Neutralization and Titrations

NEUTRALIZATION The reaction of an acid with a base is called a “neutralization” reaction. There are two types that are emphasized here. TYPE 1. Acid + Basic Hydroxide 6 H2 O + Salt (ionic compound) EXAMPLES (balance these equations:) TYPE 2. HCl + NaOH 6 H2O + NaCl HCl + Ca(OH)2 6 H2O + CaCl2 H2SO4 + KOH 6 H2O + K2SO4 H2SO4 + Ba(OH)2 6 H2O + BaSO4 H3PO4 + NaOH 6 H2O + Na3PO4 H3PO4 + Ca(OH)2 6 H2O + Ca3(PO4)2 Acid + Ammonia 6 Ammonium Salt (ionic compound) EXAMPLES (balance these equations:) HCl + NH3 6 NH4Cl H2SO4 + NH3 6 (NH4)2SO4 H3PO4 + NH3 6 (NH4)3PO4 TITRATION Titration is a process of determining how much of a solution of known concentration reacts with a given volume of solution of unknown concentration. The solution of known concentration is delivered from a buret and the needed volume measured on this device. The solution of unknown concentration is accurately pipeted into an Erlenmeyer flask. The end point of the titration can be defined in two ways. The equivalenct point, or “theoretical end point”, is that volume of titrant giving the exact moles as dictated by the balanced chemical reaction with the substance being titrated. The experimental end point of a titration, however, is that volume of titrant added in slight excess over the other reactant. The excess titrant causes a color change of the solution in the presence of an indicator dye. A common indicator dye (Ind) is a large organic molecule called phenolphthalein which itself is an acid. This dye is colorless in acid solution when only its conjugate acid is present; however, the dye becomes pink in basic solution when only its conjugate base is present: Ind.H+ conjugate acid colorless + NaOH W Ind conjugate base pink + H2O SOLVING TITRATION PROBLEMS Problem: What is the molarity of a solution of HCl if 20.00 mL of this solution requires 17.56 mL of 0.1258 M NaOH for neutralization? Steps for solving titration problems 1. Write balanced equation: HCl + NaOH 6 H2O + NaCl 2. Calculate moles of the titrant: 17.56 mL x 1 L x 0.1258 M = 2.2090 x 10-3 mol NaOH 1000 mL (Carry extra sig fig here) 3. Calculate moles of substance being titrated 2.2090 x 10-3 mol NaOH x 1 mol HCl = 2.2090x10-3 mol HCl 1 mol NaOH 4. Calculate molarity (M) of substance being titrated 2.2090 x 10-3 mol HCl x 1 x 1000 mL = 0.1104 M HCl 20.00 mL 1 L