HUE Clinical 2024 - Acid-Base Titration - Part 1 - Lec. 3-1 PDF

Pharmaceutical Analytical Chemistry I (PC 111) Level I Students (Clinical Pharm D) Acid-Base Titrations (Lecture 3) Acid-Base Titrations Remember 2  Acid-Base Titrations Equivalent weight is calculated according to the...

Pharmaceutical Analytical Chemistry I (PC 111) Level I Students (Clinical Pharm D) Acid-Base Titrations (Lecture 3) Acid-Base Titrations Remember 2  Acid-Base Titrations Equivalent weight is calculated according to the type of reaction ❑ In neutralization reactions for example: Equiv. wt. = M. wt. / n (n: no. of replaceable H+ or OH-) Acids and bases with single H+ or OH- ion: Equiv. wt.= M. wt. When the acid has two reacting H+ & these two protons involved in the reaction: Equiv. wt. = ½ M.wt. 3  Acid-Base Titrations Calculation of Equivalent weight 1- In case of Carbonic acid: H2CO3 CO2 + H2O Equiv. wt. = Mol. wt/2 H2CO3 HCO3 Equiv. wt. = Mol. wt/1 [only one hydrogen is involved in the reaction] 2- In case of Phosphoric acid: H3PO4 + OH- H2PO4- + H2O Equiv. wt. = Mol. wt/1 [only one hydrogen is involved in the reaction] H3PO4 + 2 OH- HPO42- + 2H2O Equiv. wt. = Mol. wt./2 4  Acid-Base Titrations Advantage of using normal concentrations: The main advantage of using normal concentration over molar ones is the simple 1:1 ratio involved in all reactions of the same type & that facilitates our calculations. Example: They are all equivalent in their reactive powers because they 1 equiv. wt. of NaOH ≡ 1 equiv. wt. of HCl are all referred to a common ≡ 1 equiv. wt. of H2SO4 standard [Hydrogen]. ≡ 1 equiv. wt. of H3PO4 ≡ 1 mole of H+ In other words, solutions of the same normality react in 1:1 ratio. Example: 1 L of 1 N NaOH ≡ 1 L of 1 N HCl ≡ 1 L of 1 N H2SO4 ≡ 1 L of 1 N H3PO4 5 ml of 0.2 N NaOH ≡ 5 ml of 0.2 N HCl ≡ 5 ml of 0.2 N H2SO4 ≡ 5 ml of 0.2 N H3PO4 Note: “=“ means equal (‫)يساوي‬ while “≡” means equivalent (‫)يكافئ أو يتفاعل مع‬ 5  Acid-Base Titrations REMEMBER Relation between Mole & Equivalent Weight and between Molarity & Normality of the same solution 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝑾𝒆𝒊𝒈𝒉𝒕 Equivalent Weight = 𝑵𝒖𝒎𝒃𝒆𝒓(𝒏) The number differs according to the reaction type. Normality = Molarity x Number (n) The number differs according to the reaction type. Example: For H2SO4 in acid-base reactions 1 Mole - Equiv. Weight of H2SO4 = Mol. Weight of H2SO4 / 2 H2SO4 -Normality of H2SO4 solution = Molarity x 2  2 Equiv. (Normality of 1 M H2SO4 = Molarity x 2 = 1 x 2 = 2 N) weights 6  Acid-Base Titrations Relation between concentrations of two reacting solutions OR How we calculate the concentration of the sample from the concentration of the standard? First Method (the method used in pharm. quality control): 1- Calculation of Equivalence Factor (F) {1 ml of standard  ? g of sample} 2- Calculation of Concentration in w/v {g/L, mg/L, g%, mg%} {for more details refer to the practical course} 7  Acid-Base Titrations Relation between concentrations of two reacting solutions OR How we calculate the concentration of the sample from the concentration of the standard? Second Method: 1-If we use normal concentrations: e.g. H2SO4 + 2 NaOH → Na2SO4 + 2 H2O N1 V1 = N2 V2 N1 V1 = N2 V2 2-If we use molar concentrations: H2SO4 NaOH M1 V1 M2 V2 = 𝒏𝟏 𝒏𝟐 M1 V 1 M2 V2 = 𝟏 𝟐 n1 and n2 are the number of moles in a balanced chemical equations. 8  Acid-Base Titrations ◼ Problems: 1. 30mL of 0.1 M NaOH neutralize 25mL of hydrochloric acid. Determine the concentration of the acid. Solution 1- Write the balanced chemical equation for the reaction NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l) 2- Calculate no. of moles of NaOH no. of moles(NaOH) = M x V = 0.1 x (30 x 10-3) = 3 x 10-3 moles 3. Find no. of moles of HCl -----> NaOH: HCl is 1:1 (from the balanced chemical equation) So, no. of moles(NaOH) = no. of moles(HCl) = 3 x 10-3 moles at the equivalence. 4- Calculate conc. (M) of HCl: M = no. of moles ÷ V = 3 x 10-3 ÷ (25 mL x 10-3) = 0.12 M 9  Acid-Base Titrations ◼ Problems 2: 50mL of 0.2 M NaOH neutralize 20mL of sulfuric acid. Determine the concentration of the acid. Solution 1- Write the balanced chemical equation for the reaction 2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l) 2- Calculate No. of moles of NaOH no. of moles (NaOH) = M x V = 0.2 x (50 x 10-3) = 0.01 mol. 3- Find No. of moles of H2SO4 (NaOH: H2SO4) is (2:1) (from the balanced chemical equation) So, n(H2SO4) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10-3 moles H2SO4 at the equivalence. 4- Calculate concentration of H2SO4: M = n ÷ V = 5 x 10-3 ÷ (20 mL x 10-3) = 0.25 M 10  Acid-Base Titrations ◼ Activity: 1. 25 mL of 0.05 M Ba(OH)2 neutralize 40 mL of nitric acid. Determine the concentration of the acid solution. [0.0625 M] 11  Acid-Base Titrations Thank You 12