Titrations and Titration Curves: Videos & Practice Problems

Redox titration curves illustrate the concentration changes of the analyte and titrant as titrant volume is added. The equivalence volume is calculated using the equation: $CV=CV$. At the equivalence point, equal moles of analyte and titrant form a precipitate, and the concentration is determined using the solubility product constant (Ksp). Post-equivalence, excess titrant concentration is calculated to find the analyte concentration, demonstrating the principles of redox reactions and equilibrium.

Redox Titration Curves

Redox titration curves illustrate the relationship between the concentration of either the analyte or the titrant and the volume of titrant added during a titration process. As titrant is added incrementally, the concentration of the species involved changes, allowing for the determination of the equivalence volume, which is crucial for understanding the titration's progress.

For example, consider the titration of 50 mL of 0.100 M sodium chloride (NaCl) with 0.100 M silver nitrate (AgNO₃). This reaction leads to the formation of silver chloride (AgCl) as a precipitate. In this case, the sodium ions (Na⁺) combine with nitrate ions (NO₃^-), resulting in an aqueous compound, while the silver ions (Ag⁺) react with chloride ions (Cl^-) to form the solid precipitate AgCl. The solubility product constant (K_sp) is essential in this context, as it describes the equilibrium between the solid and its ions in solution.

When discussing K_sp, it is important to note that reversing the reaction yields the formation constant (K), which is the inverse of K_sp. This means that if the original reaction has a K_sp, the formation constant for the solid is expressed as K = 1/K_sp. A K value significantly greater than 1 indicates that the formation of the solid is highly favorable, consistent with the known solubility rules for silver chloride.

To calculate the equivalence volume of the titrant, the following relationship is used: the molarity of the analyte (A) multiplied by its volume equals the molarity of the titrant (T) multiplied by the equivalence volume (V_eq) of the titrant. This can be expressed mathematically as:

M_A × V_A = M_T × V_eq

In this example, substituting the known values gives:

0.100 M × 50 mL = 0.100 M × V_eq

By solving for V_eq, we find that the equivalence volume of the titrant is 50 mL. This calculation is a fundamental step in constructing the redox titration curve, which will depict the changes in concentration as titrant is added, ultimately leading to a better understanding of the titration process.

Redox Titration Curves

In a titration process, prior to reaching the equivalence point, there exists an excess of the analyte, which in this case is chloride ion (Cl^-). To determine the concentration of the analyte at this stage, we calculate the amount that remains unreacted after adding a certain volume of titrant, which is represented by AGL⁺. The titrant reacts with the analyte to form a precipitate, and the concentration of the analyte can be expressed using the formula:

\[C_{\text{analyte}} = \frac{n_{\text{initial}} - n_{\text{titrant}}}{V_{\text{total}}}\]

Here, \(n\) represents the number of moles, which can be calculated as the product of volume (in liters or milliliters) and molarity. The term "of" in the context of moles indicates multiplication, allowing for flexibility in units. For instance, if we use milliliters, the calculation remains valid as:

\[n = V_{\text{ml}} \times C_{\text{molarity}}\]

To illustrate, if we have 50 mL of 0.100 M NaCl, this results in 5 millimoles of analyte. After adding 20 mL of 0.100 M titrant, which corresponds to 2 millimoles, the calculation for the concentration of the unreacted analyte becomes:

\[C_{\text{analyte}} = \frac{5 \, \text{mmol} - 2 \, \text{mmol}}{50 \, \text{mL} + 20 \, \text{mL}} = 0.042857 \, \text{M}\]

Next, to find the pH of the solution, we utilize the relationship between pH and the concentration of the chloride ion, expressed as:

\[pCl = -\log[C_{\text{Cl}^-}]\]

Substituting the calculated concentration yields:

\[pCl = -\log[0.042857] \approx 1.37\]

This indicates that before reaching the equivalence point, the pCl value is 1.37. As titrant is added, we can expect this value to increase, reflecting the changes in concentration as the reaction progresses. Understanding these calculations is crucial for interpreting titration curves and the behavior of solutions during titration.

Redox Titration Curves

At the equivalence point of a titration, the moles of the analyte and titrant are equal, resulting in the formation of a precipitate that exists in equilibrium with its constituent ions. To determine the concentration of the analyte, the solubility product constant (Ksp) of the precipitate, such as silver chloride (AgCl), is utilized. The dissociation of silver chloride can be represented as follows:\[\text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq)\]In this equilibrium, the solid AgCl is ignored in the Ksp expression, which focuses solely on the aqueous ions. The Ksp expression for silver chloride is given by:\[K_{sp} = [\text{Ag}^+][\text{Cl}^-]\]Assuming that at equilibrium, the concentrations of silver ions and chloride ions are both equal to \(x\), the Ksp can be expressed as:\[K_{sp} = x \cdot x = x^2\]If the stoichiometry were different, such as having two chloride ions for every silver ion, the expression would adjust accordingly:\[K_{sp} = [\text{Ag}^+][\text{Cl}^-]^2 = x \cdot (2x)^2 = 4x^3\]To find the concentration of chloride ions, \(x\), one would take the square root of the Ksp value:\[x = \sqrt{K_{sp}}\]For example, if \(K_{sp} = 1.8 \times 10^{-10}\), then:\[x = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \text{ M}\]This value represents the concentration of chloride ions. The pCl, which is the negative logarithm of the chloride ion concentration, can be calculated as:\[pCl = -\log[\text{Cl}^-] = -\log(1.34 \times 10^{-5}) \approx 4.87\]As titrant is added, the pCl value increases, reflecting the changes in chloride ion concentration. After reaching the equivalence point, further analysis is required to determine the final pCl, which will depend on the excess titrant present.

Redox Titration Curves

In a redox titration, after reaching the equivalence point, the solution contains an excess of the titrant, which is crucial for determining the concentration of the analyte. For instance, if silver ions are used as the titrant, the calculation begins by determining the moles of titrant added. This can be calculated by multiplying the volume of the titrant (in milliliters) by its molarity. For example, if 70 mL of a 0.100 M silver nitrate solution is added, it results in 7 millimoles of silver ions. To find the concentration of the analyte, the moles of the analyte must be subtracted from the total moles of titrant added.

The next step involves calculating the total volume of the solution, which is the sum of the volumes of the titrant and the analyte. In this case, the total volume is 50 mL (analyte) + 70 mL (titrant) = 120 mL. The concentration of the excess titrant can then be calculated by dividing the remaining moles of titrant by the total volume, yielding a concentration of 0.0167 M.

To find the concentration of the analyte, the solubility product constant (K_sp) is utilized. The equilibrium expression for K_sp is given by:

$$ K_{sp} = [\text{Titrant}][\text{Analyte}] $$

Substituting the known values, where K_sp is 1.8 × 10^-10 and the concentration of the titrant is 0.0167 M, allows for the calculation of the analyte concentration. Rearranging the equation to solve for the analyte concentration gives:

$$ [\text{Analyte}] = \frac{K_{sp}}{[\text{Titrant}]} $$

Plugging in the values results in an analyte concentration of approximately 1.0778 × 10^-8 M. Taking the negative logarithm of this concentration provides the pCl value, which is calculated to be 7.97. This increase in pCl reflects the changes occurring throughout the titration process.

When constructing a titration curve, the volume of titrant is plotted against the negative log of the analyte concentration. Initially, the curve shows a gradual increase, followed by a sharp rise at the equivalence point, where the moles of titrant equal the moles of analyte. After this point, the curve levels off. An indicator can be used to visually identify the endpoint of the titration, which typically occurs near the equivalence volume.

In laboratory settings, this method can also help identify unknown analytes, applying similar principles as those used in acid-base titrations. Understanding whether the system is at, before, or after the equivalence point is essential for selecting the appropriate equations to derive the final results.

Redox Titration Curves Calculations

In this scenario, we are tasked with calculating the fluoride ion concentration resulting from the titration of potassium fluoride (KF) with barium chloride (BaCl₂). The initial conditions include 130 mL of 0.120 M potassium fluoride and 150 mL of 0.100 M barium chloride. The solubility product constant (K_sp) for barium fluoride (BaF₂) is given as 1.5 × 10⁻⁶.

To determine the equivalence volume, we apply the formula for titration:

M_analyte × V_analyte = M_titrant × V_titrant

Substituting the known values:

0.120 M × 130 mL = 0.100 M × V_titrant

Solving for V_titrant gives:

V_titrant = (0.120 × 130) / 0.100 = 156 mL

Since the available volume of barium chloride is only 150 mL, we conclude that the titration is occurring before the equivalence point, indicating an excess of fluoride ions (the analyte). The fluoride ion concentration can be calculated using the formula:

[F⁻] = (moles of F⁻ - moles of Ba²⁺) / total volume

First, we calculate the millimoles of each component:

For fluoride ions:

Millimoles of F⁻ = 0.120 M × 130 mL = 15.6 mmoles

For barium ions:

Millimoles of Ba²⁺ = 0.100 M × 150 mL = 15 mmoles

Now, substituting these values into the concentration formula:

[F⁻] = (15.6 mmoles - 15 mmoles) / (130 mL + 150 mL) = 0.0014 M

Thus, the concentration of fluoride ions remaining before reaching the equivalence point is:

[F⁻] = 2.14 × 10⁻³ M

This calculation illustrates the importance of understanding titration dynamics and the relationship between reactants and products in a chemical reaction, particularly in determining concentrations before reaching the equivalence point.

Calculate the pCN from the titration of 40.0 mL of 0.060 M NaCN with 20.0 mL of 0.050 M AgC₂H_­3O₂. The solubility product constant of AgCN is 2.2 x 10^-16.

• A. 1.53
• B. 0.023
• C. 1.45
• D. 1.63
• E. 0.029

Answer: (D)

Here’s what students ask on this topic:

What is a redox titration curve and how is it different from an acid-base titration curve?

A redox titration curve tracks the concentration changes of the analyte and titrant as the titrant volume is added. Unlike acid-base titration curves, which focus on pH changes, redox titration curves focus on the redox potential or concentration of specific ions. The equivalence point in redox titrations is determined by the stoichiometric reaction between the analyte and titrant, often involving a change in oxidation states. The curve typically shows a sharp change at the equivalence point, similar to acid-base titrations, but the calculations often involve redox potentials and solubility product constants (K_sp).

How do you calculate the equivalence volume in a redox titration?

The equivalence volume in a redox titration is calculated using the equation:

$C_1{V}_1=C_2{V}_2$

where C₁ and V₁ are the concentration and volume of the analyte, and C₂ and V₂ are the concentration and volume of the titrant. By rearranging the equation to solve for V₂, you can determine the volume of titrant needed to reach the equivalence point. For example, if you have 50 mL of 0.100 M NaCl and titrate with 0.100 M AgNO₃, the equivalence volume is 50 mL.

What happens at the equivalence point in a redox titration?

At the equivalence point in a redox titration, the moles of analyte and titrant are equal, resulting in the formation of a precipitate or a change in oxidation state. For example, in the titration of NaCl with AgNO₃, Ag⁺ ions react with Cl⁻ ions to form AgCl precipitate. The concentration of the analyte is determined using the solubility product constant (K_sp) of the precipitate. The K_sp expression for AgCl is:

$K_{\mathrm{sp}}=x^2$

where x is the concentration of Ag⁺ and Cl⁻ ions at equilibrium.

How do you determine the concentration of the analyte after the equivalence point in a redox titration?

After the equivalence point, there is an excess of titrant. To determine the analyte concentration, first calculate the excess titrant concentration. For example, if 70 mL of 0.100 M AgNO₃ is added to 50 mL of 0.100 M NaCl, the excess Ag⁺ concentration is:

$\frac{7-5}{50+70}=0.0167$

Then, use the K_sp of AgCl to find the analyte concentration:

$K_{\mathrm{sp}}=C_{\mathrm{titrant}}{C}_{\mathrm{analyte}}$

Solving for the analyte concentration gives:

$C_{\mathrm{analyte}}=\frac{K_{\mathrm{sp}}}{C_{\mathrm{titrant}}}$

What is the role of the solubility product constant (Ksp) in redox titrations?

The solubility product constant (K_sp) plays a crucial role in redox titrations involving precipitates. K_sp is used to determine the concentration of ions in a saturated solution at equilibrium. For example, in the titration of NaCl with AgNO₃, Ag⁺ and Cl⁻ ions form AgCl precipitate. The K_sp expression for AgCl is:

$K_{\mathrm{sp}}=x^2$

where x is the concentration of Ag⁺ and Cl⁻ ions at equilibrium. By knowing K_sp, you can calculate the ion concentrations at the equivalence point and beyond, helping to construct the titration curve and understand the reaction dynamics.