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Acid-Base Titrations. Section 17.3. Introduction. Definition: In an acid-base titration , a solution containing a known concentration of a base is slowly added to an acid. An indicator is used to signal the equivalence point of the titration.

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## Acid-Base Titrations

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**Acid-Base Titrations**Section 17.3**Introduction**• Definition: • In an acid-base titration, a solution containing a known concentration of a base is slowly added to an acid. • An indicator is used to signal the equivalence point of the titration. • This is the point at which stoichiometrically equivalent amounts of acid and base have been mixed. • A pH meter can also be used to find the equivalence point.**Introduction**• The typical titration apparatus includes: • a buretto hold the titrant • a beaker to hold the analyte • a pH meter to measure the pH**Introduction**• In this section, we will be looking at a series of titrations in detail to understand why acids and behave the way they do. • Strong acid-strong base titration • Weak acid-strong base titration • Polyprotic acid-strong base titration**Strong Acid-Strong Base Titrations**• The titration curve of a strong acid-strong base titration has the following shape.**Strong Acid-Strong Base Titrations**• The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: 1. Initial pH 2. Initial pH to eq. point 3. Equivalence point 4. After eq. point**Strong Acid-Strong Base Titrations**• The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: 1. Initial pH The pH of the solution is determined by the concentration of the strong acid.**Strong Acid-Strong Base Titrations**• The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: 2. Initial pH to eq. point As base is added, pH increases slowly and then rapidly. The pH is determined by the concentration of the acid that is not neutralized.**Strong Acid-Strong Base Titrations**• The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: 3. Equivalence point At the equivalence point, [OH−] = [H+]. The pH = 7.00.**Strong Acid-Strong Base Titrations**• The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: 4. After eq. point As more base is added, pH increases rapidly and then slowly. pH is determined by the concentration of the excess base.**Strong Acid-Strong Base Titrations**• The titration curve of a strong acid-strong base titration has the following shape. Let’s see how this works in practice.**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL • 51.0 mL**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL This is between the initial point and the equivalence point.**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL This is between the initial point and the equivalence point. pH is determined by the amount of acid that has not been neutralized.**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL Therefore, we need to determine the number of mols of acid remaining, nacid, and the total volume, Vtotal, of the solution. (Remember, adding the NaOH increases the total volume.**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= (0.100 M)(0.0500 L)**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= 5.00 × 10−3mol**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= 5.00 × 10−3mol nbase,added= Mbase,added× Vbase,added**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= 5.00 × 10−3mol nbase,added= (0.100 M)(0.0490 L)**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= 5.00 × 10−3mol nbase,added= 4.90 × 10−3mol**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= 5.00 × 10−3mol nbase,added= 4.90 × 10−3mol nacid,remaining= nacid,i− nbase,added**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= 5.00 × 10−3mol nbase,added= 4.90 × 10−3mol nacid,remaining= (5.00 − 4.90) × 10−3mol**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= 5.00 × 10−3mol nbase,added= 4.90 × 10−3mol nacid,remaining= 0.10 × 10−3mol**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,i= Macid,i× Vacid,i= 5.00 × 10−3mol nbase,added= 4.90 × 10−3mol nacid,remaining= 1.0 × 10−4mol**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = Vacid + Vbase**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = Vacid + Vbase = 0.0500 L + 0.0490 L**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = Vacid + Vbase = 0.0990 L**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = 0.0990 L**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = 0.0990 L [H+] = nacid,remaining/Vtotal**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = 0.0990 L [H+] = (1.0 × 10−4mol)/(0.0990 L)**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = 0.0990 L [H+] = 1.0 × 10−3 M**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = 0.0990 L [H+] = 1.0 × 10−3 M pH = −log[H+]**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = 0.0990 L [H+] = 1.0 × 10−3 M pH = −log(1.0 × 10−3)**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL nacid,remaining= 1.0 × 10−4mol Vtotal = 0.0990 L [H+] = 1.0 × 10−3 M pH = 3.00**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL pH = 3.00**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 49.0 mL • 51.0 mL**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL This is beyond the equivalence point.**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL This is beyond the equivalence point. All of the strong acid has been used up.**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL This is beyond the equivalence point. All of the strong acid has been used up. The pH is determined by the excess base that has been added.**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL We already determined nacid,i = 5.00 × 10−3mol**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL nacid,i = 5.00 × 10−3mol**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL nacid,i = 5.00 × 10−3mol nbase,added= Mbase,added× Vbase,added**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL nacid,i = 5.00 × 10−3mol nbase,added= (0.100 M)(0.0510 L)**Sample Exercise 17.6 (page 731)**Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. • 51.0 mL nacid,i = 5.00 × 10−3mol nbase,added= 5.10 × 10−3mol

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