CHAPTER 6 TITRIMETRIC METHODS OF ANALYSIS Titrimetric methods

CHAPTER 6 TITRIMETRIC METHODS OF ANALYSIS

Titrimetric methods include a large and powerful group of quantitative procedures based on measuring the amount of a reagent of known concentration (standard solutiontitrant) that is consumed by the analyte. Titrimetry is a term which includes a group of analytical methods based on determining the quantity of a reagent of known concentration that is required to react completely with the analyte.

Basic principle a. A +b. B → Products A: Titrant (standard solution) B: Analyte (unknown ) a and b are number of moles of each

Requirments in. Titration 1 - stoichiometric reaction 2 - rapid rate reaction 3 - quantitative reaction (99. 9%complete at stoichiometry) 4 -have a defined end or equivalence point

Equivalence and end Points Equivalence point: Related to the amounts of reactants consumed. End point Related to the physical sign that associate with the condition of chemical equivalence. Ideally, the end point and equivalent one coincide. (variation may be to color change of indicator)

Standard solutions Standard Solution - A primary standard - A secondary standard Standardization A process in which concentration of a volumetric solution is determined by titrating it with a known mass of a primary standard.

§ Titration: This is performed by adding a standard solution from a buret or other liquid- dispensing device to a solution of the analyte until the point at which the reaction is believed to be complete. Solution of Na. OH Solution of HCl 5 m. L

PRIMARY STANDARD • • • HIGH PURITY ATMOSPHERIC STABILITY INDEPENDENT OF HUMIDITY MODEST COST LARGE MOLAR MASS NON HYROSCOPIC

Types of titrimetry 1 - Precipitation: Ag+ + Cl- → Ag. Cl (s) 2 - Acid Base titration: OH- + HA → A- + H 2 O 3 - Complex formation EDTA 2 - + Ca 2+ → EDTA Ca. x. H 2 O 4 - Oxidation Reduction Mn. O 4 - + 5 Fe 2+ + 8 H+ → 5 Fe 3+ + Mn 2++4 H 2 O

Calculations with Molarity The mole: is the formula weight of a substance expressed in grams Number of moles = weight in Grams/formula weight Number of m. moles = weight in milligrams/formula weight Molarity= Moles/Liter or mmoles/ml

Moles= Volume (L). (M) Weight (g) = Volume (L). (M). form weight (g) m. Moles= Volume (ml). (M) Weight (mg) = Volume (ml). (M). form weight (g)

Calculating the results of Titration Needs: the volume and molarity of the titrant a. A + b. B → (Titrant) Products (substance titrated) No. of m. moles of A = Volume of A titrated (ml). Molarity of A (M) Or mmoles. A = ml. A. (MA) No. of m. moles of B is obtained by multiplying the No. of mmoles of A by the combining ratio (b/a)

mmoles. B = mmoles. A. (b/a) = ml. A. (MA). (b/a)

If the weight of substance is needed: from no. of mmoles Wt (mg) = mmoles. Form wt (g) mg. B = (ml. A). (MA). (b/a). (form Wt. B) If % of B is needed in a sample = (Wt. B/Wtsample). 100 The same is B% =(mg. B/mg. S). 100 B% = (ml. A). (MA). (b/a). (form Wt. B)(100)/mg. S

Calculating the molarity of a solution from a standardizing titration This means that from the weight of primary standard substance (B )when dissolved and titrated with other solution, the molarity of the solution (A) can be calculated as follows: a. A + b. B → (Solution. A) Products (primary standard) (ml. A) (MA)/a = (ml. B) (MB)/b (ml. B) (MB) = mmoles. B = mg. B /Form Wt. B (MA)/a = mg. B/ (ml. A) (b/a) (Form Wt. B)

Exercises Excercize 1 How many ml of 0. 25 M Na. OH will react with 10. 0 ml of 0. 10 M H 2 SO 4.

Solution: H 2 SO 4 + 2 Na. OH g Na 2 SO 4 + 2 H 2 O mmol Na. OH = 2 mmol H 2 SO 4 (½) x MNa. OH x Vm. L(Na. OH) = M(H 2 SO 4) x Vm. L(H 2 SO 4) 0. 25 x Vm. L = 2 x 0. 10 x 10 Vm. L = 8 ml

Exercise 2: 0. 4671 g sample containing Na. HCO 3 (FW = 84. 01 mg/mmol) was dissolved and titrated with 0. 1067 M HCl requiring 40. 72 m. L. Find the percentage of sodium bicarbonate in the sample.

Solution write the equation in order to identify the stoichiometry Na. HCO 3 + HCl g Na. Cl + H 2 CO 3 Now it is clear that the number of mmol of sodium bicarbonate is equal to the number of mmol HCl from the equation: M 1. V 1 (HCl)=M 2. V 2 (Na. HCO 3) mmol Na. HCO 3 = mmol HCl mmol = M x Vm. L mmol Na. HCO 3 = (0. 1067 mmol/ml ) x 40. 72 m. L = 4. 345 mmol Now get mg bicarbonate by multiplying mmol times FW mg Na. HCO 3 = 4. 345 mmol x (84. 01 mg/mmol) = 365. 01 % Na. HCO 3 = (365. 01 x 10 -3 g/0. 4671 g) x 100 = 78. 14%

Exercize 3 A 0. 1876 g of pure sodium carbonate (FW = 106 mg/mmol) was titrated with approximately 0. 1 M HCl requiring 35. 86 ml. Find the molarity of HCl. Solution The first thing writing the equation of the reaction. Na 2 CO 3 + 2 HCl g 2 Na. Cl + H 2 CO 3

from the equation 2 mmol HCl react with 1 mmol carbonate mmol HCl = 2 mmol Na 2 CO 3 or MHCl X Vml 1/2 = mmol Na 2 CO 3 substitute for mmol carbonate by mg carbonate/FW carbonate. This gives MHCl x 35. 86 = 2 x 187. 6 mg/ (106 mg/mmol) MHCl = 0. 09872 M

Back-Titrations In back titration an accurately known amount of a reagent is added to analyte in such a way that some excess of the added reagent is left. This excess is then titrated to determine its amount. • a. A + b. B → Products + A(excess) (Titrant 1) (analyte) • Then back titration of A(excess) with Titrant 2 C • c. C + d. A → Products • (Back titration) (excess) Titrant 2 • •

mmol reagent(Titrant 1) taken = mmol reagent(net) reacted with analyte + mmol reagent(excess) titrated with Titrant 2 then the analyte can be determined since mmol reagent added are known, and mmol reagent titrated can be calculated from back titration. A added = A net + A excess mmol reagent taken = mmol reagent net reacted + mmol reagent back titrated mmol reagent net reacted = mmol reagent taken – mmol reagent back titrated

Finally, the number of mmol reagent net reacted with analyte can be related to the number of mmol analyte from the stoichiometry of the reaction between the two substances as normal titration up on the equation. a. A + b. B → Products

. Now net calculations will be as follows: • a. A + b. B → Products + A(excess) (Titrant 1) (analyte) • c. C + d. A → Products (Back • titration) (excess) Titrant 2 • Net reacted

Importance of back titration: Back-titrations are important especially in some situations like: 1. When the rate of reaction is slow. 2. When the titration reaction doesn't have a good end point.

Example:

Example:

Exercise 0. 1195 g of Na 2 CO 3 was mixed in 30 m. L of boiled water. Next, 40 m. L of HCl solution was added turning the solution colorless to indicate the presence of excess acid. It then required 17. 51 ml of. 1096 M Na. OH to back titrate the excess hydrogen ion in solution to faint pink phenolphthalein endpoint. Calculate the concentration of the unknown HCl solution.

First work out how many excess moles of HCl that were neutralized by the Na. OH HCl + Na. OH ------> Na. Cl + H 2 O 1 mole HCl needs 1 mole Na. OH to be neutralized, a 1: 1 ratio So moles HCl neutralized = moles Na. OH added moles Na. OH = Molarity x Litres = 0. 1096 M x 0. 01751 L = 0. 001919096 moles HCl were in excess.

Then we need moles HCl needed to react all the Na 2 CO 3 + 2 HCl ------> 2 Na. Cl + CO 2 + H 2 O 1 mole Na 2 CO 3 needs 2 moles HCl in order to fully react. moles Na 2 CO 3 = Wt / FWt = 0. 1195 g / 105. 9884 g/mol = 0. 00112748 moles Na 2 CO 3 So moles HCl used up by Na 2 CO 3 = 2 x moles Na 2 CO 3 = 2 x 0. 00112748 moles= 0. 00225496 moles

total moles HCl in solution = 0. 00225496 + 0. 001919096 = 0. 004174 moles HCl Molarity = moles / L = 0. 004174 / 0. 040 L = 0. 10435 M HCl

Calculations with Normality: The mole: is the formula weight of a substance expressed in grams The Normal or equivalent: is equivalent weight of a substance expressed in grams Normality (N) = Number of equivalents per liter = eq/liters or meq/ml Number of equivalents = weight in Grams/equivalent weight

Equivalents = Volume (L). (N) Weight (g) = Volume (L). (N). eq. weight (g) m. eq= Volume (ml). (N) Weight (mg) = Volume (ml). (N). eq. weight (g)

Calculating the results of Titration Needs: the volume and Normality of the titrant a. A + b. B → (Titrant) Products (substance titrated) Using Normality in calculations of titrations, never mind a equal b or not, at equivalence in titrations: No. of eq. of A = No. of eq. of B NA. VA = NB. VB

No. of meq. of A = No. of meq. of B NA. ml. A = NB. ml. B

How to calculate the equivalent weight: The definition of an equivalent in terms of molarity depends on the reaction a substance undergoes. 1 - In acid base-reactions, one equivalent is the number of grams of a substance that supplies or combines with a hydrogen atom. 2 - In oxidation-reduction, one equivalent is the number of grams of a substance that supplies or combines with an electron. 3 - In complex formation, one equivalent is the number of grams of a substance that supplies or react with one ion.

Examples: 1 eq of HCl = 1 mole = 35. 5 g of HCl 1 eq of H 2 SO 4 = 1/2 mole = 98/2 = 49 g of H 2 SO 4 1 eq of Fe 2+ = 1 mole (Fe 2+ → Fe 3+ + e) = 55. 85 g 1 eq of KMn. O 4 = 1/5 mole, (KMn. O 4 + 5 e+. . → Mn 2+ +…) = 158. 04/5= 31. 6 g Its seen that the equivalent weight of a substance is related to its formula weight: eq Wt of HCl = form Wt / 1=35. 5/1=35. 5 g/eq eq Wt of H 2 SO 4= form Wt/=98/2=49 g/eq eq Wt of Fe 2+=form Wt/5= 55. 85/1=55. 85 g/eq eq of KMn. O 4 =158. 04/5=31. 6 g/eq

Now we can compute generalrules for calculating the equivalent Wt from the molecular Wt: 1 - in acid base reactions or substances: eq Wt= form Wt/number of H+ replaced or denoted: i. e H 2 SO 4→ 2 H+ + SO 42 -, eq Wt of H 2 SO 4= form Wt/=98/2=49 g/eq H 3 PO 4→ 3 H+ + PO 43 -, eq Wt of H 3 PO 4 = 98/3=32. 67 g/eq

2 - Precipitate-formation and complex formation titrations: eq. Wt of a metal ion= form Wt/chagre on the ion Eq Wt of an anion = form Wt/no of metal-ion equivalents it reacts with. Examples:

3 - Redox Reactions: eq Wt = form Wt/number of electrons gained or lost in the reaction. i. e

Examples and exercises: 1 - HCl was standardized with 0. 095 N solution of Na. OH. If 25 ml of HCl required 32. 2 ml of Na. OH solution. What is the concentration of HCl.

2 - A 200 g sample of a metal alloy is dissolved and then the Tin is reduced to Tin (II). Titration of the Tin (II) required 22. 20 ml of o. 1 N K 2 Cr 2 O 7. Calculate the percentage of Tin in the alloy.

3 - A 1. 0 g sample required 28. 16 ml of 0. 1 N Na. OH for titration to a phenolphthalein-indicator. Calculate the percentage of phosphoric acid (H 3 PO 4) in the sample.

4 - A 150 mg sample of pure Na 2 CO 3 requires 30. 06 ml of HCl solution for titration: 2 H+ + CO 32 - → CO 2 (g) + H 2 O 2 HCl + Na 2 CO 3 → CO 2 (g) + H 2 O+ 2 Na. Cl Calculate the normality of HCl.