Titrations: Weak Base-Strong Acid: Videos & Practice Problems

In titrations involving a weak base and a strong acid, the pH calculation varies depending on the stage of the titration. Before the equivalence point, the weak base is in excess, creating a buffer solution, allowing the use of the Henderson-Hasselbalch equation to determine the pH. At the equivalence point, the moles of weak base and strong acid are equal, and the pH can be found using the formula:

$M_1{V}_1=M_2{V}_2$

where M and V represent molarity and volume, respectively. After the equivalence point, the strong acid is in excess, and its complete ionization simplifies the pH calculation, eliminating the need for an ICE (Initial, Change, Final) chart. Understanding these stages is crucial for accurate pH determination during acid-base titrations.

Before the Equivalence Point

In a weak base-strong acid titration, the weak base acts as the titrate while the strong acid serves as the titrant. This setup is crucial because the titrant must be a strong species to effectively neutralize the weak base. During the titration process, particularly before reaching the equivalence point, the moles of the weak base exceed those of the strong acid. As the strong acid neutralizes the weak base, a weak acid (the conjugate acid) is formed, resulting in a mixture that includes both the weak base and its conjugate acid. This combination creates a buffer solution.

To calculate the pH of the solution before the equivalence point, the Henderson-Hasselbalch equation is employed. This equation is particularly useful in buffer solutions, where the pH can be expressed as:

pH = pK_a + log₁₀([A^-]/[HA])

In this equation, pK_a represents the negative logarithm of the acid dissociation constant of the weak acid formed, [A^-] is the concentration of the conjugate base (the weak base), and [HA] is the concentration of the weak acid. By applying this equation, one can determine the pH of the solution at various points leading up to the equivalence point, allowing for a deeper understanding of the titration dynamics.

pH Before the Equivalence Point Example

To calculate the pH of a solution resulting from the titration of 25 mL of a 0.100 M chloric acid solution with 50 mL of a 0.100 M ammonia solution, we first need to understand the interaction between a strong acid and a weak base. Chloric acid (HClO₃) acts as the strong acid, while ammonia (NH₃) serves as the weak base. The reaction can be represented as follows:

HClO₃ + NH₃ → NH₄⁺ + ClO₃⁻

In this reaction, the chloric acid donates a proton (H⁺) to ammonia, forming the ammonium ion (NH₄⁺) and the chlorate ion (ClO₃⁻). To analyze the system, we can set up an Initial-Change-Final (ICF) chart, focusing on the moles of the reactants and products involved.

First, we convert the volumes of the solutions to moles using the formula:

moles = volume (L) × molarity (M)

For chloric acid:

moles of HClO₃ = 0.025 L × 0.100 M = 0.0025 moles

For ammonia:

moles of NH₃ = 0.050 L × 0.100 M = 0.0050 moles

Next, we fill in the ICF chart:

Species	Initial (mol)	Change (mol)	Final (mol)
HClO₃	0.0025	-0.0025	0
NH₃	0.0050	-0.0025	0.0025
NH₄⁺	0	+0.0025	0.0025
ClO₃⁻	0	+0.0025	0.0025

After the reaction, we are left with 0.0025 moles of the conjugate acid (NH₄⁺) and 0.0025 moles of the conjugate base (ClO₃⁻). Since we have a buffer solution, we can use the Henderson-Hasselbalch equation to find the pH:

pH = pKₐ + log\left(\frac{[A^-]}{[HA]}\right)

In this case, we can also express it in terms of Kb:

pH = pKₕ + log\left(\frac{[NH₄^+]}{[NH₃]}\right)

To find pKₐ, we first calculate pKₕ from the given Kb value of ammonia (1.75 × 10^-5):

pKₕ = -log(1.75 × 10^-5) ≈ 4.76

Now substituting the values into the equation:

pH = 4.76 + log\left(\frac{0.0025}{0.0025}\right)

Since the ratio of the conjugate acid to the weak base is 1, the log term becomes 0:

pH = 4.76 + 0 = 4.76

Thus, the pH of the solution before reaching the equivalence point is approximately 4.76.

Calculate the pH of the solution resulting from the mixing of 75.0 mL of 0.100 M NaC₂H₃O₂ and 75.0 mL of 0.30 M HC₂H₃O₂ with 0.0040 moles of HBr.

• A. 3.26
• B. 4.27
• C. 3.87
• D. 6.23

Answer: (C)

In order to create a buffer 7.321 g of potassium lactate is mix with 550.0 mL of 0.328 M lactic acid, HC₃H₅O₃. What is the pH of the buffer solution after the addition of 300.0 mL of 0.100 M hydrobromic acid, HBr? The K_a of HC₃H₅O₃ is 1.4 × 10⁻⁴.

• A. 3.16
• B. 2.96
• C. 4.74
• D. 4.35

Answer: (B)

At the Equivalence Point

In the context of titrations involving weak bases and strong acids, understanding the equivalence point is crucial. At this stage, the moles of the weak base are equal to the moles of the strong acid, indicating that neutralization has occurred. This results in the formation of a weak acid or its conjugate acid, which remains in the solution after the reaction.

To determine the equivalence volume of the titrant, the following formula is utilized:

\( M_{\text{acid}} \times V_{\text{acid}} = M_{\text{base}} \times V_{\text{base}} \)

In this equation, \( M \) represents molarity and \( V \) represents volume. This relationship is essential for calculating the volumes of both the titrant (strong acid) and the titrate (weak base) during the titration process.

Once the equivalence point is reached, calculating the pH becomes necessary. At this point, the solution contains the conjugate acid of the weak base, which can affect the pH. The pH can be calculated using the concentration of the conjugate acid and its dissociation constant, \( K_a \). The relationship between the pH and the concentration of the conjugate acid can be expressed through the following equation:

\( \text{pH} = -\log[H^+] \)

To find the concentration of \( [H^+] \), one must consider the dissociation of the conjugate acid in water, which can be represented by the equilibrium expression:

\( K_a = \frac{[H^+][A^-]}{[HA]} \)

Here, \( [HA] \) is the concentration of the conjugate acid, and \( [A^-] \) is the concentration of its conjugate base. By applying these principles, one can effectively calculate the pH at the equivalence point of a titration involving a weak base and a strong acid.

pH at the Equivalence Point Example

To calculate the pH of the solution resulting from the titration of 25 mL of 0.100 M chloric acid (HClO₃) with 50 mL of 0.050 M ammonia (NH₃), we start by recognizing that chloric acid is a strong acid and ammonia is a weak base. During the titration, the strong acid donates a proton (H⁺) to the weak base, forming ammonium ion (NH₄⁺) and chlorate ion (ClO₃^-).

First, we calculate the moles of each reactant. For HClO₃:

moles = volume (L) × molarity (M) = 0.025 L × 0.100 M = 0.0025 moles.

For NH₃:

moles = volume (L) × molarity (M) = 0.050 L × 0.050 M = 0.0025 moles.

Since both reactants are present in equal amounts, they completely neutralize each other, resulting in 0 moles of both HClO₃ and NH₃ at the end of the reaction. The product formed is 0.0025 moles of NH₄⁺.

Next, we determine the concentration of the ammonium ion in the final solution. The total volume after mixing is 25 mL + 50 mL = 75 mL = 0.075 L. Thus, the concentration of NH₄⁺ is:

[NH₄⁺] = moles / volume = 0.0025 moles / 0.075 L = 0.033 M.

To find the pH, we need to consider the dissociation of the ammonium ion in water:

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.

Using an ICE (Initial, Change, Equilibrium) table, we set up the initial concentrations:

• Initial: [NH₄⁺] = 0.033 M, [NH₃] = 0, [H₃O⁺] = 0.
• Change: [NH₄⁺] decreases by x, [NH₃] increases by x, [H₃O⁺] increases by x.
• Equilibrium: [NH₄⁺] = 0.033 - x, [NH₃] = x, [H₃O⁺] = x.

The equilibrium constant expression for the dissociation of NH₄⁺ is given by:

K_a = [NH₃][H₃O⁺] / [NH₄⁺].

To find K_a, we use the relationship between K_w and K_b:

K_a = K_w / K_b = (1.0 × 10^-14) / (1.75 × 10^-5) = 5.71 × 10^-10.

Substituting into the equilibrium expression, we have:

5.71 × 10^-10 = (x)(x) / (0.033 - x).

Since the initial concentration (0.033 M) is much greater than K_a, we can simplify to:

5.71 × 10^-10 = x² / 0.033.

Cross-multiplying gives:

x² = 5.71 × 10^-10 × 0.033 = 1.8843 × 10^-11.

Taking the square root, we find:

x = 4.34 × 10^-6 M, which represents the concentration of H₃O⁺.

Finally, we calculate the pH using the formula:

pH = -log[H₃O⁺] = -log(4.34 × 10^-6) ≈ 5.36.

Thus, the pH of the solution at the equivalence point is approximately 5.36.

Consider the titration of 100.0 mL of 0.100 M CH₃NH₂ with 0.250 M HNO₃ at the equivalence point. What would be the pH of the solution at the equivalence point? The K_b of CH₃NH₂ is 4.4 × 10⁻⁴.

• A. 8.10
• B. 11.79
• C. 2.21
• D. 5.90

Answer: (D)

After the Equivalence Point

In the context of titration involving a weak base and a strong acid, understanding the behavior of the solution after the equivalence point is crucial. At this stage, the weak base has been fully neutralized by the strong acid, resulting in an excess of the strong acid in the solution. This shift means that the moles of the weak base are now less than the moles of the strong acid, leading to a solution that no longer acts as a buffer.

Since the buffer system has been disrupted, the pH of the solution can be determined more straightforwardly. The presence of excess strong acid simplifies the calculation of pH because strong acids, such as hydrochloric acid (HCl), ionize completely in solution. This complete ionization eliminates the need for an ICE (Initial, Change, Equilibrium) chart, which is typically used for weak acids and bases where equilibrium concentrations must be calculated.

To calculate the pH after the equivalence point, one can use the concentration of the excess strong acid. The pH can be determined using the formula:

pH = -\log[H^+]

Where [H⁺] is the concentration of hydrogen ions contributed by the excess strong acid. This approach allows for a clear and efficient determination of the solution's acidity, reflecting the complete dissociation of the strong acid present.

pH After the Equivalence Point Example

To calculate the pH of a solution resulting from the titration of 125 mL of 0.100 M chloric acid (HClO₃) with 50 mL of 0.050 M ammonia (NH₃), we first identify the strong acid and the weak base involved in the reaction. According to the Bronsted-Lowry theory, the strong acid donates a proton (H⁺) to the weak base, forming the ammonium ion (NH₄⁺) and chloride ion (ClO₃^-) as a byproduct.

To analyze the reaction, we utilize an Initial-Change-Final (ICF) table, focusing on the moles of the reactants. The number of moles can be calculated using the formula:

Moles = Volume (L) × Molarity (M)

Converting the volumes from milliliters to liters, we find:

For HClO₃:
Volume = 125 mL = 0.125 L
Moles = 0.125 L × 0.100 M = 0.0125 moles

For NH₃:
Volume = 50 mL = 0.050 L
Moles = 0.050 L × 0.050 M = 0.0025 moles

In the ICF table, we start with 0.0125 moles of HClO₃ and 0.0025 moles of NH₃. Since the ammonia is the limiting reactant, we subtract its moles from the strong acid:

Final moles of HClO₃ = 0.0125 - 0.0025 = 0.010 moles
Final moles of NH₄⁺ = 0.0025 moles (produced from the reaction)

Next, we calculate the total volume of the solution:

Total Volume = 125 mL + 50 mL = 175 mL = 0.175 L

Now, we find the concentration of the remaining strong acid:

Concentration of HClO₃ = Final moles / Total volume

Concentration = 0.010 moles / 0.175 L = 0.0571 M

Since the concentration of the strong acid is equal to the concentration of H⁺ ions, we can now calculate the pH using the formula:

pH = -log[H⁺]

Substituting the concentration:

pH = -log(0.0571) ≈ 1.24

Thus, the pH of the solution after the titration is approximately 1.24, indicating a strongly acidic solution due to the presence of excess chloric acid.

A solution contains 100.0 mL of 0.550 M sodium nitrite, NaNO₂. Find the pH after the addition of 180.0 mL of 0.400 M HClO₄. The K_a of HNO₂ is 4.6 × 10⁻⁴.

• A. 11.17
• B. 1.22
• C. 3.85
• D. 12.78

Answer: (B)

Here’s what students ask on this topic:

What does the midpoint of the titration of a weak acid with a strong base correspond to?

The midpoint of the titration of a weak acid with a strong base corresponds to the point where exactly half of the weak acid has been neutralized by the strong base. At this stage, the amount of the weak acid (HA) is equal to the amount of its conjugate base (A^-) that has been formed. This is significant because it is the point where the pH of the solution is equal to the pK_a of the weak acid. The pK_a is a measure of the strength of the acid; it's the pH value at which the acid is half dissociated. Therefore, at the midpoint of the titration, the buffer capacity of the solution is at its maximum, meaning the solution can resist changes in pH upon the addition of small amounts of acid or base. This is due to the Henderson-Hasselbalch equation, which describes the relationship between pH, pK_a, and the ratio of the concentrations of the conjugate base and the weak acid.

What is the pH at the equivalence point in a strong acid-weak base titration?

In a strong acid-weak base titration, the pH at the equivalence point is typically less than 7. This is because the weak base being titrated has a conjugate acid that is a weak acid. When the strong acid (like HCl) reacts with the weak base (like NH₃), the products are a salt and water. However, the salt formed in this reaction partially dissociates in water to produce the weak conjugate acid of the base.

Since the conjugate acid is weak, it does not ionize completely in water, and it releases some hydrogen ions (H⁺) into the solution. This presence of extra H⁺ ions causes the solution to be acidic. Therefore, unlike the equivalence point in a strong acid-strong base titration, which is neutral (pH 7), the equivalence point in a strong acid-weak base titration is on the acidic side of the pH scale, typically below 7. The exact pH value will depend on the specific weak base used and its dissociation constant (K_b).

Why does the equivalence point for a weak base-strong acid titration occur at a pH < 7.00?

In a titration involving a weak base and a strong acid, the equivalence point is where the amount of acid added equals the amount of base present. A weak base does not completely dissociate in water, meaning it does not produce many hydroxide ions (OH^-). When a strong acid is added, it completely dissociates into hydrogen ions (H⁺), which are highly reactive. These hydrogen ions readily react with the hydroxide ions from the weak base to form water.

At the equivalence point, all of the weak base has been neutralized by the strong acid, but because the base is weak, the conjugate acid formed in the reaction is slightly stronger and can donate hydrogen ions to the water, creating a slightly acidic solution. This means that the pH at the equivalence point will be less than 7, reflecting the presence of excess hydrogen ions in the solution. The exact pH will depend on the strength of the weak base and the temperature of the solution.