Acid-Base Titrations Calculations

– buret to hold the titrant – beaker to hold the analyte – pH meter to measure the pH

Acid-Base Titration Types – Strong acid-strong base titration – Weak acid-strong base titration – Polyprotic acid-strong base titration

Strong Acid-Strong Base Titrations

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 We divide the curve into four regions: 1. Initial pH The pH of the solution is determined by the concentration of the strong acid.

We divide the curve into four regions: As base is added, pH increases slowly and then rapidly. The pH is determined by the concentration of the acid that is not neutralized. 2. Initial pH to eq. point Strong Acid - Strong Base Titrations

We divide the curve into four regions: At the equivalence point, [OH − ] = [H + ]. The pH = Equivalence point Strong Acid - Strong Base Titrations

We divide the curve into four regions: As more base is added, pH increases rapidly and then slowly. pH is determined by the concentration of the excess base. 4. After eq. point Strong Acid - Strong Base Titrations

Let’s see how this works in practice. Strong Acid - Strong Base Titrations

Sample Exercise Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL b)51.0 mL

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL Sample Exercise

Sample Exercise 17.6 (page 731) Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL This is between the initial point and the equivalence point.

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)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

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL Therefore, we need to determine the number of mols of acid remaining, n acid, and the total volume, V total, of the solution. (Remember, adding the NaOH increases the total volume. Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = (0.100 M)( L) Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = M base,added × V base,added Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = (0.100 M)( L) Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol n acid,remaining = n acid,i − n base,added Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol n acid,remaining = (5.00 − 4.90) × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol n acid,remaining = 0.10 × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol n acid,remaining = 1.0 × 10 −4 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = V acid + V base Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = V acid + V base = L L Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = V acid + V base = L Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = L Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = L [H + ] = n acid,remaining /V total Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = L [H + ] = (1.0 × 10 −4 mol)/( L) Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = L [H + ] = 1.0 × 10 −3 M Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = L [H + ] = 1.0 × 10 −3 M pH = −log[H + ] Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = L [H + ] = 1.0 × 10 −3 M pH = −log(1.0 × 10 −3 ) Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL n acid,remaining = 1.0 × 10 −4 mol V total = L [H + ] = 1.0 × 10 −3 M pH = 3.00 Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL pH = 3.00 Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. a)49.0 mL b)51.0 mL Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL This is beyond the equivalence point. Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL This is beyond the equivalence point. All of the strong acid has been used up. Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)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

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL We already determined n acid,i = 5.00 × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n acid,i = 5.00 × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = M base,added × V base,added Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = (0.100 M)( L) Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = 5.10 × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = 5.10 × 10 −3 mol n base,remaining = n base,added − n acid,i Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = 5.10 × 10 −3 mol n base,remaining = (5.10 − 5.00) × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = 5.10 × 10 −3 mol n base,remaining = 0.10 × 10 −3 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = V acid + V base Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = V acid + V base = L L Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = V acid + V base = L Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = n base,remaining /V total Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = (1.0 × 10 −4 mol)/( L) Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = 9.9 × 10 −4 M Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = 9.9 × 10 −4 M pOH = −log[OH − ] Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = 9.9 × 10 −4 M pOH = −log(9.9 × 10 −4 ) Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = 9.9 × 10 −4 M pOH = 3.00 Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = 9.9 × 10 −4 M pOH = 3.00 ⇒ pH = − 3.00 Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = 9.9 × 10 −4 M pOH = 3.00 ⇒ pH = − 3.00 = Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = L [OH − ] = 9.9 × 10 −4 M pH = Sample Exercise

Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution. b)51.0 mL pH = Sample Exercise

Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.

Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. 1.The initial pH is determined by the K a of the acid.

Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. 2.To determine the pH from the initial point to the eq. point, we first neutralize the weak acid and then use the Henderson- Hasselbach equation.

Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. 3.At the eq. point, we have no HX, only X −. We need to use the Kb value to find pH.

Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. 4.Beyond the eq. point, we use the excess base to calculate pH.

Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. Let’s see how this works in practice.

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = M acid × V acid Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = M acid × V acid = (0.100 M)( L) Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = M acid × V acid = 5.00 × 10 −3 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = M base × V base Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = M base × V base = (0.100 M)( L) Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = M base × V base = 4.50 × 10 −3 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = 4.50 × 10 −3 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = 4.50 × 10 −3 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial 5.0 × 10 −3 mol4.5 × 10 −3 mol0.0 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial 5.0 × 10 −3 mol4.5 × 10 −3 mol0.0 mol n change −4.5 × 10 −3 mol +4.5 × 10 −3 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial 5.0 × 10 −3 mol4.5 × 10 −3 mol0.0 mol n change −4.5 × 10 −3 mol +4.5 × 10 −3 mol n final 5.0 × 10 −4 mol0.0 mol4.5 × 10 −3 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n acid,final = 5.0 × 10 −4 mol Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n acid,final = 5.0 × 10 −4 mol [acid] = n acid,final /V total Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n acid,final = 5.0 × 10 −4 mol [acid] = (5.0 × 10 −4 mol)/( L) Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n acid,final = 5.0 × 10 −4 mol [acid] = 5.3 × 10 −3 M Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = n base,final /V total Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = (4.50 × 10 −3 mol)/( L) Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M pK a = −logK a Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M pK a = −log(1.8 × 10 −5 ) Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M pK a = 4.74 Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = pK a + log([base]/[acid) Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = log([4.74 × 10 −2 M]/[5.3 × 10 −3 M]) Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = log(9.00) Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = Sample Exercise

Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = 5.69 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. At the equivalence point, all of the weak acid is converted to its conjugate weak base. Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. At the equivalence point, all of the weak acid is converted to its conjugate weak base. This means that we will be using the base hydrolysis expression to find [OH − ], pOH, and pH. Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. The number of mols of acetate in solution at the equivalence point is equal to the number of mols of acetic acid at the start of the titration. Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. n base,eq. pt. = n initial,acid Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. n base,eq. pt. = n initial,acid = (0.100 M)( L) Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. n base,eq. pt. = n initial,acid = 5.00 × 10 −3 mol Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol The concentration of the acetate is given by: Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol The concentration of the acetate is given by: [base] = (n base,eq. pt. )/(V total ) Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol The concentration of the acetate is given by: [base] = (5.00 × 10 −3 mol)/(0.100 L) Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol The concentration of the acetate is given by: [base] = 5.00 × 10 −2 M Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M Next, we do an i-c-e table on the base hydrolysis. Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − i5.00 × 10 −2 0.0 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − i5.00 × 10 −2 0.0 c−x+x Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − i5.00 × 10 −2 0.0 c−x+x e5.00 × 10 −2 − xxx Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − i5.00 × 10 −2 0.0 c−x+x e5.00 × 10 −2 − xxx Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = K w /K a = (1.0 × 10 −14 )/(1.8 × 10 −5 ) Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = K w /K a = 5.6 × 10 −10 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = ([acid][OH − ])/[base] Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = (x 2 )/(5.00 × 10 −2 ) Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = (x 2 )/(5.00 × 10 −2 ) x 2 = (5.6 × 10 −10 )(5.00 × 10 −2 ) = 2.8 × 10 −11 x = (2.8 × 10 −11 ) ½ Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = (x 2 )/(5.00 × 10 −2 ) x 2 = (5.6 × 10 −10 )(5.00 × 10 −2 ) = 2.8 × 10 −11 x = (2.8 × 10 −11 ) ½ = 5.3 × 10 −6 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = (x 2 )/(5.00 × 10 −2 ) x 2 = (5.6 × 10 −10 )(5.00 × 10 −2 ) = 2.8 × 10 −11 x = (2.8 × 10 −11 ) ½ = 5.3 × 10 −6 = [OH − ] Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [OH − ] = 5.3 × 10 −6 M Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = −log[OH − ] Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = −log[OH − ] = −log(5.3 × 10 −6 ) Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = −log[OH − ] = −log(5.3 × 10 −6 ) = 5.28 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = 5.28 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = 5.28 pH = – pOH Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = 5.28 pH = – pOH = − 5.28 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = 5.28 pH = – pOH = − 5.28 = 8.72 Sample Exercise

Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH 3 COOH with M NaOH. pH = 8.72 Sample Exercise

The pH of the titration at the equivalence point depends on the type of acids and bases being titrated. – Strong acid-Strong base: eq. pt. = 7.0 – Weak acid-Strong base: eq. pt. > 7.0 – Strong acid-Weak base: eq. pt. < 7.0

Titrations of Polyprotic Acids When weak acid contain more than one ionizable H atom, as in phosporous acid, H 3 PO 3, reaction with OH − occurs in a series of steps. – H 3 PO 3 (aq) + H 2 O(l) ➙ H 2 PO 3 − (aq) + H 3 O + (aq) – H 2 PO 3 − (aq) + H 2 O(l) ➙ HPO 3 2− (aq) + H 3 O + (aq) – HPO 3 2− (aq) + H 2 O(l) ➙ PO 3 3− (aq) + H 3 O + (aq)

When the neutralization steps are sufficiently separated, the substance exhibits multiple equivalence points. Titrations of Polyprotic Acids