Conductometric Titration

The titrations in which the end point detection of acid-base reaction is measured using the conductance, are known as conductometric titrations. Another name of conductometric titrations is conductometry in which the conductivity of a solution is measured with the help of electrodes dipped in that solution to determine the concentration of analyte.

Principle of Conductometric Titration

Principle of Conductometric titration relies on Kohlrausch's law, that states that the ions available in a solution independently involve in the electrical conductivity of that solution. During the titration, alteration in the number and identity of the ions available in a solution affects the conductivity of that solution. Thus, during the experiment the conductivity of the solution can be determined.

In conductometry, there are two types of ions (one ion from titrant and second ion from analyte) one of the ions is replaced by another ion during the titration.

These two ions alter the conductivity of solution during the entire titration process as they differ from each other in ionic conductivity. In conductometry, ions can come from both titrants and the analyte (the substance being analyzed).

Titrants and Analyte

1. Titrants: The titrant is the solution of known concentration that is added during the titration. It contains ions that contribute to the conductivity of the solution. For example:

• If the titrant is a strong acid like hydrochloric acid (HCl), it will release H⁺ ions.
• If the titrant is a strong base like sodium hydroxide (NaOH), it will release OH⁻ ions.
• If the titrant is a salt, it will dissociate into its corresponding cations and anions.

2. Analyte: The analyte is the substance being titrated. It also contains ions that contribute to the conductivity. For example:

• If the analyte is a salt, it will dissociate into cations and anions.
• If the analyte is an acid, it will dissociate into hydrogen ions (H⁺) and the corresponding anion.
• If the analyte is a base, it will dissociate into hydroxide ions (OH⁻) and the corresponding cation.

As the titration progresses, the ions from both the titrant and the analyte interact, and changes in conductivity are measured to determine the endpoint of the titration. The key to conductometric titration is monitoring how the concentration of ions in the solution changes as the titrant is added.

End Point Determination

End point determination is possible graphically when the change in conductance is plotted against the consumed titrant volume.

The ions present in a solution in a chemical cell are responsible for the flow of electric current. When electrodes are applied with potential difference, current flows through the solution depending on the concentration and types of ions present in the solution.

The ease with which the current flows through the solution is based on conductance.

It is known that at infinite dilutions or in very dilute solutions, ions act independent of each other and they contribute to the conductance of the solution.

Both cations and anions have varying degree of ionic mobilities (or conductance values). Thus, when a solution of one electrolyte is added (as a titrant) to the solution of another electrolyte the overall conductance (after addition) will depend whether a reaction occurs or not.

If no chemical reaction occurs between the electrolyte solution and another added to it, the overall conductance of the solution will increase.

All ions will contribute to the conductance of the solution for example addition of sodium nitrate solution to the sodium chloride solution.

In conductometric titration, the endpoint is the point at which the reaction between the titrant and the analyte is complete, and there is a sharp change in the conductivity of the solution.

To calculate the endpoint, the process involves careful monitoring of the conductivity during the titration and analyzing the data. Below are the steps to calculate the endpoint:

1. Graph of Conductivity vs. Volume of Titrant:

• First, the conductivity of the solution is measured continuously as the titrant is added.
• A graph is plotted with conductivity (y-axis) against the volume of titrant added (x-axis).
• The curve typically shows a gradual change in conductivity as the titrant is added, followed by a sharp inflection or steep slope at the endpoint.

2. Identifying the Endpoint:

• The endpoint corresponds to the point of sharp change in the slope of the curve. This is where the rate of conductivity change is maximum, and it corresponds to the equivalence point of the titration.

Strong Acid with Strong Base

When hydrochloric acid (HCl) is added to a solution of sodium hydroxide (NaOH), a neutralization reaction takes place.

The change in conductivity during this reaction can be explained step-by-step.

1. Before Adding HCl (Initial State):

• Sodium hydroxide (NaOH) is a strong base and dissociates completely in water:
   NaOH (aq)→Na+(aq) + OH−(aq)
• The solution will have high conductivity due to the presence of the ions Na⁺ and OH⁻.

2. As HCl is Added (During the Titration):

• HCl is a strong acid and dissociates completely in water:
    HCl (aq)→H+(aq)+Cl−(aq)
• Initially, as small amounts of HCl are added to the NaOH solution, the OH⁻ ions from NaOH will neutralize the H⁺ ions from HCl:
    OH−(aq)+H+(aq)→H2O (l)
• In this phase, the conductivity will decrease because the OH⁻ ions (which contribute significantly to conductivity) are being neutralized and removed from the solution. The only ions left in the solution are Na⁺ and Cl⁻, both of which contribute less to conductivity compared to OH⁻.
• So, the overall conductivity decreases as the titration progresses.>

3. At the Equivalence Point:

• At the equivalence point, all the OH⁻ ions have reacted with H⁺ ions, forming water:
    NaOH (aq)+HCl (aq)→NaCl (aq)+H2O (l)
• The solution now contains only Na⁺ and Cl⁻ ions, which are spectator ions and do not significantly contribute to conductivity.
• As a result, the conductivity will be at its lowest at the equivalence point, because there is no strong ion (like OH⁻ or H⁺) left in the solution to conduct electricity.

4. After the Equivalence Point (Excess HCl):

• After the equivalence point, if excess HCl is added, the concentration of H⁺ ions will increase.
• This will lead to an increase in the number of H⁺ ions in the solution, which will increase the conductivity because H⁺ is highly conductive.
• As more HCl is added, the conductivity increases again due to the increased concentration of H⁺ ions.

When hydrochloric acid (HCl) is added to a solution of sodium hydroxide (NaOH), a neutralization reaction takes place. The change in conductivity during this reaction can be explained step-by-step:

1. Before Adding HCl: NaOH dissociates into Na⁺ and OH⁻ → high conductivity.
2. During Addition: HCl dissociates into H⁺ and Cl⁻, OH⁻ is neutralized by H⁺ → conductivity decreases.
3. At Equivalence Point: Only Na⁺ and Cl⁻ remain → lowest conductivity.
4. After Equivalence Point: Excess HCl adds more H⁺ → conductivity increases again.

Strong Acid with Weak Base

Example: HCl + NH₄OH → NH₄Cl + H₂O. Conductivity initially increases due to H⁺ ions, then decreases when neutralized by OH⁻ ions. After equivalence, no significant change.

Same as the titration of the strong acid with strong base, it initially shows the increase in the conductivity because of the H ions.

This conductivity is decreased by the addition of the weak base that is with the NH₂OH that neutralises the H ions with the OH ions and decreases the conductivity.

The excess addition of the NH OH does not show the change in the conductivity.

Then the plot between the conductivity and the volume of the titrant shows the plateau.

Weak Acid with Strong Base

The weak acid such as acetic acid is titrated with the strong base such as sodium hydroxide.

Example: CH₃COOH + NaOH → CH₃COONa + H₂O.

The acetic acid dissociates to produce the H ions which shows the high conductivity and is titrated with the sodium hydroxide, which is dissociated to produce the OH ions which shows slight increase in the conductivity by the formation of the CH3COONa at the equivalence point.

Then it shows the gradual increase in the conductivity by the addition of excess titrant.

Then plot the graph between the conductivity and the volume of the titrant which shows the plateau.

Weak Acid with Weak Base

The weak acid such as the acetic acid is titrated with the weak base such as ammonium hydroxide.

Example: CH₃COOH + NH₄OH → CH₃COONH₄ + H₂O. Conductivity increases until equivalence,

The acetic acid is dissociated and it combines with the ammonium ion after dissociation of the ammonium hydroxide.

This forms the ammonium acetate salt which shows the increase in the conductivity.

After attaining the equivalence point, the addition of the titrant does not show the conductivity change. The plot between the conductivity and the volume of the titrant shows the plateau.

Calculate Concentration of Analyte

If you know the concentration of the titrant, you can calculate the concentration of the analyte (the substance being titrated) at the endpoint using the concept of stoichiometry.

Formula: C₁V₁ = C₂V₂

Where:

• C₁ = concentration of the titrant,
• V₁ = volume of the titrant added at the endpoint,
• C₂ = concentration of the analyte (which we need to calculate),
• V₂ = volume of the analyte solution.

Calculation Example (Strong Acid-Strong Base):

Suppose you are titrating NaOH (strong base) with HCl (strong acid).

• The equation for the neutralization reaction is: NaOH (aq)+HCl (aq)→NaCl (aq)+H2O (l)

• You know the concentration of HCl (C₁), and you monitor the conductivity of the NaOH solution as you add HCl.

• At the endpoint, note the volume of HCl (V₁) added.

• If you know the volume of NaOH (V₂), use the above formula to calculate the concentration of NaOH (C₂) at the equivalence point.

• Suppose you are titrating 50 mL of NaOH with 0.1 M HCl.

• The volume of HCl required to reach the endpoint (from the graph or first derivative) is 60 mL.

• Using the equation:
  
  Example: 50 mL NaOH titrated with 0.1 M HCl, endpoint at 60 mL HCl → concentration of NaOH = 0.12 M.