Volumetric and Gravimetric Methods of Analysis

Port Said University - Chemistry Department - Volumetric and Gravimetric Methods of Analysis

• Course: CHE12r2 (2 hours/week)
• For: First Level Students – Faculty of Education (General Chemistry Section- Special Program)
• Prepared By: Prof. Dr. Sayed Nour El Din Moalla (Professor of Analytical Chemistry)
• Academic Year: 2024/2025

Contents

• Chapter 1: Volumetric Titrations
  • 1.1 Classification of Analytical Methods
  • 1.2 Principles of Volumetric Analysis
    • Standard Solutions
    • Requirements of a Good Primary Standard
    • Properties of a Standard Solution
  • 1.3 Classification of Methods of Volumetric Analysis
    • (i) Acid-Base Titrations
    • (ii) Precipitation Titrations
    • (iii) Complexometric Titrations
    • (iv) Reduction – Oxidation (Redox) Titrations
  • 1.4 Review of Fundamental Concepts
    • Atomic, Molecular, and formula Weights
  • 1.5 Express Concentrations of Solutions
  • 1.6 Acid Base Titrations
    • Acid-Base Indicators
    • Theory of Indicator Function
    • Indicator pH Range
    • Titration (Neutralization) Curves
    • Questions and Problems
  • 1.7 Reduction-Oxidation (Redox) Titrations
    • Introduction
    • Detection of the End Point
    • Titration with Potassium Permanganate
    • Potassium Dichromate Titrations
    • Titrations Involving Iodine
  • 1.8 Precipitation Titrations
    • Introduction
    • Detection of the End Point: Indicators
    • Questions and Problems
  • 1.9 Complexometric titrations
    • Chelates: EDTA - The Ultimate Titrating Agent for Metals
    • Detection of the End Point: Metal Indicators
    • Application of Complexometric Titrations
    • Types of Complexometric titrations
    • Questions and Problems
• Chapter 2: Gravimetric Methods of Analysis
  • 2.1 Principle of Gravimetric Analysis
  • 2.2 Types of Gravimetric Methods
    • (i) Precipitation Method
    • (ii) Electrogravimetric Method
    • (iii) Volatilization Method
    • (iv) Particulate Method
  • 2.3 Precipitation Method
    • Weight Relationships
    • How to Perform a Successful Gravimetric Analysis
    • Requirements for Precipitates
    • Requirements for the Weighed form
    • Questions and Problems
• Chapter 3: Practical Applications
• References

CHAPTER 1 Volumetric Titrations

• 1.1 Classification of Analytical Methods

  • Analytical methods are classified as classical (wet chemical) or instrumental.
  • Classical methods preceded instrumental methods.

• 1.2 Principles of Volumetric Analysis

  • Volumetric analysis involves adding a standard solution (titrant) from a burette to the analyte until the reaction is complete.
  • The titrant's volume is carefully measured to calculate the analyte's quantity.
  • Titration:
    • A: titrant
    • B: substance titrated (analyte)
    • a & b: moles of each
    • $aA + bB <br /> ightharpoonup products$
  • Equivalence point: Titrant amount is stoichiometrically equal to the analyte.
  • End point: A sudden change in physical property marks the end point.
  • Titration error: The difference between the end point and the equivalence point.

• Standard Solutions

  • Reagents with exactly known concentrations.
  • Accuracy is limited by the accuracy of the standard solution's concentration.
  • Prepared by dissolving a primary standard (highly pure material) in a volumetric flask.
  • Alternatively, a solution can be standardized against a primary standard if the material isn't a primary standard itself.
  • Standardization: Determining the concentration of a standard solution.

• Requirements of a Good Primary Standard

  • (i) Highest purity.
  • (ii) Stable to drying.
  • (iii) Absence of hydrate water.
  • (iv) Readily available at reasonable cost.
  • (v) Reasonable high formula weight.
  • (vi) Possess the properties required for a titration.

• Properties of a Standard Solution

  • (i) Concentration should remain constant for a long period.
  • (ii) Reaction with analyte should be rapid.
  • (iii) Reaction with the analyte should be reasonably complete.
  • (iv) Reaction should be described by a balanced chemical equation.
  • (v) Method must exist for detecting the equivalence point.

• 1.3 Classification of Methods of Volumetric Analysis

  • Four general classes of titrimetric methods:
    • (i) Acid-Base Titrations
      • Acids or bases titrated with a standard solution of a strong base or acid.
      • End points detected by indicators or pH meters.
      • $OH^- + HA <br /> ightharpoonup A^- + H_2O$ (base titrant + acid titrated)
      • $H^+ + B^- <br /> ightharpoonup BH^+$ (acid titrant + base titrated)
    • (ii) Precipitation Titrations
      • Titrant forms an insoluble product with the analyte.
      • Example: Titration of chloride ion with silver nitrate.
      • $Ag^+ + Cl^- <br /> ightharpoonup AgCl(s)$ (titrant + analyte)
      • End point detected by visual-indicator methods.
    • (iii) Complexometric Titrations
      • Titrant is a complexing agent forming a water-soluble complex with the analyte (metal ion).
      • Often uses a chelating agent like Ethylene di-amine tetraacetic acid (EDTA).
      • Reaction controlled by adjusting pH.
      • Indicators form colored complexes with the metal ion.
    • (iv) Reduction – Oxidation (Redox) Titrations
      • Oxidizing agent titrated with a reducing agent, or vice versa.
      • Oxidizing agent gains electrons; reducing agent loses electrons.
      • Example: Titration of ferrous ion with permanganate ion.
      • $MnO<em>4^- + 5Fe^{2+} + 8H^+ ightharpoonup 5Fe^{3+} + Mn^{2+} + 4H</em>2O$
      • End point marked by a permanent violet color from excess permanganate.
      • End point determined by visual indicators or potentiometric titration.

• 1.4 Review of fundamental Concepts

  • Quantitative analysis relies on atomic and molecular concepts.

• Atomic, Molecular, and formula Weights

  • Gram-atomic weight: Weight of a specified number of atoms of an element.
  • Avogadro’s number: $6.022 × 10^{23}$, atoms in 1 g-at wt. of any element.
  • Molecular weight (Mol. wt): Sum of atomic weights of atoms in a compound.
  • Formula weight (F. wt): Accurate description for ionic compounds (strong electrolytes).
  • Molar mass: Term sometimes used in place of formula weight.

• Mole

  • Atoms and molecules react in definite proportions.
  • Mole: Avogadro’s number ($6.022 × 10^{23}$) of atoms, molecules, ions, or other species.
  • Numerically, atomic, molecular, or formula weight of a substance expressed in grams.
  • Atoms react in the same mole ratio as their atom ratio in the reaction.
  • Example: One silver ion reacts with one chloride ion; each mole of silver ion reacts with one mole of chloride ion.
  • Number of moles:
    • grams of substance / formula weight
  • Number of millimoles:
    • mg of substance / formula weight
  • Grams of material:
    • moles × formula weight
  • Milligrams of material:
    • millimoles × formula weight
  • $g/mol = mg/mmol$, $g/L = mg/mL$, $mol/L = mmol/mL$
  • Example 1.1: Calculate moles in 500 mg $Na<em>2WO</em>4$ (sodium tungstate).
    • Solution: $moles = 500 mg / (293.8 g/mol) = 1.70 mmol$
  • Example 1.2: Weight in milligrams of 0.250 mmol $Fe<em>2O</em>3$ (ferric oxide)?
    • Solution: $0.250 mmol × 159.7 mg/mmol = 39.9 mg$

• 1.5 Express Concentrations of Solutions

  • Chemists use various ways to express solution concentrations.

• Molarity

  • One-molar solution: One mole of substance in each liter of solution.
  • Molarity (M): Moles per liter or millimoles per milliliter.
    • $M = moles/liter = millimoles/milliliter$
  • Example: A one-molar $AgNO_3$ (silver nitrate) solution reacts equally with a one-molar $NaCl$ (sodium chloride) solution.
    • $Ag^+ + Cl^- <br /> ightharpoonup AgCl$
  • Moles of substance in any volume of solution: Molarity × Volume (in liters).
  • Millimoles of substance in any volume of solution: Molarity × Volume (in milliliters).
  • Example 1.3: Solution prepared by dissolving 1.26 g $AgNO<em>3$ in a 250-mL volumetric flask. Calculate the molarity and millimoles of dissolved $AgNO</em>3$.
    • Solution:
      • $Molarity = (1.26 g / 169.9 g/mol) / (0.250 L) = 0.0297 M$
      • $millimoles = (0.0297 mmol/mL) (250 mL) = 7.42 mmol$
  • Example 1.4: Grams per milliliter of $NaCl$ in a 0.250 M solution?
    • Solution:
      • $0.250 mol/L = 0.250 mmol/mL$
      • $0.250 mmol/mL × 58.4 mg/mmol × 0.001 g/mg = 0.0146 g/mL$
  • Example 1.5: Grams of $Na<em>2SO</em>4$ to prepare 500 mL of a 0.100 M solution?
    • Solution:
      • $500 mL × 0.100 mmol/mL = 50.0 mmol$
      • $50.0 mmol × 142 mg/mmol × 0.001 g/mg = 7.10 g$
  • Example 1.6: Concentration of potassium ion in grams per liter after mixing 100 mL of 0.250 M $KCl$ and 200 mL of 0.100 M $K<em>2SO</em>4$.
    • Solution:
      • $mmol K^+ = mmol KCl + 2 × mmol K<em>2SO</em>4$
      • $= 100 mL × 0.250 mmol/mL + 2 × 200 mL × 0.100 mmol/mL = 65.0 mmol in 300 mL$
      • $[K^+] = (65.0 mmol / 300 mL) × (39.1 mg/mmol) × (1 g/1000 mg) × (1000 mL/L) = 8.47 g/L$

• Normality

  • Normality (N): Equivalents per liter.
  • Equivalents = moles × number of reacting units per molecule or atom.
  • Equivalent weight: Formula weight divided by the number of reacting units.
  • Acids and bases: Reacting units = number of protons an acid furnishes or a base reacts with.
  • Redox reactions: Reacting units = number of electrons an oxidizing or reducing agent takes on or supplies.
  • Example: Sulfuric acid ($H<em>2SO</em>4$) has two reacting units (protons); normality is twice its molarity.
  • $N = (g/eq wt)/L$
  • Equivalents: $g / eq wt$
  • Milliequivalents (meq): Typically used instead of equivalents.

• Molality

  • Molality (m): Moles per 1000 g of solvent.
  • Convenient for physicochemical measurements (colligative properties).
  • Does not change with temperature.

• Density Calculations

  • Concentrations of commercial acids and bases often given as percent by weight.
  • Density needed to calculate molarity.
  • Density: Weight per unit volume (g/mL or g/cm³).
  • Specific gravity: Sometimes listed instead of density.
  • Example 1.7: Milliliters of concentrated sulfuric acid (94.0%, density 1.831 g/cm³) to prepare 1 liter of a 0.100 M solution?
    • Solution:
      • Concentrated acid contains 0.940 g $H<em>2SO</em>4$ per gram of solution, and the solution weighs 1.831 g/mL.
      • Density × concentration = $\frac{1.831 g solution}{mL} * \frac{0.940 g H<em>2SO</em>4}{g solution} = \frac{1.72 g H<em>2SO</em>4}{mL}$

• $M = \frac{\frac{1.72 g H<em>2SO</em>4}{mL}}{\frac{98.1 g H<em>2SO</em>4}{Mol}} * \frac{1000 mL}{Liter} = 17.5 mol H<em>2SO</em>4 / L solution$

• Must dilute this solution to prepare 1 liter of a 0.100 M solution.
  * Since mmol = M × mL and mmol dilute acid = mmol concentrated acid, 0.100 M × 1000 mL = 17.5 M × y mL
  * y = 5.71 mL concentrated acid to be diluted to 1000 mL

  • Mass is conserved: $M<em>1V</em>1 = M<em>2V</em>2$
  • Stock × mL stock = M diluted × mL diluted
  • (for concentrated acid) = 10 x % x density / Mol. wt
  • You wish to prepare 500 mL of a 0.100 M $K<em>2Cr</em>2O_7$ solution from a 0.250 M solution. What volume of the 0.250 M solution must be diluted to 500 mL?
    • final × mL final = M original × mL original
    • $0.100 mmol/mL × 500 mL = 0.250 mmol/mL × mL original$
    • mL original = 200 mL
  • Acid-base neutralization: Acids and bases react in equivalent amounts to produce salts and water.
  • Equivalent weight of an acid: Molecular weight divided by its basicity (number of replaceable hydrogen ions).
  • Equivalent weight of a base: Molecular weight divided by the number of replaceable $OH^-$ ions.
  • Normality: Number of gram equivalents in one liter of solution.
  • Strength (S): Number of grams of an acid or base in one liter of solution.
  • Strength (g/L) = N × Eq. wt

• 1.6 Acid Base Titrations

  • Acid-base titration: Gradual neutralization of an acid or base until the equivalent point is reached, indicated by a color change.
  • $N<em>1 × V</em>1 = N<em>2 × V</em>2$ (acid and base)

• Acid-Base Indicators

  • Organic dyes indicating the equivalence point.
  • Equivalent point: Acid or base added is equivalent to the amount being titrated.
  • Specific pH value signals the equivalence point.
  • Indicators change color over a narrow pH range.
  • Color change occurs near the equivalence point with a small indicator error.

• Theory of Indicator Function

  • Ostwald's theory: Acid-base indicator is a weak organic acid with different colors in undissociated and dissociated forms.
  • Equilibrium in aqueous solution: HIn ⇄ In- + H+
  • Acidic medium: Excess $H^+$ shifts equilibrium towards undissociated form (HIn), color of HIn predominates.
  • Alkaline medium: Excess $OH^-$ combines with $H^+$, shifts equilibrium towards dissociated form (In-), color of In- predominates.

• Indicator pH Range

  • Assume indicator is a weak acid (HIn), unionized form is red, and ionized form is blue:

  • $HIn <br /> ightharpoonup H^+ + In^-$

  • unionized form (red) ionized form (blue)

  • $K<em>a (or K</em>{In}) = \frac{[H^+][In^-]}{[HIn]}$

  • $pH = pK_{In} + log \frac{[In^-]}{[HIn]}$

    • pH transition range depends on the ability to detect color changes.
    • Typically only one color is observed if the concentration ratio of the two indicator forms is 10:1. Therefore, most indicators have a transition range of ~ 2 pH units.
    • When only the color of the unionized form is seen, $[In^-]/[HIn] = 1/10$, therefore, $pH = pK<em>{In} + log (0.1) = pK</em>{In} – 1$
    • When only the color of the ionized form is observed, $[In^-]/[HIn] = 10/1$, and $pH = pK<em>{In} + log 10 = pK</em>{In} + 1$

  • Indicator pH range (ΔpH) = $pK_{In} + ± 1$

    • During transition, observed color is a mixture of two colors. Midway in transition, the concentration of the two forms are equal and pH = $pK_{In}$.
    • Therefore, the $pK_{In}$ should be close to the pH of the equivalence point.

• Titration (Neutralization) Curves

  • When neutralizing an acid with an alkali, the pH increases.
  • When neutralizing an alkali with an acid, the pH decreases.
  • Variation in pH isn't linear due to volume changes.
    • Initial pH of a solution can be calculated assuming complete dissociation of strong acids and bases or using Ostwald’s Law for weak acids and bases.
    • At the equivalence point where only salt and water remain relations are used to determine the pH.
    • Titration of Strong acid versus strong bases and weak acids against weak bases, the solution at the equivalence point has pH = 7.
    • Titration of weak acid against strong bases, or strong acids against weak bases have pH > 7, and pH < 7 respectively.

• Strong acid versus strong bases

  • Titration curves show how end points are detected.
  • Titration curve: Plot of pH vs. volume of titrant added.
  • In the case of a strong acid against strong base, both the titrant and analyte are completely ionized.
  • Example: Titration of hydrochloric acid with sodium hydroxide.
    • $H^+ + Cl^- + Na^+ + OH^- <br /> ightharpoonup Na^+ + Cl^- + H_2O$

• Example 1.9

  • Derive a curve for the titration of 50.0 ml of 0.05 M HCl with 0.10 M NaOH.

  • Solution

    • Initial point: only HCl present: mmol of HCl = 50.0 x 0.050 = 2.5 mmol

    • pH = -log (0.05) = 1.30

    • After addition of 10.0 ml of NaOH:

    • HCl + NaOH $H_2O$ + NaCl

    • mmol of NaOH added = 10.0 x 0.10 = 1 mmol

    • HCl remining = 2.5 – 1.0 = 1.5 mmol MHCl = 1.5/(50+10) = 1.5 / 60 = 2.5 x 10-2 M

    • pH = -log (2.5 x 10-2) = 1.60

    • At the equivalence point:

    • First, note the procedure for determining equivalence point volume:

    • V1 x N1 = V2 x N2 acid base

    • VNaOH = (50.0 x 0.050) / 0.10 = 25.0 ml

    • Next, remember the equilibrium equation

    • pH + pOH = 14.0Since at equivalence point, pH = pOH

    • Then pH = 7.0 at equivalence point.

    • After addition of 25.10 ml of NaOH:

    • mmol of NaOH = 25.1 x 0.10 = 5.51 mmol added

    • The excess of NaOH = 2.51– 2.50 = 0.01 mmol

    • MNaOH = 0.01 / 75.1 = 1.33 x 10-4 M

    • pOH = - log (1.33 x 10-4) = 3.88

    • pH = 14.0 – 3.88 = 10.12

• Example 1.10

  • Calculate the pH at 0, 10, 90, 100, and 110% titration (% of the equivalence point volume) for the titration of 50.0 mL of 0.100 M HCl with 0.100 M NaOH.
  • Solution
    • At 0 %: pH = - log 0.100 = 1.00
    • At 10%: 5.0 ml NaOH is added. We start with 0.100 M x 50.0 ml = 5.00 mmol $H^+$.
    • Calculate the concentration of $H^+$ after adding the NaOH:
    • mmol at start = 5.00 mmol $H^+$.
    • mmol OH^-$ added = 0.100 M x 5.00 = 0.500 mmol OH-$.
    • mmol H+ left 4.5 mmol H+ (n 55.0 ml)
    • Then: [H^+] = 4.50 mmol/55.0 ml = 0.0818 M.$</li> <li>pH = - log (0.0818) = 1.09.</li> <li>At 90$H^+$</li> <li>mmol OH- added = 0.100 M x 45.0 ml = 4.50 mmol OH- .</li> <li>mmol H+ left = 0.50 mmol H+ (in 95.0 ml).</li> <li>$[H^+] = 0.50 mmol / 95.0 ml = 0.00526 M.$</li> <li>pH = -log (= 0.00526) = 2.28.</li> <li>At 100<li>all the$H^+$has been reacted with OH- , and we have a 0.0500 M solution of NaCl.</li> <li>Therefore, the pH is 7.00 (due to$H_2O$dissociation).</li> <li>At 110<li>we now have a solution consisting of NaCl and excess added NaOH.</li> <li>Mmol OH- = 0.100 x 5.00 ml = 0.50 mmol OH- in 105 ml.</li> <li>$[OH^-] = 0.50 mmol/105 ml = 0.00476 M.$</li> <li>pOH = -log (0.00476) = 2.32</li> <li>pH = 14.0 – 2.32 = 11.68.<br /> *Before the equivalence point, when there is excess acid, the relationship is$[H^+] = (M{acid} × V{acid} - M{base} × V{base})/V_{total}$, where V is the volume.</li></ul></li> <li>Likewise, beyond the equivalence point when there is excess base,$[OH^-] = (M{base} × V{base} - M{acid} × V{acid})/V_{total}$.</li> <li>Note that$V{total}$is always$V{acid} + V_{base}$.</li></ul></li> <li><p>Weak Acid versus Strong Base</p> <ul> <li>The neutralization reaction is HOAc + Na+ + OH- →$H_2O$+ Na+ + OAc-</li> <li>To convert [H+] to pH we take logarithms of the above equation and reverse the signs. We then have: - Log [H] = - ½ log$Ka$– ½ log$Ca$<ul> <li>or pH = ½$pKa$- ½ log$Ca$</li></ul></li> <li>As soon as the titration is started, some of HOAc is converted to NaOAc, and buffer system is setup.</li> <li>As the titration proceeds, the pH slowly increases as the ratio [OAc-] / [HOAc] changes.</li> <li>At the midpoint of the titration, [OAc-] = [HOAc], and pH is equal to$pK_a$.</li> <li>At the equivalence point, we have a solution of NaOAc. Since this is a Bronsted base (it hydrolyzes), the pH at the equivalence point will be alkaline. The pH will depend on the concentration of NaOAc. The greater the concentration, the higher the pH.</li> <li>As excess NaOH is added beyond the equivalence point, the ionization of the base OAc- is suppressed to a negligible amount, and the pH is determined only by the concentration of excess OH- . Therefore, the titration curve beyond the equivalence point follows that for the titration of a strong acid.</li></ul></li> <li><p>Example 1.11</p> <ul> <li>Calculate the pH at 0, 10.0, 25.0, 50.0-, and 60.0-mL titrant in the titration of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH.</li> <li>Solution: At 0 mL:</li></ul></li> <li><p>HA ⇌ H+ + A-<br /> * 0.1-x M x M x M<br /> *$Ka = x^2 / (0.1 - x) = [H^+]^2 / C{HA}$<br /> *$[H^+] = \sqrt{Ka C{HA}} = \sqrt{1.75 x 10^{-5} x 0.100} = 1.32 x 10^{-3} M$<br /> *$pH = -log (1.32 x 10^{-3}) = 2.88$</p> <ul> <li>At 10.0 mL: We started with 0.100 M × 50.0 mL = 5.00 mmol HOAc; part has reacted with OH- and has been converted to OAc- :<ul> <li>We have a buffer. Since volumes cancel, use millimoles:</li> <li>$pH = pK_a + log ([A^-] / [HA])$</li> <li>We started with 5.00 mmol HA; part has reacted with OHand ahs been converted to A-::</li> <li>$mmol HA = 5.00 – 1.00 = 4.00 mmol$,$ mmol A^- = 1.00 mmpl (V= 60 ml)$</li> <li>$pH = 4.76 + log (1.00 / 4.00) = 4.76 - 0.60 = 4.16$</li></ul></li> <li>At 25.0 mL: One-half the HOAc has been converted to OAc-, so pH = pKa:<ul> <li>pH = 4.76</li></ul></li> <li>At 50.0 mL: All the HOAc has been converted to OAc- (5.00 mmol in 100 mL, or 0.0500 M):<ul> <li>$OAc^- + H_2O \rightleftharpoons HOAc + OH^-$$
    • Hydrolysis Equation:
    • Kb = \[HOAc].\[OH-\] /\[OAc-\]

= X.X/(C-X) = X2 / C.

Where, Kb= Kw / Ka

Or Kb. KA = KW

[ H+ ][OH-] = 10-14 at 25°C. Hence KB: KB = KW/ KA = 10-14 / 1.75 X 10-5 = 5.7 X 10-10.

Then, X2 / 0.05m= 5.7 X 10-10 ⇒ x= 5.3 X10-6mol / L=\[OH-\]then pOH = 5.3 ⇒pH 14-5.3= 8.7
* All OAc , has been converted To 0.025 mmol/ ml
* KB= Kw/Ka=
* Kb of acetate = 5.71 10-10 M
* x2= Kh (Concentration of acetate
acetate Conc.= 510-2
x= √5.71 x10-10 X 5 x 10 -2
x[ OH-\]= 1.69 x 10(pOH= 5.774
pH=(14-5-774)= 8.23

     mmol (V= 60 ml) \

• We added 10 ml of base so Conc. of A(OH).= 1.69 x 10( mmol conc. so X = 1.29 mmel X100 X10
  X 2O=KaKb = 10-14\[OH-\][H] -= -[A-\] /\[OH-]-
• At 60.0 mL: We have a solution of NaOAc, and excess added NaOH. The hydrolysis of the acetate is negligible in the presence of added OH-. So, the pH is determined by the concentration of excess OH-:
  * Excess OAc = 0.100 mol/L *0.010 L
  * 1. 4x10 M Naoh X= 1.29-x1-4 x 10 M(1.4 x 10 OH17.
  * [HA)=3-X=3X 2003M =273x X2(1 69 X 10-4 ) -X4 13pH=-LOH10. 16= 2 7328 -I + [1-9-3 - =11.22 + og -2 + log = 11.745 3 74 / -+ [OH124/1 loggOH =poh a B: K - 2355 +1 OH = 2 355 407 46
• • The slowly rising region before the equivalence point is called the buffer region.
    *It is flattest at the midpoint, that is, where the ratio [OAc-] / [HOAc] is unity and so the buffer intensity is greatest at a pH corresponding to pKa.
    *Note:
    *Strong acids are actually good buffers, except their pH changes with dilution.
    *Weak-acid titrations require careful selection of the indicator.
• Questions and Problems
  • 1- What is the equivalence point and the end point of a titration?
  • 2- What is a standard solution? How is it prepared?
  • 3- What are the requirements of a primary standard?
  • 4- Why should a primary standard have a high formula weight?
  • 5- What is the minimum pH change required for a sharp indicator color change at the end point and why?
  • 7- Is the pH at the equivalence point for the titration of a weak acid with strong base neutral, alkaline, or acidic? Why?
  • 8- What would be a suitable indicator for the titration of ammonia with HCl acid and of acetic acid with sodium hydroxide?
  • 9- What is principle of volumetric analysis?
  • 10- What are the requirements of a good primary standard?
  • 11- What are the properties of a standard solution?
  • 12- Write a Henderson equation for a weak acid indicator, and calculation required pH-change to go from one color of the indicator to the other. Around what Ph is the transition?
  • 13- Calculate the pH at 0.0, 10.0, 25.0, and 30.0 mL of titrant in the titration of 50.0 mL of 0.100 M NaOH with 0.200M HCl.
  • 14- Calculate the pH at 0.0, 10.0, 25.0, 50.0, and 60.0 mL