06. Volumetric Methods Flashcards by Ray Mark

Define volumetric analysis

Controlled addition of titrant (known concentration, inside the burette) to a sample until it reaches an endpoint (color change or instrumental signal)

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Differentiate endpoint and equivalence point

Endpoint
→ usually indicated by color change of indicator or instrumental signal

Equivalence point
→ Region where analyte and titrant are stoichiometrically equal

Ideally, endpoint is near the equivalence point but realistically endpoint occurs after slight excess of titrant is added

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Describe the titration curve of a strong acid/strong base vs weak acid/weak base

Volume of titrant added (x-axis) vs pH of solution (y-axis)

Strong Acid/base
→ very sharp change in pH at the equivalence point

Weak acid/base
→ less define change in pH
→ more than one inflection points if polyprotic acids or multi-step reactions

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How to know if there is a buffer region in titration curve?

Plateau before equivalence (mainly in weak acid/base titrations)

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Species present and equilibrium/equation to use for different regions in the titration curve of weak acid with strong base titrant

Before titration
→ analyte only
→ ICE → Ka of analyte → [H+] → pH
Before half equivalence point
→ analyte in excess
→ ICE → HHE
At half equivalence point
→ analyte = conjugate
→ pH = pKa
At equivalence point
→ [conjugate] + H2O
→ ICE₁ → ICE₂ → Kb conj. → x = [OH-] → pOH → pH
After equivalence point
→ excess titrant
→ ICE → excess [OH-] → pOH → pH

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Ionization of the analyte and reactions of analyte and titrant during titration during different regions in the titration curve

Before titration
Before half equivalence point
At half equivalence point
At equivalence point
After equivalence point

Region 1:
WA: HA + H2O ⇌ A- + H3O+
WB: B + H2O ⇌ BH+ + OH-

Regions 2,3,5:
HA + OH- ⇌ A- + H2O
B + H3O+ ⇌ BH+ + H2O

Region 4
A- + H2O ⇌ HA + OH-
BH+ + H2O ⇌ B + H3O+

titrant

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Calculate the pH at the equivalence point in the titration of 50.00 mL of 0.100 M HCl with 0.100 M NaOH.

HCl + NaOH → NaCl + H2O
0.005 mol | 0.005 mol
- 0.005 mol | -0.005 mol
0 mol | 0 mol

2H2O ⇌ H3O+ + OH-
Kw = [H3O+] [OH-]
10⁻¹⁴ = x²
x = 10⁻⁷ = [H3O+] = [OH-]
pH = -log(10⁻⁷)
pH = 7.00

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Calculate the pH at 0.00, 10.00, 25.00, 50.00, and 60.00 mL of titrant in the titration of 50.00 mL of 0.100 M HCl with 0.100 M NaOH.

0.00 mL: pH = 1.00
10.00 mL: pH = 1.18
25.00 mL: pH = 1.47
50.00 mL: pH = 7.00
60.00 mL: pH = 11.96

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Calculate the pH at the equivalence point in the titration of 50.00 mL of 0.100 M CH3COOH with 0.100 M NaOH.

Ka CH3COOH = 1.8x10⁻⁵

CH3COOH + NaOH → NaCH3COO + H2O
0.005 mol | 0.005 mol | 0 mol
-0.005 mol | -0.005 mol | +0.005 mol
0 mol | 0 mol | 0.005 mol

[CH3COO-] = 0.005mol/(0.05mL+0.05mL)
[CH3COO-] = 0.05 M

Kw = KaKb
Kb = Kw/Ka = 10⁻¹⁴/1.8x10⁻⁵ = 5.56x10⁻¹⁰

CH3COO- + H2O ⇌ CH3COOH + OH-
0.05 | - | 0 | 0
-x | - | +x | +x
0.05-x | - | x | x

Kb = [CH3COOH][OH-]/[CH3COO-]
5.56x10⁻¹⁰ = x²/0.05-x
5.56x10⁻¹⁰ (0.05-x) = x²
2.78x10⁻¹¹ - 5.56x10⁻¹⁰x = x²
0 = x² + 5.56x10⁻¹⁰x - 2.78x10⁻¹¹
x = 5.27x10⁻⁶ = [OH-]

pOH = -log [OH-] = -log [5.27x10⁻⁶] = 5.28
pH = 14 - 5.28 = 8.72

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Calculate the pH at 0.00, 10.00, 25.00, 50.00, and 60.00 mL of titrant in the titration of 50.00 mL of 0.100 M CH3COOH with 0.100 M NaOH

Ka = 1.8 x 10⁻⁵

0.00 mL: pH = 2.88
10.00 mL: pH = 4.14
25.00 mL: pH = 4.74
50.00 mL: pH = 8.72
60.00 mL: pH = 11.96

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A 50.00-mL portion of HCl solution required 29.71 mL of 0.01963 M Ba(OH)2 to reach an end point with bromocresol green (BCG). Calculate the molar concentration of the HCl solution.

2HCl + Ba(OH)2 → BaCl2 + 2H2O

mol Ba(OH)2 = 29.71x10⁻³L (0.01963 mol Ba(OH)2/L
mol Ba(OH)2 = 5.8321x10⁻⁴ mol

mol/L HCl = 5.8321x10⁻⁴ mol Ba(OH)2 (2 mol HCl/1 mol Ba(OH)2) (1/0.0500L)
mol/L HCl = 0.0233 M

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Sodium carbonate (Na₂CO₃) is one of the most commonly used reagents for standardizing acids. It occurs naturally in large deposits as ____ and _____. Primary standard grade is produced by extensively purifying these natural sources.

It needs how many moles of acid to standardized?

washing soda (Na₂CO₃-10H₂O) and trona (Na₂CO₃-NaHCO₃-2H₂O)

CO₃²⁻ + 2H₃O⁺ → H₂CO₃ + 2H₂O

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Calculate the molar concentration of a dilute HCI solution if the titration of 0.2459 g of primary standard Na₂CO₃ required 36.52 mL of the acid (products: CO₂ and H₂O).

Na₂CO₃ + 2HCl → H₂CO₃ + 2NaCl

mol/L HCl = 0.2459 g Na₂CO₃ (mol Na₂CO₃/ 106 g) (2 mol HCl/1mol Na₂CO₃) (1/0.03652 L)
mol/L HCl = 0.1270 M

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Potassium hydrogen phthalate (KHP) is a nearly ideal primary standard. It is a non-hygroscopic crystalline solid with a relatively large molar mass. For most purposes, the commercial analytical-grade salt can be used without further purification.

What is its chemical formula, molar mass, and equation with NaOH?

KHC₈H₄O₄ + NaOH → KNaC₈H₄O₄ + H₂O
(204.2 g/mol)
→ only one acidic proton

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Calculate the molar concentration of a dilute Ba(OH)2 solution if titration of 0.4512 g of primary standard potassium hydrogen phthalate (KHP) required 26.46 mL of the base.

2KHC₈H₄O₄ + Ba(OH)₂ → Ba(C₈H₄O₄)₂ + 2H₂O

mol/L Ba(OH)₂ = 0.4512 g KHP (mol KHP/204.2 g) (1mol Ba(OH)₂/2 mol KHP) (1/0.02646 L)
mol/L Ba(OH)₂ = 0.0418 M

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Why bases should be stored in plastic containers and not in glass?

The concentration will decrease due to the reaction of the base with the glass to form silicates.

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What is a carbonate error? Explain the difference of using acid-range and basic-range indicator.

Hydroxides of Na, K, and Ba react rapidly with atmospheric CO2 to produce carbonate:
CO₂ + 2OH- → CO₃²⁻ + H₂O

Using acid-range indicator (e.g., bromocresol green):
CO₃²⁻ + 2H₃O⁺ → H₂CO₃ + 2H₂O
→ 2H₃O⁺ matches the 2OH- and no error occur

Using basic-range indicator (e.g., phenolphthalein):
CO₃²⁻ + H₃O⁺ → HCO₃⁻ + 2H₂O
→ systematic error occur

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The hydroxide concentration of a freshly prepared, carbonate-free NaOH solution was 0.05118 M. If 1.000 L of this solution absorbs 0.1962 g of CO₂ from the air, calculate the relative carbonate error in acetic acid titration using this contaminated solution with phenolphthalein as the indicator.

2NaOH + CO₂ → Na₂CO₃ + H₂O

mol/L Na₂CO₃ = 0.1962 g CO₂ (mol CO₂/44g) (mol Na₂CO₃/mol CO₂) (1/1.000L)
mol/L Na₂CO₃ = 4.459x10⁻³ M

NaOH + CH₃COOH → NaCH₃COO + H₂O
Na₂CO₃ + CH₃COOH → NaHCO₃ + NaCH₃COO
[CO₃²⁻ + H₃O⁺ → HCO₃⁻ + 2H₂O]

effective concentration of NaOH:
mol/L NaOH = 0.05118 mol NaOH/L − [4.459x10⁻³ mol Na₂CO₃ (mol CH₃COOH/mol Na₂CO₃) (mol NaOH/mol CH₃COOH)
mol/L NaOH = 0.04672 M

Er = (0.04672 - 0.05118)/0.05118
Er = 8.7%

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If 1.000 L of 0.1500 M NaOH was unprotected from the air after standardization and absorbed 11.2 mmol of CO₂, what is its new molar concentration when it is standardized against a standard solution of HCI using phenolphthalein.

2NaOH + CO₂ → Na₂CO₃ + H₂O

mol/L Na₂CO₃ = 11.2x10⁻³ mol CO₂ (mol Na₂CO₃/mol CO₂) (1/1.000L)
mol/L Na₂CO₃ = 0.0112 M Na₂CO₃

NaOH + HCl → NaCl + H₂O
Na₂CO₃ + HCl → NaHCO₃ + NaCl
[CO₃²⁻ + H₃O⁺ → HCO₃⁻ + 2H₂O]

effective concentration of NaOH:
mol/L NaOH = 0.1500 mol NaOH/L − (0.0112 mol Na₂CO₃/L)(mol HCl/mol Na₂CO₃)(mol NaOH/mol HCl)
mol/L NaOH = 0.1388 M

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A NaOH solution was 0.1019 M immediately after standardization. Exactly 500.0 mL of the reagent was left exposed to air for several days and absorbed 0.652 g of CO₂. Calculate the relative carbonate error in the determination of acetic acid with this solution if the titrations were performed with phenolphthalein.

2NaOH + CO₂ → Na₂CO₃ + H₂O

mol/L Na₂CO₃ = 0.652 g CO₂ (mol CO₂/44g)(mol Na₂CO₃/mol CO₂) (1/0.500 L)
mol/L Na₂CO₃ = 0.02964 mol/L

NaOH + CH₃COOH → NaCH₃COO + H₂O
Na₂CO₃ + CH₃COOH → NaHCO₃ + NaCH₃COO
[CO₃²⁻ + H₃O⁺ → HCO₃⁻ + 2H₂O]

effective concentration of NaOH:
mol/L NaOH = 0.1019 mol NaOH/L − (0.02964 mol Na₂CO₃/L)(mol CH₃COOH/mol Na₂CO₃)(mol NaOH/mol CH₃COOH)
mol/L NaOH = 0.0723 M

Er = (0.0723 − 0.1019)/0.1019
Er = 29.1%

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Titration method used to identify and quantify carbonate species in water

double indicator

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Titration method used for substances that react slowly or are insoluble. What is the principle behind?

back titration
→ excess acid/base, then titrated with opposite reagent

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Titration method used for the determination of nitrogen (and protein by extension). What is the principle?

Kjeldahl
→ ammonium formed from digestion is titrated after distillation

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Automated or precise titration method using pH or voltage readings

Potentiometric
→ signal change rather than color change

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Steps involved in the steps in Kjeldahl method

1. Digestion sample + H₂SO₄ → (NH₄)₂SO₄ + CO₂(g) + SO₂(g) + H₂O(g) 2. Distillation (NH₄)₂SO₄ + NaOH → Na₂SO₄ + 2H₂O(l) + 2NH₃(g) 3. Capture of NH₃ (measured excess HCl) NH₃(g) + HCl → NaCl + 2H₂O(l) 4. Titration (back titration) HCl + NaOH → NaCl + 2H₂O(l)

A 0.7121 g sample of wheat flour was analyzed using the Kjeldahl method. Ammonia, formed by adding concentrated base after digestion with sulfuric acid, was distilled into 25.00 mL of 0.04977 M HCl. The excess HCl was then back titrated with 3.97 mL of 0.04012 M NaOH. Using a 5.70 conversion factor for cereals, calculate the percent protein in the wheat flour sample.

mol NH₃ → %N → %protein mol NH₃ = (MV)_HCl − (MV)_NaOH mol NH₃ = [25x10⁻³ L (0.04977 mol HCl/L)(mol NH₃/mol HCl)] − [3.97x10⁻³ L (0.04012 mol NaOH/L)(mol NH₃/mol NaOH)] mol NH₃ = 1.085x10⁻³ %N = 1.085x10⁻³ mol NH₃(mol N/mol NH₃)(14.01 g/mol N)(1/0.7121g sample) x 100 %N = 2.13 %N %protein = 2.13x5.70 = **12.17% protein**

_____ is commonly used to determine the alkalinity of water samples when the primary contributors are limited to hydroxide (OH⁻), bicarbonate (HCO₃⁻), and carbonate (CO₃²⁻) ions.

Double indicator titration

____ represents the acid-neutralizing capacity of a water sample, and while it is mainly due to these carbonate species, other weak bases like phosphates may also contribute.

Alkalinity

pH range of phenolphthalein and methyl orange

pH 8.3-10 → phenolphthalein (acidic colorless/ basic pink) pH 3-4.5 → methyl orange (acidic red/orange basic) bromocresol green is same range with methyl orange

Relationship between end point volumes of phenolphthalein and bromocresol green when testing for hydroxide, OH⁻

OH⁻ ↓+ H⁺ H₂O Vph = Vbcg

Relationship between end point volumes of phenolphthalein and bromocresol green when testing for bicarbonate, HCO₃⁻

HCO₃⁻ ↓ + H⁺ H₂CO₃ Vph = 0 → supposed to turn pink if the solution is basic → colorless already

Relationship between end point volumes of phenolphthalein and bromocresol green when testing for carbonate, CO₃²⁻

CO₃²⁻ ↓+ H⁺ HCO₃⁻ ↓+ H⁺ H₂CO₃ Vph = 1/2 Vbcg

Relationship between end point volumes of phenolphthalein and bromocresol green when testing for hydroxide, OH⁻ and carbonate, CO₃²⁻

OH⁻..........CO₃²⁻ ↓+ H⁺........↓+ H⁺ H₂O......... HCO₃⁻ .................↓+ H⁺ .................H₂CO₃ Vph > 1/2 Vbcg

Relationship between end point volumes of phenolphthalein and bromocresol green when testing for carbonate, CO₃²⁻ and bicarbonate, HCO₃⁻

CO₃²⁻ ↓+ H⁺ HCO₃⁻ HCO₃⁻ ↓+ H⁺ ↓+ H⁺ H₂CO₃ H₂CO₃ Vph < 1/2 Vbcg

A solid mixture contains NaOH, NaHCO₃, and Na₂CO₃, either individually or in a permissible combination. The sample was titrated with 0.250 M HCl, requiring 26.20 mL to reach the phenolphthalein endpoint, and an additional 15.20 mL to reach the bromocresol green endpoint. Calculate the mass (mg) of each component present in the mixture.

Vph = 26.20 mL Vbcg = 26.20+15.20 mL = 41.4 mL 1/2 Vbcg = 20.7 mL Vph > 1/2 Vbcg → OH⁻ and CO₃²⁻ OH⁻ + H⁺ → H₂O CO₃²⁻ + H⁺ → HCO₃⁻ + H⁺ → H₂CO₃ For Na₂CO₃: V_HCl = (41.4 - 26.2) x 2 = 30.4 mL For NaOH: V_HCl = 41.4 - 30.4 = 11.0 mL mass NaOH = 11.0 mL (0.250 mmol HCl/mL)(mmol NaOH/mmol HCl)(40 mg/mmol NaOH) = **110 mg NaOH** mass Na₂CO₃ = 30.4 mL (0.250 mmol HCl/mL)(mmol Na₂CO₃/2 mmol HCl)(106 mg/mmol Na₂CO₃) = **403 mg Na₂CO₃**

A solution contains NaHCO₃, Na₂CO₃, and NaOH, either alone or in a permissible combination. Titration of a 50.0-mL portion to a phenolphthalein end point requires 22.1 mL of 0.100 M HCl. A second 50.0-mL aliquot requires 48.4 mL of the HCI when titrated to a bromocresol green end point. Determine the composition and the molar solute concentrations of the original solution.

mol/L Na₂CO₃ = 0.0442L (0.100 mol HCl/L)(mol Na₂CO₃/2 mol HCl) (1/0.05L) = **0.0442 M** mol/L NaHCO₃ = 0.0042L (0.100 mol HCl/L)(mol NaHCO₃/mol HCl) (1/0.05L) = **0.0084 M** For Na₂CO₃: V = 22.1 mL + 22.1 mL = 44.2 mL HCl For NaHCO₃: V = 48.4 mL - 44.2 mL = 4.2 mL

Organic compounds that contain both amino (-NH₂) and carboxylate (-COO⁻) groups, which act as ligands in titrations. These are the most commonly used titrating agents in complexometric analysis.

Aminocarboxylic acids

An organic compound with two or more binding sites (functional groups) that can simultaneously coordinate to a metal ion.

Chelating agent

A complex formed when a metal ion binds with a chelating agent at multiple sites, creating a ring-like structure.

Chelate

A titration method involving a chelating agent to determine metal ion concentration. It is a highly practical and widely used analytical technique.

Chelometric Titration

The most widely used chelating agent in complexometric titrations. It can bind metal ions through multiple (six) coordination sites, usually forming stable 1:1 complexes.

Ethylenediaminetetraacetic acid (EDTA)

Color indicator and effective pH range of Eriochrome Black T

Metal-EBT complex (wine red) → free EBT (blue) → unstable in solution (must be freshly prepared) → effective pH range only: 7-11 → buffered around pH 10 by ammonia-ammonium chloride buffer

A 100.0 mL water sample required 23.63 mL of 0.0109 M EDTA for hardness analysis. Calculate the hardness in mg CaCO₃ per liter.

mg CaCO₃/L = 23.63mL (0.0109 mmol EDTA/mL)(mmol CaCO₃/mmol EDTA)( 100 mg/mmol CaCO₃)(1/0.1000L) **mg CaCO₃/L = 258 ppm**

In a 9.57-g sample of rodenticide, thallium (Tl) was oxidized to TI³⁺ and treated with an excess of Mg/EDTA solution. The liberated Mg²⁺ was titrated with 12.77 mL of 0.03610 M EDTA. Calculate the percent of thallium (TI) in the sample.

TI³⁺ + MgY²⁻ ⇌ TlY⁻² + Mg²⁺ Mg²⁺ + Y⁴⁻ ⇌ MgY²⁻ %Tl = 12.77x10⁻³L (0.03610 mol EDTA/L)(mol Mg²⁺/mol EDTA)(mol TI³⁺/mol Mg²⁺)(204.4g/mol Tl)(1/9.57g) x 100 **%Tl = 0.98%Tl**

Titration of Ca²⁺ and Mg²⁺ in a 50.00-mL sample of hard water required 23.65 mL of 0.01205 M EDTA. A second 50.00-mL aliquot was made strongly basic with NaOH to precipitate Mg²⁺ as Mg(OH)₂. The supernatant liquid was then titrated with 14.53 mL of the same EDTA solution. Calculate the concentrations of CaCO₃ and MgCO₃ in the sample, expressed in ppm.

mg/L CaCO₃ = 14.53 mL (0.01205 mmol EDTA/mL)(mmol CaCO₃/mmol EDTA)(100mg/mmol CaCO₃)(1/0.05000L) **mg/L CaCO₃ = 350 ppm CaCO₃** mg/L MgCO₃ = [[23.65 mL (0.01205 mmol EDTA/mL)] − [14.53 mL (0.01205 mmol EDTA/mL)]](mmol MgCO₃/mmol EDTA)(84.32mg/mmol MgCO₃)(1/0.05000L) **mg/L MgCO₃ = 185 ppm**

One of the most widely used complexometric titrations involving a monodentate ligand that involves the titration of cyanide with AgNO₃. What is the indicator and equation involved?

Liebig Method Ag⁺ + 2CN⁻ ⇌ [Ag(CN)₂]⁻ Iodide indicator → formation of white ppt AgI when Ag⁺ is in slight excess Ag⁺ + I⁻ → AgI(s)

A 0.4482 g sample of impure sodium cyanide is titrated to the endpoint using 39.68 mL of 0.1018 M AgNO₃. Calculate the purity of the sample as %w/w, NaCN.

Ag⁺ + 2CN⁻ ⇌ [Ag(CN)₂]⁻ %g/g NaCN = 39.68x10⁻³ L (0.1018 mol AgNO₃/L)(1 mol Ag⁺/1mol AgNO₃)(2 mol CN⁻/1 mol Ag⁺)(1 mol NaCN/1 mol CN⁻)(49 g/mol NaCN)(1/0.4482 g) x 100 **%g/g NaCN = 88.32%**

Complexometric titration equations for (a) cyanide analyte with NiSO₄ titrant and (b) Cu²⁺, Hg²⁺, and Ni²⁺ analyte with KCN titrant

(a) Ni²⁺ + 4CN⁻ ⇌ [Ni(CN)₄]²⁻ (b) Cu²⁺ + 4CN⁻ ⇌ [Cu(CN)₄]²⁻ Hg²⁺ + 2CN⁻ ⇌ [Hg(CN)] Ni²⁺ + 4CN⁻ ⇌ [Ni(CN)₄]²⁻

When silver nitrate is used as titrant, the method is specifically referred to as an _____

argentometric titration

Precipitimetric method where the titrant is KSCN and involves precipitation of AgX, and the excess Ag⁺ is back titrated with SCN⁻ What is the medium, indicator used and application?

Volhard method → acidic medium (indirect method) → ferric ammonium sulfate (red FeSCN²⁺ after endpoint) → determination of halide ions in solution

Precipitimetric method where the titrant is AgNO₃ and involves precipitation of AgCl What is the indicator and application of this method?

Mohr → slightly basic medium (direct method) → potassium chromate (red color after the end point, Ag₂CrO₄) → determination of chloride content in water or soils

Precipitimetric method where the titrant is AgNO₃ and involves precipitation of anionic complexes on the surface of solid precipitates What is the indicator and application of this method?

Fajans → fluorescent or adsorption indicators (dichlorofluorescein, DCF (green) adsorbs on AgCl precipitate) → for chloride determination in complex matrices

Discuss the equations involved in Volhard method

1. Addition of AgNO₃ X⁻ (analyte) + Ag⁺ m. excess→ AgX(s) precipitate + Ag⁺ excess 2. During titration Ag⁺ excess + SCN⁻ (titrant) → AgSCN(s) 3. At end point Fe³⁺ indicator + SCN⁻ (titrant) → FeSCN²⁺ (blood-red complex)

Discuss the equations involved in Mohr method

1. During titration Cl⁻ (analyte) + Ag⁺ (titrant) → AgCl(s) 2. At end point CrO₄²⁻ (indicator) + Ag⁺ (titrant) → Ag₂CrO₄(s) (red precipitate)

Although the Mohr method for chloride determination is named after Karl Friedrich Mohr, the basic idea of using silver nitrate to precipitate chloride was actually first proposed by ______ in the early 1800s. Why it is named after Mohr?

Joseph Louis Gay-Lussac → Mohr refined the approach by introducing chromate ions as indicator

Discuss the equation involved in Fajans method using DCF as indicator

1. Before end point AgCl(s) (white) | Cl⁻ (excess) | DCF⁻ (green) [no adsorption reactions, negative charges] 2. After end point AgCl(s) (white) | Ag⁺ (excess) | DCF⁻ (pink) → DCF is adsorbed on the ppt [opposite charges]

A 50.00 mL sample was titrated using the Mohr method to determine chloride concentration. After adding a few drops of K₂Cr₂O₄ indicator, 15.70 mL of 0.0800 M AgNO₃ was required. Calculate the chloride content in mg/L.

mg/L Cl⁻ = 15.70mL (0.0800 mmol AgNO₃/mL)(mmol Ag⁺/mmol AgNO₃)(mmol Cl⁻/mmol Ag⁺)(35.45 mg/mmol Cl⁻)(1/0.05000L) **mg/L Cl⁻ = 891 ppm**

lodide in a 0.6712 g sample is determined using the Volhard method. After adding 50.00 mL of 0.05619 M AgNO₃ to the sample and allowing the precipitate to form, the excess Ag⁺ is back titrated with 35.14 mL of 0.05322 M KSCN. Calculate the percentage of iodide in the sample.

I⁻ + Ag⁺ m. excess→ AgI(s) + Ag⁺ excess Ag⁺ excess + SCN⁻ → AgSCN(s) mol I⁻ = [0.05000L (0.05619 mol AgNO₃/L)(mol I⁻/mol AgNO₃)] − [0.03514 L(0.05322 mol KSCN/L)(mol I⁻/mol KSCN)] mol I⁻ = 9.3945x10⁻⁴ mol I⁻ %I = 9.3945x10⁻⁴ mol I⁻ (126.9 g/mol I⁻)(1/0.6712g) x 100 **%I = 17.76%**

A NaOH solution is to be standardized by titrating it against a known mass of potassium hydrogen phthalate (KHP). Which of the following errors would result in calculating a NaOH molarity that is too low? A. Weighing only half of the recommended amount of KHP. B. Dissolving KHP in more water than is recommended. C. Losing some of the KHP solution from the flask before titrating. D. Not filling the buret tip with NaOH solution before starting the titration.

D. Not filling the buret tip with NaOH solution before starting the titration. mol/L NaOH = g KHP (mol KHP/g)(mol NaOH/mol KHP)(1/L NaOH solution) A and C → mol KHP = mol NaOH B → more water does not affect the calculation of NaOH molarity D → higher volume of NaOH will be recorded, ↑M NaOH

The molar mass of a solid carboxylic acid is determined by titrating a known mass of the acid with a standardized solution of NaOH solution to a phenolphthalein endpoint. Which of the following errors cause the calculated molar mass to be smaller than the true value? I. Some of the acid is spilled during transfer into the titration flask. II. The endpoint is recorded when the solution turns dark pink instead of the correct light pink. A. I only B. II only C. Both I and II D. Neither I nor II

B. II only I → at ep: mol acid = mol base

The protein content in a cheese sample is determined by Kjeldahl analysis. → A 0.9814 g sample is digested, converting nitrogen to ammonium ion. → Ammonium ion is then converted to ammonia by adding NaOH and distilled into 50.00 mL of 0.1047 M HCI. → The excess HCl is back titrated with 22.85 mL of 0.1183 M NaOH. Calculate the percentage of protein in the cheese sample. Use a Jones factor of 6.38 for dairy products. A. 3.614% B. 4.394% C. 23.06% D. 28.04%

C. 23.06% %N = [0.050L(0.1047molHCl/L) - 0.02285L(0.1183molNaOH/L)(molHCl/molNaOH)] (molNH3/molHCl)(molN/molNH3)(14g/molN)(100/0.9814g) %N = 3.61% % protein = 3.61 x 6.38 = 23.04%

Alkalinity is a commonly measured property of water. Which of the following statements about alkalinity are true? I. It is the capacity of a natural water sample to neutralize H⁺ ions until reaching pH 4.5, corresponding to the second equivalence point in the carbonate titration. II. The ions OH⁻, CO₃²⁻, and HCO₃⁻ contribute most significantly to alkalinity in natural waters. III. Both HCI and H₂SO₄ can be used as titrants to determine alkalinity. A. I and III B. I and II C. II and III D. I, II, and III

D. I, II, and III

Which of the following is not a desirable property of an indicator to be used in a complexometric titration that involves EDTA? A. The indicator should be a Lewis base. B. The indicator should bind more tightly to the analyte metal than does EDTA. C. The uncomplexed form of the indicator should be a different color than the Min complex. D. The complexation reaction between the indicator and the analyte metal should be reversible.

B. The indicator should bind more tightly to the analyte metal than does EDTA.

A 0.7457-g sample of foot powder contains zinc, which was titrated with 22.57 mL of 0.01639 M EDTA. Calculate the weight percent of zinc in the sample. A. 3.24 B. 3.63 C. 32.4 D. 36.3

A. 3.24 %Zn = 0.02257 L (0.01639molEDTA/L)(molZn/molEDTA)(65.4g/molZn)(100/0.7457g) %Zn = 3.24 %

A 0.4723-g sample of potassium cyanide is titrated with 34.95 mL of 0.1025 M AgNO₃. Calculate the percent purity of the potassium cyanide sample. A. 24.70% B. 49.39% C. 98.79% D. 99.99%

C. 98.79% Ag⁺ + 2CN⁻ ⇌ [Ag(CN)₂]⁻ %KCN = 0.03495L(0.1025molAgNO₃/L)(2molKCN/molAgNO₃)(65.12gKCN/mol)(100/0.4723g) %KCN = 95.79%

Which of the following statements accurately describe Volhard titration method? I. It is performed in acid solution. II. It is performed in slightly alkaline solution. III. It uses an indicator that forms a colored precipitate with the titrant. IV. A known excess of AgNO₃ is added to the analyte solution to precipitate the anion and the remaining Ag⁺ is then determined by back-titration with standard KSCN solution. A. I only B. I and II only C. I and III only D. I and IV only

D. I and IV only II → Mohr method III → red complex, not precipitate

Chloride concentration in a brine solution is determined using the Volhard method. A 10.00-mL aliquot of the solution is treated with 15.00 mL of standard 0.1182 M AgNO₃ to precipitate chloride. The excess silver is then back-titrated with 0.1010 M KSCN solution, requiring 2.38 mL to reach the red Fe(SCN)²⁺ end point. Calculate the concentration of chloride in the original brine solution. A. 5.43 g/L B. 6.29 g/L C. 7.48 g/L D. 9.57 g/L

A. 5.43 g/L g/L Cl = 0.015L(0.1182molAgNO₃/L) - 0.00238L(0.1010molKSCN/L)(molAgNO₃/molKSCN)] (mol AgCl/molAgNO₃)(molCl/molAgCl)(35.45g/molCl)(1/0.010L) g/L Cl = 5.43 g/L

Chloride concentration in a brine solution is determined using the Fajans method. A 10.00-mL aliquot of the brine solution is titrated with 15.00 mL of a standard 0.1100 M AgNO₃. A blank titration shows that 0.50 mL of AgNO₃ is consumed by impurities or reagents. Calculate the concentration of chloride in the original brine solution. A. 0.1450 M B. 0.1595 M C. 0.1650 M D. 0.5195 M

B. 0.1595 M mol/L Cl = 0.015L-0.0005L(0.11molAgNO₃/L)(molAgCl/molAgNO₃)(molCl/molAgCl)(1/0.010L) mol/L Cl = 0.1595 M

A standard solution of barium hydroxide is 0.250 M. What volume of 0.200 M nitric acid would be required to neutralize 10.0 mL of the barium hydroxide solution? A. 30.0 mL B. 25.0 mL C. 20.0 mL D. 10.0 mL

B. 25.0 mL

During the titration of a weak base with a strong acid, which type of acid-base indicator should be used? A. One that changes color in the buffer region. B. One that changes color in the basic pH range. C. One that changes color near neutral pH. D. One that changes color in the acidic pH range.

D. One that changes color in the acidic pH range.

Which of the following best distinguishes a complexing agent from a chelating agent? A. A complexing agent is used only in acidic solutions, while a chelating agent is used only in basic solutions. B. A complexing agent forms ionic bonds, while a chelating agent forms covalent bonds. C. A complexing agent is organic, while a chelating agent is always inorganic. D. A complexing agent binds to a metal ion at a single site, while a chelating agent binds at multiple sites forming a ring structure.

D. A complexing agent binds to a metal ion at a single site, while a chelating agent binds at multiple sites forming a ring structure.

Why is a small amount of magnesium salt added to the EDTA solution used for the titration of calcium with an Eriochrome Black T indicator? A. To increase the pH of the solution B. To form a stable Mg-EDTA complex that helps in producing a clear endpoint C. To prevent the precipitation of calcium ions D. To enhance the color change of the indicator at the beginning of the titration

B. To form a stable Mg-EDTA complex that helps in producing a clear endpoint

In the Mohr method for determining chloride, which of the following correctly describes the role of the indicator and its reaction during the titration? A. Methyl orange is added as the indicator, changing from red to yellow at the endpoint. B. A chromate salt is used as the indicator, forming a yellow solution and a red precipitate with excess silver ions after all chloride has reacted. C. A chromate salt is used to form a complex with chloride ions, resulting in a colorless endpoint. D. Phenolphthalein is used as the indicator, producing a pink color in acidic solution.

B. A chromate salt is used as the indicator, forming a yellow solution and a red precipitate with excess silver ions after all chloride has reacted.

Which indicator is commonly used in iodine titrations and forms a dark blue complex with iodine? A. ferroin B. dichlorofluorescein C. methylene blue D. starch

D. starch

06. Volumetric Methods Flashcards

(74 cards)