Determining pH and Titrations Section 15.2

Acid-Base Indicators acid-base indicators: compounds whose colors are sensitive to pH Indicators change colors because they are either weak acids or weak bases

Indicators Indicators come in many different colors There are specific pH ranges over which an indicator changes color Transition interval: the pH range over which an indicator changes color

pH Meter and Paper pH meter: determines the pH of a solution by measuring the voltage between the two electrodes that are placed in the solution pH paper is used by comparing the color the paper turns when it is in contact with the solution

Titration Titration: the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration

Equivalence Point Equivalence point: the point at which the two solutions used in a titration are present in chemically equivalent amounts Indicators and/or pH meters can be used to determine the equivalence point

More about Titrations End point (of the indicator): the point in a titration at which an indicator changes color Indicators that undergo transition at about pH 7 are used to determine the equivalence point of strong acid/strong base titrations

Standard Solution Standard solution: (AKA known solution) the solution that contains the precisely known concentration of a solute See pages 518-519 in the book at school for the procedure for carrying out an acid-base titration (pages 500-501 in your book at home)

Sample Problem for a Titration Suppose that 10.1 mL HNO3 is neutralized by 71.4 mL of a 4.2 x 10-3 M solution of KOH in a titration. Calculate the concentration of the HNO3 solution. Step 1: write and balance the equation. HNO3 + KOH  H2O + KNO3

Problem Continued Step 2: define what you are given in the units required V of KOH = 71.4 mL = 0.0714 L [KOH] = 4.2 x 10-3 M V of HNO3 = 10.1 mL = 0.0101 L [HNO3] = ?

Continued Step 3: find the mol of the known volume Mol = [ KOH] (volume in liters) Mol = 4.2 x 10-3 mol/L KOH  0.0714 L Mol = 2.9988 x 10-4 mol KOH Step 4: find the mol of the unknown (HNO3) 2.9988 x 10-4 mol KOH  1 mol HNO3 / 1 mol KOH 2.9988 x 10-4 mol HNO3

Continued Step 5 calculate [unknown] 2.9988 x 10-4 mol HNO3 / 0.0101 L = 3.0 x 10-2 M HNO3

Calculations See pages 520-521 in the book at school and pages 502-503 in the book at home for sample calculations