Diprotic Acid Titrations: Videos & Practice Problems

In the titration of a diprotic acid, two equivalence points are crucial. The first equivalence volume can be calculated using the equation: $C^{1}V=C^{2}V$. The Henderson-Hasselbalch equation helps determine pH before reaching the equivalence point, while after, the pH increases significantly due to the formation of conjugate bases. Understanding these concepts is essential for mastering acid-base reactions and their calculations.

Diprotic Acid Titrations

In the titration of a diprotic acid, such as sulfuric acid, it is essential to recognize the presence of two equivalence points, which necessitates careful calculations to determine the equivalence volumes. The diprotic acid acts as the analyte, while a strong base, like potassium hydroxide (KOH), serves as the titrant. To find the equivalence volumes, we can use the relationship between the molarity and volume of the acid and base.

The first equivalence volume can be calculated using the formula:

M_a \cdot V_a = M_b \cdot V_{eq1}

where M_a is the molarity of the acid, V_a is the volume of the acid, M_b is the molarity of the base, and V_{eq1} is the first equivalence volume. For example, if we have a 0.100 M sulfuric acid solution with a volume of 100 mL, we can set up the equation:

0.100 \, \text{M} \cdot 100 \, \text{mL} = 0.050 \, \text{M} \cdot V_{eq1}

Solving for V_{eq1} gives us 200 mL. To reach the second equivalence point, we would need an additional 200 mL, resulting in a total of 400 mL of titrant.

Before any strong base is added, the solution contains a weak diprotic acid. To analyze the equilibrium state, we can use an ICE (Initial, Change, Equilibrium) chart. The dissociation of the first acidic hydrogen from sulfuric acid can be represented as:

H_2SO_4 + H_2O \rightleftharpoons HSO_4^- + H_3O^+

For this reaction, we utilize the first dissociation constant, K_{a1}, which for sulfuric acid is 1.6 \times 10^{-2}. To determine if we can apply the 5% approximation, we compare the initial concentration of the acid (0.100 M) to K_{a1}. The ratio:

\frac{0.100}{1.6 \times 10^{-2}} = 6.25

indicates that we cannot ignore the change in concentration (denoted as -x) in our equilibrium expression. Thus, we must solve for x using the quadratic formula, yielding x = 0.032792 \, \text{M}. This value represents the concentration of H^+ ions.

To find the pH of the solution, we take the negative logarithm of the H^+ concentration:

pH = -\log(0.032792) \approx 1.48

At this stage, the titration has not yet commenced. As we begin to add KOH to the sulfuric acid solution, it is crucial to identify the specific point in the titration process—whether it is before, at, or after the equivalence points—since the calculations will vary accordingly. Understanding these dynamics is vital for accurately interpreting the titration results.

Diprotic Acid Titrations

In the titration of 100 mL of 0.100 M sulfuric acid (H₂SO₄) with 75 mL of 0.050 M potassium hydroxide (KOH), we focus on the calculations prior to reaching the first equivalence point. At this stage, a buffer solution is formed, necessitating the use of the Henderson-Hasselbalch equation to determine the pH.

To begin, we convert the volumes of the solutions into moles. Moles can be calculated using the formula:

moles = volume (L) × molarity (mol/L)

For sulfuric acid:

Volume = 100 mL = 0.100 L

Moles of H₂SO₄ = 0.100 L × 0.100 mol/L = 0.0100 moles

For potassium hydroxide:

Volume = 75 mL = 0.075 L

Moles of KOH = 0.075 L × 0.050 mol/L = 0.00375 moles

Initially, we have no bisulfite ions (HSO₄^-), so its initial amount is 0. As the reaction proceeds, we apply the law of conservation of mass, where the moles of reactants decrease and the moles of products increase. The moles of KOH will react with an equal amount of H₂SO₄:

Remaining moles of H₂SO₄ = 0.0100 moles - 0.00375 moles = 0.00625 moles

Produced moles of bisulfite = 0.00375 moles

Now, we can use the Henderson-Hasselbalch equation to find the pH of the buffer solution:

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

Here, pK_a for the first dissociation of sulfuric acid is approximately -log(1.6 × 10^-2) = 1.79.

Substituting the values into the equation:

pH = 1.79 + log(0.00375 / 0.00625)

Calculating the log term:

log(0.00375 / 0.00625) = log(0.6) ≈ -0.2218

Thus, the pH = 1.79 - 0.2218 = 1.57.

As potassium hydroxide is added, the pH gradually increases due to the buffer's ability to resist significant changes in pH. However, once the first equivalence point is reached, the buffer capacity is exceeded, leading to a sharp increase in pH. Understanding this behavior is crucial for predicting the outcomes of titration reactions.

Diprotic Acid Titrations

During a titration involving a weak acid and a strong base, reaching the first equivalence point signifies that equal moles of both reactants have been mixed, resulting in complete neutralization. At this stage, the weak acid is transformed into its conjugate base, adhering to the law of conservation of mass. Although the weak acid is no longer present, it is essential to calculate the concentration of the conjugate base formed.

To determine the formal concentration of the conjugate base, the initial concentration of the weak acid (0.100 M) is multiplied by the volume of the acid used (100 mL), and then divided by the total volume of the solution (100 mL + 200 mL). This calculation yields a formal concentration of:

\[ \text{Concentration of conjugate base} = \frac{0.100 \, \text{mol/L} \times 100 \, \text{mL}}{100 \, \text{mL} + 200 \, \text{mL}} = 0.0333 \, \text{mol/L} \]

Next, to find the concentration of hydrogen ions (\(H^+\)), we utilize the equilibrium expression related to the dissociation of the conjugate base. For a diprotic acid, the first dissociation constant (\(K_1\)) is relevant, which for sulfuric acid is given as:

\[ K_1 = 1.6 \times 10^{-2} \]

Additionally, the second dissociation constant (\(K_2\)) is:

\[ K_2 = 6.4 \times 10^{-8} \]

And the ion product of water (\(K_w\)) is:

\[ K_w = 1.0 \times 10^{-14} \]

By substituting these values into the appropriate equilibrium expression, the concentration of \(H^+\) can be calculated, resulting in:

\[ [H^+] = 2.63 \times 10^{-5} \, \text{mol/L} \]

To find the pH, the negative logarithm of the hydrogen ion concentration is taken:

\[ \text{pH} = -\log(2.63 \times 10^{-5}) \approx 4.58 \]

This pH value indicates a significant jump at the first equivalence point, as the buffer system is disrupted, leading to a rapid increase in pH. It is important to note that since we are dealing with a diprotic acid, further titration points exist, and additional calculations will be necessary as we progress toward the second equivalence point.

Diprotic Acid Titrations

In the titration of 100 mL of 0.100 M sulfuric acid (H₂SO₄) with 250 mL of 0.050 M potassium hydroxide (KOH), we observe that the first equivalence point is reached after adding 200 mL of KOH. However, with an additional 50 mL of KOH, we have surpassed this point, leading to an excess of the strong base.

To calculate the excess moles of KOH, we first determine the volume of excess base added, which is 50 mL. Converting this volume to liters gives us 0.050 L. The moles of KOH can be calculated using the formula:

$\left(0.050\right)\left(0.050\right)=0.00250$

At the first equivalence point, all of the sulfuric acid has reacted with KOH, resulting in the formation of bisulfite ions (HSO₃^-) and sulfide ions (SO₃^2-). The remaining moles of bisulfite ions after the reaction can be calculated as follows:

$\left(0.0100\right)-\left(0.00250\right)=0.00750$

As we move past the first equivalence point, we transition to discussing the second dissociation constant, K_a2, since we are now removing the second hydrogen ion from the bisulfite ion to form the conjugate base. The pH can be calculated using the formula:

$pH=-log(6.4\times 10^(-8)+log(0.00250/0.00750)$

Upon calculating, the resulting pH is approximately 6.72. This increase in pH illustrates the effect of adding a strong base to a weak diprotic acid solution, demonstrating the shift in equilibrium as we progress towards the second equivalence point.

Diprotic Acid Titrations

At the second equivalence point in a titration involving bisulfide ions and KOH, the reaction continues until both the bisulfide ion and the hydroxide ions are completely neutralized, resulting in the formation of sulfide ions and water. At this stage, equal moles of the bisulfide ion and hydroxide ion have reacted, leaving behind the sulfite ion, which acts as a weak conjugate base.

To understand the concentration of the sulfite ion, we start with the initial concentration of the weak acid, which is 0.100 M, and the total volume of the solution after mixing 100 mL of the acid with 400 mL of the strong base. The formal concentration of the sulfite ion can be calculated as follows:

Formal concentration of sulfite ion = \[ \frac{0.100 \, \text{mol/L} \times 100 \, \text{mL}}{100 \, \text{mL} + 400 \, \text{mL}} = 0.020 \, \text{mol/L} \]

Since we are dealing with the removal of the second hydrogen ion, we refer to the second dissociation constant, \( K_{a2} \). To find the base dissociation constant \( K_{b1} \) for the sulfite ion, we use the relationship between \( K_{a2} \) and \( K_{b1} \):

\[ K_{a2} \times K_{b1} = K_w \]

Where \( K_w \) is the ion product of water, equal to \( 1.0 \times 10^{-14} \). Rearranging gives:

\[ K_{b1} = \frac{K_w}{K_{a2}} = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-8}} = 1.56 \times 10^{-7} \]

Next, we apply the 5% approximation method to determine if we can ignore the change in concentration (denoted as \( x \)). We compare the initial concentration of the sulfite ion to \( K_{b1} \) to ensure the ratio is greater than 500. This confirms that the change is negligible, allowing us to simplify our calculations.

After calculating \( x \), which represents the hydroxide ion concentration, we find:

\[ x = 5.59 \times 10^{-5} \]

To find the pOH, we take the negative logarithm of the hydroxide concentration:

\[ pOH = -\log(5.59 \times 10^{-5}) = 4.25 \]

Finally, we can determine the pH by subtracting the pOH from 14:

\[ pH = 14 - pOH = 14 - 4.25 = 9.75 \]

This indicates that as more strong base is added, the pH increases, particularly around the equivalence points where the most significant changes occur. The second equivalence point also results in a noticeable jump in pH, although it may not be as pronounced as the first equivalence point.

Diprotic Acid Titrations

In the titration of 100 mL of 0.100 M sulfuric acid (H₂SO₄) with 420 mL of 0.050 M potassium hydroxide (KOH), we analyze the scenario after surpassing the second equivalence point. To reach the second equivalence point, 400 mL of KOH is required, but with 420 mL used, there is an excess of 20 mL of KOH.

At this stage, the reaction between bisulfite ions (HSO₄^-) and KOH has been completed, resulting in the formation of sulfide ions (S^2-). However, since we have exceeded the equivalence point, the remaining KOH contributes to the overall pH as a strong base. To determine the concentration of the excess base, we use the formula:

$C=\left(0.050\right)\left(20/(100+420)\right)$

This calculation yields a concentration of approximately 0.001923 M for the excess KOH. To find the pOH, we take the negative logarithm of the concentration:

$pOH=-log\left(0.001923\right)=2.72$

Finally, to find the pH, we subtract the pOH from 14:

$pH=14-2.72=11.28$

This pH of 11.28 represents the final pH after the second equivalence point. Understanding the steps involved in titrating a diprotic acid is crucial, as it requires careful consideration of the calculations at various stages—before, at, or after the equivalence points. This approach ensures accurate determination of pH throughout the titration process.

Diprotic Acid Titrations Calculations 1

To calculate the pH of a solution when titrating carbonic acid (H₂CO₃) with sodium hydroxide (NaOH), we first need to determine the equivalence volume of the titrant. In this case, we have 100 mL of 0.25 M carbonic acid and are adding 70 mL of 0.25 M sodium hydroxide. The relationship between the molarity and volume of the acid and base can be expressed as:

M_a \cdot V_a = M_b \cdot V_e

Where M_a and V_a are the molarity and volume of the acid, and M_b and V_e are the molarity and equivalence volume of the base. Plugging in the values:

0.25 \, \text{M} \cdot 100 \, \text{mL} = 0.25 \, \text{M} \cdot V_e

Solving for V_e gives us 100 mL for the first equivalence point. Since carbonic acid is a diprotic acid, reaching the second equivalence point would require 200 mL of NaOH. However, with only 70 mL of NaOH added, we are still before the first equivalence point.

In this scenario, we can analyze the stoichiometric relationship between the diprotic acid and the strong base. The reaction can be represented as:

H_2CO_3 + NaOH \rightarrow NaHCO_3 + H_2O

To find the moles of each reactant, we convert the volumes of the solutions to moles:

For carbonic acid:

0.100 \, \text{L} \cdot 0.25 \, \text{mol/L} = 0.0250 \, \text{mol}

For sodium hydroxide:

0.070 \, \text{L} \cdot 0.25 \, \text{mol/L} = 0.0175 \, \text{mol}

Next, we determine the remaining moles of the weak acid and the conjugate base after the reaction:

Subtracting the moles of NaOH from the moles of H₂CO₃ gives:

0.0250 \, \text{mol} - 0.0175 \, \text{mol} = 0.0075 \, \text{mol} \, \text{(H}_2\text{CO}_3\text{ remaining)}

And the moles of the conjugate base (sodium bicarbonate, NaHCO₃) produced is:

0.0175 \, \text{mol} \, \text{(NaHCO}_3\text{)}

Since we are before the first equivalence point, a buffer solution is formed. The pH can be calculated using the Henderson-Hasselbalch equation:

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

Here, pK_a corresponds to the first dissociation constant of carbonic acid, which is approximately 6.37 (derived from K_{a1} = 4.3 \times 10^{-7}). Thus, we can calculate:

pH = 6.37 + \log\left(\frac{0.0175}{0.0075}\right)

Calculating the logarithm and adding it to the pK_a gives a final pH of approximately 6.73. This demonstrates the importance of understanding the stoichiometry and the buffer system in titrations involving diprotic acids, as well as the need to identify which dissociation constant to use based on the stage of the titration.

Diprotic Acid Titrations Calculations 1

To calculate the pH of a solution resulting from the titration of phosphorus acid (H₃PO₃) with sodium hydroxide (NaOH), we start by determining the equivalence volumes. In this scenario, we have 75 mL of 0.10 M phosphorus acid and 80 mL of 0.15 M sodium hydroxide. The first step involves using the formula for titration:

m_acid × V_acid = m_base × V_equivalence

Substituting the known values, we find:

0.10 M × 75 mL = 0.15 M × V_equivalence

Solving for V_equivalence gives us:

V_equivalence = (0.10 × 75) / 0.15 = 50 mL

To find the second equivalence volume, we note that it would require 100 mL of NaOH. Since we are using 80 mL, we have surpassed the first equivalence point but have not yet reached the second.

At the first equivalence point, the reaction produces dihydrogen phosphide (H₂PO₃^-). The moles of H₂PO₃^- can be calculated as follows:

Initial moles of H₃PO₃ = (0.10 M × 75 mL) / 1000 = 0.0075 moles

With 50 mL of NaOH needed to reach the first equivalence point, we have an excess of 30 mL of NaOH remaining. The moles of excess NaOH are calculated as:

Excess moles of NaOH = (0.15 M × 30 mL) / 1000 = 0.0045 moles

After the first equivalence point, we subtract the moles of the weak acid from the moles of the strong base:

Remaining moles of NaOH = 0.0045 moles (excess) - 0.0075 moles (H₂PO₃^-) = 0.0030 moles

Now, we can apply the Henderson-Hasselbalch equation to find the pH, as we are dealing with the second equivalence point:

pH = pK_a2 + log(_[A^-]/_[HA])

Where pK_a2 is calculated from the given Ka₂ value of 2.0 × 10^-7:

pK_a2 = -log(2.0 × 10^-7) ≈ 6.70

Substituting into the Henderson-Hasselbalch equation, we find:

pH = 6.70 + log(0.0030/0.0075) = 6.70 - 0.176 = 6.88

Thus, the final pH of the solution after the titration is approximately 6.88. Understanding the equivalence points and the application of the Henderson-Hasselbalch equation is crucial for accurately determining the pH in titration scenarios.

Diprotic Acid Titrations Calculations 2

In this scenario, we are tasked with finding the pH after titrating 100 mL of a 0.1 M dibasic compound B with 11 mL of a 1 M hydrochloric acid (HCl) solution. A dibasic compound can donate two protons (H⁺) in a titration, and understanding the equivalence points is crucial for determining the pH at various stages of the titration.

To begin, we calculate the equivalence volumes using the relationship between the molarity and volume of the acid and the base. The formula used is:

$\left(M\right)\left(V\right)=\left(M\right)\left(V\right)$

Here, the molarity of HCl is 1 M, and the volume of the dibasic compound is 100 mL (0.1 M). The equivalence volume for the first equivalence point is calculated to be 10 mL, while the second equivalence point would require 20 mL of HCl. Since we are using 11 mL of HCl, we are past the first equivalence point but have not yet reached the second.

At the first equivalence point, the dibasic compound B reacts with HCl to form B^H+ and Cl^-. The reaction can be represented as:

B + HCl → B^H+ + Cl^-

After reaching the first equivalence point, we have equal moles of the reactants consumed, which is 10 millimoles each. With 11 mL of HCl used, we have 1 mL of excess HCl, which corresponds to 1 millimole of HCl remaining. This excess HCl will react with the B^H+ to form B^H2+:

B^H+ + HCl → B^H2+ + Cl^-

At this stage, we have 9 millimoles of B^H+ and 1 millimole of B^H2+, creating a buffer solution. To find the pH, we apply the Henderson-Hasselbalch equation:

pH = pK_a1 + log([conjugate base]/[weak acid])

To use this equation, we first need to determine pK_a1. The relationship between the base and acid dissociation constants is given by:

K_b2 + K_a1 = K_w

Taking the negative logarithm of both sides leads to:

pK_b2 + pK_a1 = 14

Given that pK_b2 is 8, we find:

pK_a1 = 14 - 8 = 6

Substituting this value into the Henderson-Hasselbalch equation, we have:

pH = 6 + log(9/1) = 6 + 0.954 = 6.95

Thus, the final pH after the titration is approximately 6.95. This example illustrates the application of titration principles to dibasic compounds, emphasizing the importance of understanding equivalence points and the resulting buffer systems formed during the titration process.

Diprotic Acid Titrations Calculations 2

Carbonic acid (H₂CO₃) is a diprotic acid that can lose two protons, resulting in the formation of bicarbonate (HCO₃^-) and carbonate ions (CO₃^2-). The dissociation occurs in two steps: first, carbonic acid donates a proton to water, forming bicarbonate and hydronium ions (H₃O⁺), and then bicarbonate can further dissociate to produce carbonate ions and additional hydronium ions. Each dissociation step is characterized by its own dissociation constant, represented as pK_a1 and pK_a2.

In a titration scenario, a 50 mL solution of 0.50 M carbonic acid is titrated with a 1 M sodium hydroxide (NaOH) solution. To determine the pH after the addition of 35 mL of NaOH, we first calculate the equivalence volumes. The equivalence volume for the first dissociation (pK_a1) is 25 mL, while the second (pK_a2) is 50 mL. Since 35 mL of NaOH has been added, this indicates that we are beyond the first equivalence point, where bicarbonate ions are present and can react with the added NaOH to form carbonate ions.

To find the excess moles of NaOH, we note that 10 mL of NaOH is in excess (35 mL added - 25 mL required for the first equivalence point). This corresponds to 10 millimoles of NaOH. The moles of bicarbonate remaining after the first equivalence point can be calculated from the initial moles of carbonic acid, which is 25 millimoles. After the reaction with NaOH, we have 15 millimoles of bicarbonate left and 10 millimoles of carbonate formed.

At this stage, we have a buffer solution consisting of bicarbonate (weak acid) and carbonate (conjugate base). To calculate the pH, we apply the Henderson-Hasselbalch equation:

pH = pK_a2 + log( [A^-] / [HA] )

Substituting the values, we have:

pH = 10.30 + log(10 millimoles / 15 millimoles)

This results in a pH of approximately 10.12. Understanding the stoichiometric relationships and the buffer system formed after the first equivalence point is crucial for accurately determining the pH during titration processes.

Here’s what students ask on this topic:

What are the two equivalence points in a diprotic acid titration?

In a diprotic acid titration, there are two equivalence points. The first equivalence point occurs when the first proton (H⁺) is completely neutralized by the titrant, resulting in the formation of the intermediate conjugate base. The second equivalence point is reached when the second proton is neutralized, leading to the formation of the final conjugate base. These points are crucial for calculating the titration curve and understanding the pH changes during the titration process.

How do you calculate the first equivalence volume in a diprotic acid titration?

The first equivalence volume in a diprotic acid titration can be calculated using the equation:

$C_{1}V=C_{2}V$

Here, $C_1$ and $C_2$ are the molarities of the acid and base, respectively, and $V$ represents the volumes. By rearranging the equation and solving for the volume of the titrant, you can determine the first equivalence volume.

How is the Henderson-Hasselbalch equation used in diprotic acid titrations?

The Henderson-Hasselbalch equation is used to determine the pH of a buffer solution before reaching the equivalence point in a diprotic acid titration. The equation is:

$\mathrm{pH}=\mathrm{pK}{a}_1+\frac{\log (\frac{\mathrm{[A}-]}{\mathrm{[HA]}}}{}$

Here, $\mathrm{pK}{a}_1$ is the negative logarithm of the first dissociation constant, $\mathrm{[A}-$ is the concentration of the conjugate base, and $\mathrm{[HA]}$ is the concentration of the weak acid. This equation helps in calculating the pH of the solution as the titration progresses.

What happens to the pH at the first equivalence point in a diprotic acid titration?

At the first equivalence point in a diprotic acid titration, the pH experiences a significant jump. This is because the weak acid and the strong base have completely neutralized each other, forming the intermediate conjugate base. The buffer system is destroyed, leading to a noticeable increase in pH. The pH at this point can be calculated using the concentration of the conjugate base formed and the relevant equilibrium expressions.

How do you calculate the pH after the second equivalence point in a diprotic acid titration?

After the second equivalence point in a diprotic acid titration, the pH is determined by the excess strong base. To calculate the pH, first find the concentration of the excess base using:

$\mathrm{Concentration}=\frac{\mathrm{Initial}\mathrm{Concentration}\mathrm{of}\mathrm{Base}\times \mathrm{Excess}\mathrm{Volume}}{\mathrm{Total}\mathrm{Volume}}$

Then, calculate the pOH by taking the negative logarithm of the hydroxide ion concentration and convert it to pH by subtracting from 14:

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