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

Understanding the pH during various stages of a weak acid-strong base titration is crucial for predicting the outcome of the reaction. Initially, the moles of weak acid exceed those of the strong base, leading to the formation of a conjugate base as neutralization occurs. The ICE (Initial, Change, Final) chart is a valuable tool for calculating the final concentrations in moles, using the relationship:

moles=liters×molarity

At the equivalence point, the moles of weak acid are equal to the moles of strong base, indicating complete neutralization with only the conjugate base remaining. The volume of titrant can be determined using the equation:

M_acid×V_acid=M_base×V_base

if not provided. Beyond the equivalence point, the reaction features an excess of strong base, simplifying the process of finding the pH as the system no longer behaves like a buffer. This understanding is essential for accurate titration analysis and pH calculation in chemical reactions.

Before the Equivalence Point

In a titration involving a weak acid and a strong base, the weak acid serves as the titrate, while the strong base is the titrant. This setup is common in acid-base titrations, where the strong species (the base) is used to determine the concentration of the weak species (the acid). To analyze the reaction quantitatively, an ICF chart—standing for Initial, Change, and Final—can be employed. This chart helps in calculating the final amounts of the reactants and products involved in the titration.

The units used in the ICF chart are expressed in moles, which can be calculated using the formula: \( \text{moles} = \text{liters} \times \text{molarity} \). During the titration process, particularly before reaching the equivalence point, the moles of the weak acid will be greater than the moles of the strong base. As the strong base neutralizes the weak acid, a conjugate base is formed. This is a crucial aspect to remember when analyzing titration problems, as it influences the resulting concentrations and pH of the solution.

pH Before the Equivalence Point Example

In this scenario, we are examining the titration of 75 mL of 0.0300 M pyruvic acid, a weak acid with a dissociation constant (K_a) of 4.1 × 10^-3, with 12 mL of 0.0450 M potassium hydroxide (KOH), a strong base. The interaction between a weak acid and a strong base necessitates the use of an ICF (Initial, Change, Final) chart to track the changes in concentrations throughout the reaction.

Initially, we calculate the moles of pyruvic acid and potassium hydroxide. The number of moles is determined using the formula:

moles = volume (L) × molarity (M)

Converting the volumes from mL to L, we find:

• For pyruvic acid: 0.075 L × 0.0300 M = 0.00225 moles
• For KOH: 0.012 L × 0.0450 M = 0.00054 moles

In the ICF chart, we focus on the weak acid (pyruvic acid) and its conjugate base (pyruvate), while the strong base (KOH) reacts with the weak acid. According to Brønsted-Lowry definitions, the acid donates a proton (H⁺) to the hydroxide ion (OH^-) from KOH, forming water (H₂O) and the conjugate base (C₃H₃O₃^-).

Next, we apply the law of conservation of mass. The moles of the strong base (0.00054) are subtracted from both reactants. This results in:

• Remaining moles of pyruvic acid: 0.00225 - 0.00054 = 0.00171 moles
• Moles of KOH: 0 (since it is completely reacted)
• Moles of conjugate base (pyruvate): 0 + 0.00054 = 0.00054 moles

To find the pH of the resulting buffer solution, we utilize the Henderson-Hasselbalch equation:

pH = pK_a + log( [A^-] / [HA] )

Here, pK_a is calculated as:

pK_a = -log(K_a) = -log(4.1 × 10^-3) ≈ 2.39

Substituting the values into the equation:

pH = 2.39 + log(0.00054 / 0.00171)

Calculating the logarithmic term gives:

pH ≈ 2.39 - 0.25 ≈ 2.14

Thus, the final pH of the buffer solution after the titration is approximately 2.14, indicating the acidic nature of the solution due to the presence of the weak acid and its conjugate base.

In order to create a buffer 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 12.0 mL of 0.300 M NaH? K_a = 4.9 × 10⁻¹⁰.

• A. 6.82
• B. 10.01
• C. 8.52
• D. 10.17

Answer: (D)

At the Equivalence Point

In the titration of a weak acid with a strong base, the equivalence point is reached when the moles of the weak acid equal the moles of the strong base. At this stage, the weak acid has been completely neutralized, resulting in the formation of its conjugate base. This conjugate base is crucial for determining the pH at the equivalence point.

To calculate the pH at this point, it is essential to recognize that the solution will primarily consist of the conjugate base, which can hydrolyze in water. The hydrolysis reaction can be represented as follows:

\[ \text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^- \]

Here, A^- represents the conjugate base, and HA is the weak acid. The presence of hydroxide ions (OH^-) from this reaction will affect the pH, making it basic.

To find the pH, one can use the equilibrium expression for the hydrolysis of the conjugate base. The base dissociation constant (K_b) can be calculated using the relationship:

\[ K_b = \frac{K_w}{K_a} \]

where K_w is the ion product of water (1.0 x 10^-14 at 25°C) and K_a is the acid dissociation constant of the weak acid.

Additionally, the equivalence volume of the titrant can be determined using the equation:

\[ m_{\text{acid}} \times V_{\text{acid}} = m_{\text{base}} \times V_{\text{base}} \]

This equation allows for the calculation of the volume of the titrant needed to reach the equivalence point if it is not provided. Understanding these concepts is essential for accurately determining the pH at the equivalence point in a titration involving a weak acid and a strong base.

pH at the Equivalence Point Example

In the titration of 75 mL of 0.0300 M pyruvic acid (with a dissociation constant, K_a, of 4.1 × 10^-3) with 50 mL of 0.0450 M potassium hydroxide, we analyze the reaction between a weak acid and a strong base. The potassium hydroxide acts as the strong base, reacting with the pyruvic acid, which donates a proton (H⁺) to form water and the conjugate base of pyruvic acid.

To begin, we set up an ICF (Initial, Change, Final) chart to track the moles of each species involved. First, we convert the volumes from milliliters to liters and calculate the moles of the acid and base:

Moles = Volume (L) × Molarity (M)

For pyruvic acid: 0.075 L × 0.0300 M = 0.00225 moles

For potassium hydroxide: 0.050 L × 0.0450 M = 0.00225 moles

Since both reactants have equal moles, they completely neutralize each other, resulting in no remaining acid or base, but producing the conjugate base. The total volume of the solution after the reaction is 125 mL (0.125 L), leading to a concentration of the conjugate base:

Concentration = Moles / Total Volume = 0.00225 moles / 0.125 L = 0.018 M

Next, we set up an ICE chart for the conjugate base reacting with water. The reaction can be represented as:

Conjugate Base + H₂O ⇌ Pyruvic Acid + OH^-

In this case, we ignore the neutral potassium ions. The initial concentration of the conjugate base is 0.018 M, and the initial concentration of hydroxide ions is 0. The change in concentration will lead to the formation of hydroxide ions. We can use the equilibrium constant expression for the base:

K_b = &frac{x²}{0.018 - x}

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

K_w = K_a × K_b

Given K_w = 1.0 × 10^-14, we calculate:

K_b = K_w / K_a = 1.0 × 10^-14 / 4.1 × 10^-3 = 2.44 × 10^-12

Using the approximation method, since the initial concentration (0.018 M) is much greater than K_b, we can simplify the equilibrium expression:

K_b ≈ &frac{x²}{0.018}

Solving for x gives:

x² = K_b × 0.018 = 4.32 × 10^-14

Taking the square root, we find:

x = 2.08 × 10^-7 M

This value represents the concentration of hydroxide ions (OH^-). To find the pOH:

pOH = -log[OH^-] = -log(2.08 × 10^-7) ≈ 6.68

Finally, we can calculate the pH using the relationship between pH and pOH:

pH + pOH = 14

pH = 14 - pOH = 14 - 6.68 = 7.32

Thus, the pH of the solution after the titration is approximately 7.32.

Consider the titration of 75.0 mL of 0.60 M HNO₂ with 0.100 M NaOH at the equivalence point. What would be the pH of the solution at the equivalence point? The K_a of HNO₂ is 4.6 × 10⁻⁴.

• A. 11.73
• B. 8.13
• C. 5.87
• D. 1.35

Answer: (B)

After the Equivalence Point

In a weak acid-strong base titration, once the equivalence point is surpassed, the dynamics of the solution change significantly. At this stage, the moles of weak acid present are less than the moles of strong base added. This indicates that the weak acid has been completely neutralized, and any additional strong base contributes to an excess in the solution.Beyond the equivalence point, the solution no longer functions as a buffer. This is crucial because buffers are typically effective in resisting changes in pH, but with the complete neutralization of the weak acid, the remaining strong base dictates the pH level. The presence of excess strong base simplifies the calculation of pH, as it allows for a straightforward determination based on the concentration of the unreacted strong base.To calculate the pH in this scenario, one can use the formula for pH derived from the concentration of the strong base. If \( [OH^-] \) represents the concentration of hydroxide ions from the excess strong base, the pH can be calculated using the relationship:\[pH = 14 - pOH\]where \( pOH \) is calculated as:\[pOH = -\log[OH^-]\]This approach highlights the importance of understanding the behavior of acids and bases in titration processes, particularly when transitioning from a buffered solution to one dominated by a strong base.

pH After the Equivalence Point Example

In this scenario, we are examining the titration of 75 mL of 0.0300 M pyruvic acid, a weak acid, with 75 mL of 0.0450 M potassium hydroxide (KOH), a strong base. The acid dissociation constant (K_a) for pyruvic acid is 4.1 × 10^-3. To determine the pH of the resulting solution, we will utilize an Initial-Change-Final (ICF) chart to track the changes in concentrations of the reactants and products throughout the reaction.

First, we convert the volumes of the solutions from milliliters to liters to calculate the number of moles. For pyruvic acid:

Number of moles of pyruvic acid = Volume (L) × Molarity (M) = 0.075 L × 0.0300 M = 0.00225 moles.

For potassium hydroxide:

Number of moles of KOH = Volume (L) × Molarity (M) = 0.075 L × 0.0450 M = 0.003375 moles.

Next, we set up the ICF chart:

Species	Initial (moles)	Change (moles)	Final (moles)
Pyruvic Acid	0.00225	-0.00225	0
Potassium Hydroxide	0.003375	-0.00225	0.001125
Potassium Pyruvate	0	+0.00225	0.00225

After the reaction, we find that all the pyruvic acid has been consumed, leaving us with 0.001125 moles of KOH. The total volume of the solution after mixing is 150 mL (75 mL + 75 mL), which is equivalent to 0.150 L.

To find the concentration of the remaining strong base (KOH), we calculate:

Concentration of KOH = Number of moles / Total volume = 0.001125 moles / 0.150 L = 0.0075 M.

Since KOH completely dissociates in solution, the concentration of hydroxide ions (OH^-) is also 0.0075 M. We can now calculate the pOH:

pOH = -log[OH^-] = -log(0.0075) ≈ 2.12.

Finally, we can determine the pH using the relationship between pH and pOH:

pH = 14 - pOH = 14 - 2.12 = 11.88.

Thus, the pH of the solution after the equivalence point of the titration is approximately 11.88, indicating a basic solution due to the presence of excess strong base.

Calculate the pH of the solution resulting from the mixing of 55.0 mL of 0.100 M NaCN and 75.0 mL of 0.100 M HCN with 0.0090 moles of NaOH. 

• A. 12.06
• B. 11.74
• C. 1.89
• D. 2.26

Answer: (A)

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 weak acid is equal to the amount of its conjugate base formed in the solution. This is significant because it is the point at which the pH of the solution equals the p_aK_a of the weak acid. The p_aK_a is a measure of the acid's strength; it is the pH at which the acid is half dissociated. Therefore, at the midpoint of the titration, the solution acts as a buffer, and the pH changes very little with the addition of small amounts of strong base or acid. This is also the point where the buffer capacity is at its highest. Understanding this concept is crucial for accurately determining the p_aK_a of the weak acid and for preparing effective buffer solutions.

What is the relative pH at the equivalence point of the titration of a weak acid with a strong base?

The pH at the equivalence point of a titration involving a weak acid and a strong base will be greater than 7. This is because the weak acid (HA) reacts with the strong base (like NaOH) to form a salt (such as NaA) and water. The salt produced in this reaction typically dissociates into its ions (Na⁺ and A^-). The A^- ion, which is the conjugate base of the weak acid, undergoes hydrolysis with water to form OH^- (hydroxide ions) and HA (the weak acid).

Since the hydrolysis of the conjugate base (A^-) produces OH^- ions, the solution becomes basic. Therefore, the resulting solution at the equivalence point has a pH that is above 7, indicating a basic solution. The exact pH value depends on the strength of the weak acid and the concentrations of the acid and base used in the titration.

In the titration of a weak acid with a strong base with a 1:1 ratio, when does pH equal the pKa?

In a titration of a weak acid with a strong base, the pH equals the pK_a at the point known as the half-equivalence point. This is the stage where exactly half of the amount of the weak acid has been neutralized by the strong base. At this point, the concentration of the weak acid (HA) is equal to the concentration of its conjugate base (A^-). This creates a buffer system, and according to the Henderson-Hasselbalch equation, when [HA] = [A^-], the pH of the solution is equal to the pK_a of the weak acid. This is a key concept in understanding buffer systems and acid-base titrations, as the pK_a gives you insight into the acid strength and the pH at which the buffer system can effectively resist changes in pH.