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What is the pH of 0.001 M Aniline Solution?

Understanding the pH of chemical solutions is crucial, particularly in the chemical industry, where the behavior of substances in different environments is essential for various applications. One such compound of interest is aniline, a weak base commonly used in the production of dyes, drugs, and other chemicals. In this article, we will explore what the pH of a 0.001 M aniline solution is, delving into the principles behind its calculation and the factors that influence it.

1. Aniline as a Weak Base

Aniline, with the chemical formula ( C6H5NH_2 ), is an aromatic amine. Unlike strong bases, which fully dissociate in water, aniline is a weak base. This means that in an aqueous solution, aniline only partially ionizes, establishing an equilibrium between the unionized aniline molecules and the ions it produces. The equation for the ionization of aniline in water is as follows:

[ C6H5NH2 + H2O \rightleftharpoons C6H5NH_3^+ + OH^- ]

The extent of this ionization is characterized by the base dissociation constant ( K_b ), a value that is much smaller than that of strong bases, indicating a low degree of ionization.

2. Calculating the pH of 0.001 M Aniline Solution

To calculate the pH of a 0.001 M aniline solution, it is essential first to determine the concentration of hydroxide ions (OH⁻) produced in the solution. The dissociation constant ( Kb ) for aniline is typically around ( 4.3 \times 10^{-10} ). The expression for ( Kb ) can be written as:

[ Kb = \frac{[C6H5NH3^+][OH^-]}{[C6H5NH_2]} ]

For a 0.001 M aniline solution, let ( x ) represent the concentration of ( OH^- ) ions at equilibrium. Since the concentration of aniline, ( [C6H5NH_2] ), will decrease slightly due to ionization, it can be approximated as ( 0.001 - x ). Given that ( x ) is small, we simplify this to ( 0.001 ) M. Substituting into the equation gives:

[ K_b = \frac{x^2}{0.001} ]

Solving for ( x ), we find:

[ x = \sqrt{K_b \times 0.001} = \sqrt{4.3 \times 10^{-10} \times 0.001} = 2.07 \times 10^{-7} \, M ]

The concentration of ( OH^- ) ions is therefore ( 2.07 \times 10^{-7} ) M. To find the pH, we first calculate the pOH:

[ pOH = -\log[OH^-] = -\log(2.07 \times 10^{-7}) \approx 6.68 ]

Since ( pH + pOH = 14 ), the pH of the solution is:

[ pH = 14 - 6.68 = 7.32 ]

Thus, the pH of a 0.001 M aniline solution is approximately 7.32, which is slightly basic, consistent with aniline being a weak base.

3. Factors Influencing the pH of Aniline Solutions

While the pH of a 0.001 M aniline solution is approximately 7.32 under standard conditions, several factors can influence this value. Temperature is one such factor, as ( K_b ) is temperature-dependent. An increase in temperature typically increases the ionization of weak bases, which could slightly raise the pH of the solution. Additionally, the presence of other substances in the solution, such as salts or acids, could impact the equilibrium, leading to variations in pH.

In conclusion, understanding the pH of a 0.001 M aniline solution requires a clear grasp of the principles of weak base ionization. By carefully calculating the hydroxide ion concentration and understanding the factors that can affect it, we can determine that the pH is approximately 7.32, a value that reflects the weakly basic nature of aniline.