PH Calculation Of Ethylammonium Bromide Solution A Step-by-Step Guide

Introduction

In this comprehensive article, we will delve into the process of calculating the pH of a 0.36 M solution of ethylammonium bromide ($C_2H_5NH_3Br$) at a temperature of 25°C. Ethylammonium bromide is the salt of a weak base, ethylamine ($C_2H_5NH_2$), and a strong acid, hydrobromic acid (HBr). This characteristic makes it an acidic salt, as it will undergo hydrolysis in water, leading to the formation of hydronium ions ($H_3O^+$) and a decrease in pH. To accurately determine the pH, we'll utilize the given $pK_b$ value of ethylamine, which is 3.19, and apply equilibrium principles. This calculation is crucial in various fields, including chemistry, biology, and environmental science, where understanding the acidity or basicity of solutions is essential. The pH value influences chemical reactions, biological processes, and the behavior of substances in different environments. Therefore, mastering the calculation of pH for solutions like ethylammonium bromide is a fundamental skill for anyone working in these disciplines.

Understanding the Chemistry Behind the Calculation

To begin, it's essential to grasp the underlying chemistry involved. Ethylammonium bromide ($C_2H_5NH_3Br$) dissociates completely in water to form ethylammonium ions ($C_2H_5NH_3^+$) and bromide ions ($Br^−$). The ethylammonium ion, being the conjugate acid of the weak base ethylamine ($C_2H_5NH_2$), will react with water in a hydrolysis reaction. This reaction leads to the formation of ethylamine and hydronium ions ($H_3O^+$), which contribute to the acidity of the solution. The bromide ion, being the conjugate base of a strong acid (HBr), does not significantly affect the pH.

The equilibrium reaction for the hydrolysis of the ethylammonium ion can be represented as follows:

$C_2H_5NH_3^+(aq) + H_2O(l) ightleftharpoons C_2H_5NH_2(aq) + H_3O^+(aq)$

To calculate the pH, we need to determine the concentration of hydronium ions ($[H_3O^+]$) at equilibrium. This requires the use of the acid dissociation constant ($K_a$) for the ethylammonium ion. However, we are given the $pK_b$ value for ethylamine. We can use the relationship between $K_a$, $K_b$, and the ion product of water ($K_w$) to find $K_a$. The relationship is:

$K_a imes K_b = K_w$

At 25°C, $K_w$ is approximately $1.0 imes 10^{-14}$. We can calculate $K_b$ from the given $pK_b$ using the formula:

$K_b = 10^{-pK_b} = 10^{-3.19}$

Once we have $K_b$, we can calculate $K_a$ and then set up an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of the species involved in the hydrolysis reaction. This will allow us to find $[H_3O^+]$ and subsequently calculate the pH.

Step-by-Step Calculation of pH

Let's proceed with the step-by-step calculation to determine the pH of the 0.36 M ethylammonium bromide solution at 25°C.

1. Calculate $K_b$ from $pK_b$

Given that the $pK_b$ of ethylamine ($C_2H_5NH_2$) is 3.19, we can calculate $K_b$ using the formula:

$K_b = 10^{-pK_b} = 10^{-3.19}

K_b ext{ ≈ } 6.46 imes 10^{-4}$

2. Calculate $K_a$ from $K_b$

We know that $K_a imes K_b = K_w$, where $K_w = 1.0 imes 10^{-14}$ at 25°C. Therefore, we can calculate $K_a$ as follows:

$K_a = rac{K_w}{K_b} = rac{1.0 imes 10^{-14}}{6.46 imes 10^{-4}}

K_a ext{ ≈ } 1.55 imes 10^{-11}$

3. Set up an ICE Table

The hydrolysis reaction is:

$C_2H_5NH_3^+(aq) + H_2O(l) ightleftharpoons C_2H_5NH_2(aq) + H_3O^+(aq)$

We can set up an ICE table to track the concentrations:

4. Write the $K_a$ Expression and Solve for x

The acid dissociation constant $K_a$ expression is:

K_a = rac{[C_2H_5NH_2][H_3O^+]}{[C_2H_5NH_3^+]} = rac{x^2}{0.36 - x}

Since $K_a$ is very small, we can assume that x is much smaller than 0.36, so 0.36 - x ≈ 0.36. This simplifies the equation:

1.55 imes 10^{-11} = rac{x^2}{0.36}

Solving for x:

$x^2 = 1.55 imes 10^{-11} imes 0.36

x^2 = 5.58 imes 10^{-12}

x = ext{√}(5.58 imes 10^{-12})

x ext{ ≈ } 2.36 imes 10^{-6}$

5. Calculate $[H_3O^+]$

The value of x represents the equilibrium concentration of hydronium ions, $[H_3O^+]$:

$[H_3O^+] = x ext{ ≈ } 2.36 imes 10^{-6} ext{ M}$

6. Calculate pH

Now we can calculate the pH using the formula:

$pH = -log[H_3O^+]

pH = -log(2.36 imes 10^{-6})

pH ext{ ≈ } 5.63$

7. Round the Answer to 1 Decimal Place

Rounding the pH value to 1 decimal place, we get:

$pH ext{ ≈ } 5.6$

Conclusion

In this detailed calculation, we determined the pH of a 0.36 M solution of ethylammonium bromide at 25°C. By understanding the hydrolysis of the ethylammonium ion and utilizing the given $pK_b$ value of ethylamine, we were able to calculate the $K_a$, set up an ICE table, and find the equilibrium concentration of hydronium ions. This allowed us to accurately calculate the pH of the solution, which was found to be approximately 5.6. This result highlights the acidic nature of ethylammonium bromide solutions due to the hydrolysis of the ethylammonium ion. This process demonstrates a fundamental aspect of acid-base chemistry and its practical application in determining solution acidity. Understanding these principles is crucial for various scientific disciplines, ensuring accurate predictions and interpretations of chemical behavior in aqueous solutions. This comprehensive guide not only provides a step-by-step solution but also reinforces the theoretical framework necessary for solving similar problems in chemistry and related fields. Mastering these calculations enables a deeper understanding of chemical reactions and their implications in real-world scenarios.