Acid Base Reactions HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Acids and bases react to produce a salt and water. Salts are ionic compounds consisting of an anion from an acid and a cation from a base. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

This is special type of double replacement reaction called neutralization

Titration A laboratory method for determining the concentration of an unknown acid or base using a neutralization reaction.

Neutralization occurs when there is an equal number of moles of hydroxide ions and hydronium ions This is called the equivalence point.

Equivalence points are impossible to see so when completing a titration, an acid-base indicator is chosen that will change color at or near the pH of the equivalence point

An indicator is a solution that changes color as the pH changes Shown below are the three indicators at different pH values. Methyl Red Bromothymol Blue Phenolphthalein

Titration Steps Add base with known concentration to the burette. A measured volume of an acid solution of unknown concentration is added to a flask. Several drops of an indicator are added to the solution in the flask. A flask with a known volume of acids (and an indicator) is placed beneath a buret that is filled with a base of known concentration.

Drop by drop add the base of known concentration to the acid until the indicator just barely changes color and stays that color. This is known as the end point. The base is slowly added from the buret to the acid. A change in the color of the solution is the signal that neutralization has occurred.

At the completion of a titration, you have three pieces of data…… 1. Volume of solution added to the flask from the burette to reach the equivalence point/end point 2. Concentration of the solution in the burette 3. Volume of solution in the flask These three numbers can be used to solve for the unknown concentration.

When a neutralization reaction is complete, the following is true…. nBMAVA = nAMBVB MA = molarity of acid MB = molarity of base VA = volume of acid VB = volume of base nA = moles of acid (from balanced equation) nB = moles of base (from balanced equation)

Titration Problem (1:1 ratio) Consider the titration of acetic acid with sodium hydroxide. A 10.0 mL sample of acetic acid requires 37.55 mL of 0.223 M NaOH. What is the concentration of the acetic acid? HC2H3O2(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(l) Using MAVA = MBVB MA = MBVB / VA MA = (0.223 M )(37.55 mL) / 10.0 mL MA = 0.837M

Titration Problem (not 1:1 ratio) A volume of 50.0 mL of 1.20 M H3PO4 neutralizes 45.5 mL of a Ca(OH)2 solution. What is the molarity of the Ca(OH)2 solution? Balanced equation: 2 H3PO4 (aq) + 3Ca(OH)2 (aq) → Ca3(PO4 )2 (aq) + 6H2O (l) nBMAVA = nAMBVB MB = nBMAVA / nAVB MB = 3(1.20 M)(50.0 mL)/2(45.5 mL) MB = 1.98 M