak stability; thus, they need to be standardized regularly
75
Q

Indicator that end point is reach in iodimetry

A

Iodide ion is dissolved in iodine solution to form I^3-, which creates dark blue complex with starch

76
Q

Redox couple for Iodometry

A

S2O3^2- — S4O6^2-
(S +2). — (S +2.5)

77
Q

Redox Couple for Ferrous Ion

A

Fe2+ — Fe3+

Day 1 (4 decks)

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