Polyprotic Titrations: Videos & Practice Problems

In titrations involving polyprotic acids like phosphoric acid, multiple equivalence points are crucial for understanding acid-base reactions. The first equivalence point occurs when equal moles of weak acid and strong base react, forming a conjugate base. The Henderson-Hasselbalch equation is used to calculate pH at various stages, reflecting the acid ionization constants (K_a) and concentrations of the acid and its conjugate base. As strong base is added, pH increases, demonstrating the transition from weak acid to its conjugate base, ultimately leading to a strong base excess beyond the second equivalence point.

Polyprotic Acid Titrations

In the study of polyprotic acids, such as phosphoric acid, it is essential to understand the concept of equivalence points during titration. Polyprotic acids can donate more than one proton (H⁺), leading to multiple equivalence points in a titration with a strong base. For phosphoric acid, there are three equivalence points, which means three distinct volumes of strong base are required to fully neutralize the acid.

The relationship between the molarity and volume of the acid and base can be expressed with the equation:

M_a \cdot V_a = M_b \cdot V_b

where M_a and V_a are the molarity and volume of the acid, and M_b and V_b are the molarity and volume of the base. For example, if the molarity of the strong base is 0.100 M, the first equivalence volume needed to neutralize phosphoric acid is 50 mL. Consequently, the total volumes required for each equivalence point are 50 mL for the first, 100 mL for the second, and 150 mL for the third.

Before any strong base is added, the solution contains only the weak acid. To analyze the equilibrium state, an ICE (Initial, Change, Equilibrium) chart is utilized. For the first dissociation of phosphoric acid, the equilibrium expression involves the first dissociation constant, K_a1, which is defined as:

K_a1 = \frac{[H_2PO_4^-][H_3O^+]}{[H_3PO_4]}

In this case, the initial concentration of phosphoric acid is 0.100 M, and since no products are present initially, the change in concentration will reflect the loss of reactants as products form. The 5% approximation method can help determine if the change in concentration (denoted as x) can be ignored. However, for phosphoric acid, the calculated value of K_a1 is 7.5 x 10^-3, which indicates that the approximation cannot be applied, necessitating the use of the quadratic formula to solve for x.

Upon applying the quadratic formula, the concentration of hydrogen ions, [H^+] , is found to be approximately 0.0239 M. The pH of the solution can then be calculated using the formula:

pH = -\log[H^+]

This results in a pH of approximately 1.62, indicating a strongly acidic solution before any base is added. As titration begins and strong base is introduced, the pH will increase, reflecting the neutralization process. Understanding these concepts is crucial for mastering the behavior of polyprotic acids in titration scenarios.

Polyprotic Acid Titrations

In the process of titrating a weak acid with a strong base, the formation of a buffer occurs before reaching the first equivalence point. To understand this, we utilize the Henderson-Hasselbalch equation, which is essential for calculating the pH of a buffer solution. At the first equivalence point, 50 milliliters of the titrant, sodium hydroxide (NaOH), is required, but initially, only 30 milliliters is present.

To analyze the stoichiometric relationship between the weak acid and the strong base, we start by writing a balanced chemical equation. The initial concentrations of the weak acid and the strong base are converted from milliliters to liters and then multiplied by their respective molarities to determine the number of moles. Initially, the conjugate base is absent, represented as 0 moles.

As the reaction proceeds, the smaller mole total of the reactants will be subtracted from the larger total. Consequently, all of the strong base will be consumed, leaving some weak acid remaining. According to the law of conservation of mass, this reaction will produce a conjugate base, resulting in a buffer solution composed of both the weak acid and its conjugate base.

Since we have not yet reached the first equivalence point, we focus on the first dissociation constant (K_a1) of phosphoric acid, which is relevant for the removal of the first acidic hydrogen. The Henderson-Hasselbalch equation is expressed as:

pH = pK_a1 + log\left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right)

Here, pK_a1 is calculated as the negative logarithm of K_a1, which is 7.5 × 10^-3. Plugging in the values, we find:

pH = -\log(7.5 \times 10^{-3}) + \log\left(\frac{0.003}{[\text{weak acid}]}\right)

After performing the calculations, the resulting pH at this stage is determined to be 2.30. This increase in pH is a direct consequence of the addition of the strong base during the titration process. Understanding these concepts is crucial as we progress towards the first equivalence point in the titration.

Polyprotic Acid Titrations

During a titration involving a weak acid and a strong base, reaching the first equivalence point signifies that the moles of the acid and base are equal, resulting in their complete neutralization. At this stage, while the acid and base are consumed, the conservation of mass ensures that a conjugate base is formed, which remains in the solution. To calculate the pH at this point, it is essential to determine the formal concentration of the conjugate base.

The formal concentration (f) can be calculated using the formula:

f = \frac{C_a \times V_a}{V_{total}}

where:

• C_a = initial concentration of the acid (0.100 M)
• V_a = volume of the acid (50 mL)
• V_total = total volume of the solution (50 mL acid + 50 mL base = 100 mL)

Substituting the values gives:

f = \frac{0.100 \, \text{mol/L} \times 50 \, \text{mL}}{100 \, \text{mL}} = 0.050 \, \text{mol/L}

Next, to find the concentration of hydrogen ions (H⁺), we utilize the dissociation constants for the polyprotic acid. The first dissociation constant (K_a1) is 7.5 × 10^-3 and the second dissociation constant (K_a2) is 6.2 × 10^-8. The ion product of water (K_w) is 1.0 × 10^-14. By plugging these values into the appropriate equilibrium expressions, we can determine the concentration of H⁺ ions:

[H⁺] = 2.01 × 10^-5 mol/L

With the concentration of H⁺ known, the pH can be calculated using the formula:

pH = -\log[H⁺]

Thus, the pH at the first equivalence point is:

pH = 4.70

As the titration progresses, the addition of strong base continues to increase the pH, leading to further equivalence points and additional changes in the solution's acidity. Understanding these dynamics is crucial for analyzing the behavior of polyprotic acids throughout the titration process.

Polyprotic Acid Titrations

In this titration scenario, we are examining the reaction between 50 mL of 0.100 M phosphoric acid (H₃PO₄) and 70 mL of 0.100 M sodium hydroxide (NaOH). At the first equivalence point, 50 mL of NaOH completely neutralizes the phosphoric acid, resulting in the formation of dihydrogen phosphate (H₂PO₄^-). However, since we have added an additional 20 mL of NaOH, we need to calculate the moles of excess NaOH and how it affects the pH.

To find the moles of excess NaOH, we multiply the volume of the excess NaOH (20 mL) by its concentration (0.100 M), which gives us:

Excess moles of NaOH = 0.020 L × 0.100 mol/L = 0.002 moles.

At this point, we still have some dihydrogen phosphate remaining, and no hydrogen phosphate (HPO₄^2-) has been formed yet. The moles of dihydrogen phosphate can be calculated by subtracting the moles of NaOH used to reach the first equivalence point from the initial moles of phosphoric acid. This results in a mixture of weak acid and its conjugate base.

Since we are dealing with the removal of the second hydrogen ion from the polyprotic acid, we utilize the second dissociation constant, K_a2. The pH can be calculated using the Henderson-Hasselbalch equation:

pH = -log(K_a2) + log([A^-]/[HA])

Here, K_a2 is given as 6.2 × 10^-8. Plugging this value into the equation, we find:

pH = -log(6.2 × 10^-8) + log(moles of conjugate base/moles of weak acid).

After performing the calculations, we arrive at a pH of 7.03. This indicates that we are in a buffered region of the titration, where both the weak acid and its conjugate base are present, allowing for a relatively stable pH despite the addition of more base.

As we progress to the second equivalence point, further analysis will reveal how the pH changes as we continue to add NaOH. Understanding these concepts is crucial for mastering acid-base titrations and the behavior of polyprotic acids.

Polyprotic Acid Titrations

At the second equivalence point in a titration involving dihydrogen phosphate and sodium hydroxide (NaOH), all dihydrogen phosphate ions have been completely neutralized. This process adheres to the principle of conservation of mass, where the loss of dihydrogen phosphate results in the formation of hydrogen phosphate, which acts as the conjugate base in this scenario.

To determine the formal concentration of hydrogen phosphate, we start with the initial concentration of the weak acid, which is 0.100 M. By applying the dilution formula, we calculate the concentration as follows:

$$C = \frac{C_1 \times V_1}{V_1 + V_2}$$

Here, \(C_1\) is the initial concentration (0.100 M), \(V_1\) is the volume of the acid (50 mL), and \(V_2\) is the volume of the base added (100 mL). The total volume of the solution is 150 mL, leading to a formal concentration of:

$$C = \frac{0.100 \, \text{mol/L} \times 50 \, \text{mL}}{150 \, \text{mL}} = 0.0333 \, \text{mol/L}$$

Next, we can find the concentration of hydrogen ions (\(H^+\)) using the equilibrium constants \(K_2\) and \(K_3\), along with the ion product of water (\(K_w\)). The formula used to find the concentration of the intermediate species from a polyprotic acid is applied here. After substituting the necessary values, we arrive at an \(H^+\) concentration of:

$$[H^+] = 2.20 \times 10^{-10} \, \text{mol/L}$$

With this concentration, we can calculate the pH of the solution using the formula:

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

This results in a pH of 9.66, indicating that the pH increases as more strong base is added during the titration process. This behavior is characteristic of the titration of weak acids with strong bases, where the pH rises significantly as the equivalence point is approached and surpassed.

Polyprotic Acid Titrations

When titrating a weak acid like dihydrogen phosphate with a strong base such as sodium hydroxide (NaOH), understanding the behavior beyond the second equivalence point is crucial. At this stage, all the dihydrogen phosphate has reacted, and we are left with an excess of NaOH, which significantly influences the pH due to its strong nature.To calculate the concentration of the excess strong base, we start with the initial concentration of NaOH, which is 0.100 M. After reaching the second equivalence point, we have added an additional 15 mL of NaOH beyond the required 100 mL. The total volume of the solution is now 165 mL (50 mL of the original solution plus 115 mL of NaOH added).The concentration of the excess NaOH can be calculated using the formula:\[\text{Concentration of NaOH} = \frac{\text{Initial concentration} \times \text{Volume of excess}}{\text{Total volume}} = \frac{0.100 \, \text{mol/L} \times 0.015 \, \text{L}}{0.165 \, \text{L}} = 0.009091 \, \text{mol/L}\]Next, to find the pOH, we use the formula:\[\text{pOH} = -\log[\text{OH}^-] = -\log(0.009091) \approx 2.04\]Since pH and pOH are related by the equation \( \text{pH} + \text{pOH} = 14 \), we can find the pH:\[\text{pH} = 14 - \text{pOH} = 14 - 2.04 \approx 11.96\]Thus, after surpassing the second equivalence point, the pH of the solution is approximately 11.96. As we approach the third equivalence point, the pH will continue to change, reflecting the ongoing reaction dynamics and the presence of excess strong base.

Here’s what students ask on this topic:

What are polyprotic acids and how do they differ from monoprotic acids in titrations?

Polyprotic acids are acids that can donate more than one proton (H⁺) per molecule, such as phosphoric acid (H₃PO₄). In titrations, they differ from monoprotic acids, which donate only one proton, by having multiple equivalence points. Each equivalence point corresponds to the donation of a proton. For example, phosphoric acid has three equivalence points, each representing the removal of one proton. This results in more complex calculations and the need to consider multiple acid dissociation constants (K_a1, K_a2, K_a3).

How do you calculate the pH at the first equivalence point in a polyprotic titration?

At the first equivalence point, the moles of the weak acid equal the moles of the strong base, forming the conjugate base. To calculate the pH, first determine the formal concentration of the conjugate base. Use the formula:

$F=\frac{C}{}$₀ V₀V₀ + V_t

where C₀ is the initial concentration of the acid, V₀ is the volume of the acid, and V_t is the volume of the titrant. Then, use the equilibrium expression involving K_a1 and K_a2 to find [H⁺]. Finally, calculate pH as:

$\mathrm{pH}=\; -\log ([H+])$

What is the Henderson-Hasselbalch equation and how is it used in polyprotic titrations?

The Henderson-Hasselbalch equation is used to estimate the pH of a buffer solution. It is given by:

$\mathrm{pH}=\mathrm{pK}$_a + log[A^-][HA]

where pK_a is the negative logarithm of the acid dissociation constant (K_a), [A^-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. In polyprotic titrations, this equation is used at various stages to calculate the pH, particularly before reaching each equivalence point, where a buffer system is present.

How do you determine the pH after the second equivalence point in a polyprotic titration?

After the second equivalence point, the solution contains an excess of the strong base (e.g., NaOH). To determine the pH, first calculate the concentration of the excess strong base using:

$\mathrm{[OH^-]}=\frac{C}{}$_base V_excessV_total

where C_base is the concentration of the strong base, V_excess is the volume of excess base, and V_total is the total volume of the solution. Then, find the pOH:

$\mathrm{pOH}=\; -\log ([OH-])$

Finally, convert pOH to pH:

$\mathrm{pH}=\; 14\; -\mathrm{pOH}$

What are the key steps in performing a polyprotic titration involving phosphoric acid?

Key steps in performing a polyprotic titration involving phosphoric acid (H₃PO₄) include:

1. Determine the initial pH of the weak acid solution using its initial concentration and K_a1.
2. Add the strong base (e.g., NaOH) incrementally and calculate the pH at various points using the Henderson-Hasselbalch equation before reaching the first equivalence point.
3. At the first equivalence point, calculate the pH using the formal concentration of the conjugate base and the equilibrium expression involving K_a1 and K_a2.
4. Continue adding the strong base and use the Henderson-Hasselbalch equation for the second buffer region involving K_a2.
5. At the second equivalence point, calculate the pH using the formal concentration of the second conjugate base and the equilibrium expression involving K_a2 and K_a3.
6. After the second equivalence point, calculate the pH based on the excess strong base concentration.