Chapter 14 Acids and Bases 14.8 Acid−Base Titration Learning Goal Calculate the molarity or volume of an acid or a base from titration information. © 2014 Pearson Education, Inc.

is a laboratory procedure used to determine the molarity of an acid Acid−Base Titration Titration is a laboratory procedure used to determine the molarity of an acid uses a base such as NaOH in the buret to neutralize an acid A measured amount of acid is placed in a flask with an indicator. © 2014 Pearson Education, Inc.

is added to the acid in the flask Acid−Base Titration An indicator is added to the acid in the flask changes the color of the solution when the acid is neutralized The acid solution briefly turns pink when the base is added. The pink quickly disappears once the solution is mixed. © 2014 Pearson Education, Inc.

the indicator has a permanent pink color Acid−Base Titration At the endpoint, the indicator has a permanent pink color the volume of the base used is measured the molarity of the acid is calculated using the neutralization equation for the reaction The acid solution turns a permanent pink when the endpoint is reached. © 2014 Pearson Education, Inc.

Guide to Calculations for Acid−Base Titrations © 2014 Pearson Education, Inc.

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Learning Check What is the molarity of 10.0 mL of HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize the HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) © 2014 Pearson Education, Inc.

Step 1 State the given and needed quantities and concentrations. Solution What is the molarity of 10.0 mL of HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize the HCl? Step 1 State the given and needed quantities and concentrations. Given Need 10.0 mL (0.010 L) of HCl solution Molarity of HCl solution 18.5 mL (0.0185 L) of 0.225 M NaOH solution HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) © 2014 Pearson Education, Inc.

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Solution What is the molarity of 10.0 mL of HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize the HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Step 2 Write a plan to calculate molarity or volume. L of NaOH Molarity moles of NaOH Mole-Mole factor moles of HCl Divide by liters molarity of HCl © 2014 Pearson Education, Inc.

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Solution What is the molarity of 10.0 mL of HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize the HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Step 3 State equalities and conversion factors, including concentrations. © 2014 Pearson Education, Inc.

Solution What is the molarity of 10.0 mL of HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize the HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Step 4 Set up the problem to calculate the needed quantity. © 2014 Pearson Education, Inc.

Learning Check Calculate the milliliters of 2.00 M H2SO4 needed to completely neutralize 0.0500 L of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) © 2014 Pearson Education, Inc.

Solution Calculate the milliliters of 2.00 M H2SO4 needed to completely neutralize 0.0500 L of 1.00 M KOH. Step 1 State the given and needed quantities and concentrations. Given Need 2.00 M HCl solution Volume of HCl solution 0.0500 L of 1.00 M KOH solution H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) © 2014 Pearson Education, Inc.

H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Solution Calculate the milliliters of 2.00 M H2SO4 needed to completely neutralize 0.0500 L of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Step 2 Write a plan to calculate molarity or volume. L of KOH Molarity Mole-Mole factor moles of KOH moles of H2SO4 Molarity volume of H2SO4 © 2014 Pearson Education, Inc.

H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Solution Calculate the milliliters of 2.00 M H2SO4 needed to completely neutralize 0.0500 L of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Step 3 State equalities and conversion factors, including concentrations. © 2014 Pearson Education, Inc.

H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Solution Calculate the milliliters of 2.00 M H2SO4 needed to completely neutralize 0.0500 L of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Step 3 State equalities and conversion factors, including concentrations. © 2014 Pearson Education, Inc.

H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Solution Calculate the milliliters of 2.00 M H2SO4 needed to completely neutralize 0.0500 L of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Step 4 Set up the problem to calculate the needed quantity. © 2014 Pearson Education, Inc.