11.8 Acid–Base Titration The titration of an acid. A known volume of an acid is placed in a flask with an indicator and titrated with a measured volume of a base solution, such as NaOH, to the neutralization endpoint. Learning Goal Calculate the molarity or volume of an acid or base from titration information.

Acid–Base Titration Titration is a laboratory procedure used to determine the molarity of an acid. uses a base such as NaOH to neutralize a measured volume of an acid. requires a few drops of an indicator such as phenolphthalein to identify the endpoint. Core Chemistry Skill Calculating Molarity or Volume of an Acid or Base in a Titration

Acid–Base Titration In the following titration, a specific volume of acidic solution is titrated to the endpoint with a known concentration of NaOH. Base  NaOH Acid  Solution

Indicator The indicator phenolphthalein is added to identify the endpoint. turns pink when the solution is neutralized.

Endpoint of Titration At the endpoint of the titration, the moles of base are equal to the moles of acid in the solution. the concentration of the base is known. the volume of the base used to reach the endpoint is measured. the molarity of the acid is calculated using the neutralization equation for the reaction.

Guide to Calculating Boiling Point Elevation, Freezing Point Lowering General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc.

Acid–Base Titration Calculations What is the molarity of an HCl solution if 18.5 mL of 0.225 M NaOH is required to neutralize 0.0100 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 1 State given and needed quantities and concentrations. ANALYZE Given Need Equation THE 18.5 mL of molarity of HCl(aq) + NaOH(aq)  PROBLEM 0.225 M NaOH HCl solution NaCl(aq) + H2O(l) 0.0100 L HCl

Acid–Base Titration Calculations What is the molarity of an HCl solution if 18.5 mL of 0.225 M NaOH is required to neutralize 0.0100 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 2 Write a plan to calculate molarity or volume. mL NaOH solution Metric factor L NaOH solution Molarity moles of NaOH moles of NaOH Mole–Mole factor moles of HCl Divide by liters molarity of HCl solution

Acid–Base Titration Calculations What is the molarity of an HCl solution if 18.5 mL of 0.225 M NaOH is required to neutralize 0.0100 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 3 State equalities and conversion factors including concentration.

Acid–Base Titration Calculations What is the molarity of an HCl solution if 18.5 mL of 0.225 M NaOH is required to neutralize 0.0100 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 4 Set up the problem to calculate the needed quantity.

Study Check What is the molarity of an HCl solution if 25.5 mL of 0.438 M NaOH is required to neutralize 0.0250 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Solution What is the molarity of an HCl solution if 25.5 mL of 0.438 M NaOH is required to neutralize 0.0250 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 1 State given and needed quantities and concentrations. ANALYZE Given Need Equation THE 25.5 mL of molarity of HCl(aq) + NaOH(aq)  PROBLEM 0.438 M NaOH HCl solution NaCl(aq) + H2O(l) 0.0250 L HCl

Solution What is the molarity of an HCl solution if 25.5 mL of 0.438 M NaOH is required to neutralize 0.0250 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 2 Write a plan to calculate molarity or volume. Metric factor L NaOH solution Molarity mL NaOH solution moles of NaOH Mole–Mole factor moles of HCl Divide by liters molarity of HCl solution

Solution What is the molarity of an HCl solution if 25.5 mL of 0.438 M NaOH is required to neutralize 0.0250 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 3 State equalities and conversion factors including concentration.

Solution What is the molarity of an HCl solution if 25.5 mL of 0.438 M NaOH is required to neutralize 0.0250 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 4 Set up the problem to calculate the needed quantity.