PH Calculation Of Sodium Propionate Solution A Step-by-Step Guide

Introduction

In this comprehensive article, we will delve into the intricate process of calculating the pH of a 0.68 M solution of sodium propionate ($NaC_2H_5CO_2$) at a temperature of 25°C. This calculation requires a thorough understanding of acid-base chemistry, equilibrium constants, and the behavior of weak acids and their conjugate bases in aqueous solutions. Sodium propionate, being the salt of a weak acid (propionic acid, $HC_2H_5CO_2$), undergoes hydrolysis in water, leading to the formation of hydroxide ions and, consequently, a basic solution. To accurately determine the pH, we must consider the equilibrium established between propionate ions and water, utilizing the given $pK_a$ value of propionic acid (4.89) to calculate the base dissociation constant ($K_b$) for the propionate ion. This step is crucial as it allows us to quantify the extent of hydrolysis and the concentration of hydroxide ions in the solution. The subsequent calculations will involve setting up an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of the species involved, followed by the calculation of the $pOH$ and finally, the $pH$ of the solution. Throughout this article, we will provide a step-by-step explanation of each calculation, ensuring clarity and a deep understanding of the underlying principles. This detailed approach will not only help in solving this specific problem but also in tackling similar acid-base equilibrium problems with confidence. Understanding the nuances of such calculations is fundamental in various scientific disciplines, including chemistry, biology, and environmental science, where pH plays a critical role in many chemical and biological processes. By the end of this article, readers will have a solid grasp of how to calculate the pH of a solution containing the salt of a weak acid, a skill that is invaluable in both academic and practical settings.

Understanding the Chemistry

To accurately calculate the pH of a 0.68 M solution of sodium propionate ($NaC_2H_5CO_2$), it is essential to first understand the underlying chemical principles. Sodium propionate is the salt of propionic acid ($HC_2H_5CO_2$), a weak acid with a given $pK_a$ of 4.89. When sodium propionate dissolves in water, it dissociates completely into sodium ions ($Na^+$) and propionate ions ($C_2H_5CO_2^-$). The propionate ion, being the conjugate base of a weak acid, can accept a proton from water, leading to the formation of propionic acid and hydroxide ions ($OH^-$). This process, known as hydrolysis, is the key to understanding why the solution becomes basic. The equilibrium reaction for this process can be represented as follows:

$C_2H_5CO_2^-(aq) + H_2O(l) ightleftharpoons HC_2H_5CO_2(aq) + OH^-(aq)$

This equilibrium reaction dictates the concentration of hydroxide ions in the solution, which directly affects the pH. To quantify the extent of this hydrolysis, we need to determine the base dissociation constant ($K_b$) for the propionate ion. The $K_b$ is related to the acid dissociation constant ($K_a$) of propionic acid through the ion product of water ($K_w$):

$K_w = K_a imes K_b$

At 25°C, $K_w$ is approximately $1.0 imes 10^{-14}$. The $K_a$ can be calculated from the given $pK_a$ using the following relationship:

$K_a = 10^{-pK_a}$

Therefore, by calculating $K_a$ from the given $pK_a$ and using the $K_w$ value, we can determine the $K_b$ for the propionate ion. This $K_b$ value is crucial for setting up an ICE (Initial, Change, Equilibrium) table, which helps us determine the equilibrium concentrations of the species involved in the hydrolysis reaction. The ICE table allows us to track the changes in concentration as the reaction reaches equilibrium, providing a clear picture of the hydroxide ion concentration. Understanding these fundamental concepts and relationships is essential for accurately calculating the pH of the sodium propionate solution. The subsequent steps will involve applying these principles to set up the ICE table and calculate the equilibrium concentrations, ultimately leading to the determination of the pH.

Calculating the Base Dissociation Constant ($K_b$)

The calculation of the pH for a 0.68 M sodium propionate ($NaC_2H_5CO_2$) solution crucially depends on determining the base dissociation constant ($K_b$) for the propionate ion ($C_2H_5CO_2^-$). As previously mentioned, the propionate ion acts as a base in water, accepting a proton to form propionic acid ($HC_2H_5CO_2$) and hydroxide ions ($OH^-$). The extent of this reaction is quantified by the $K_b$ value. To find $K_b$, we first need to calculate the acid dissociation constant ($K_a$) for propionic acid, given its $pK_a$ value of 4.89. The relationship between $K_a$ and $pK_a$ is defined as:

$K_a = 10^{-pK_a}$

Substituting the given $pK_a$ value:

$K_a = 10^{-4.89}$

Calculating this value gives us:

$K_a imes 1.29 imes 10^{-5}$

Now that we have the $K_a$ value, we can calculate the $K_b$ using the relationship between $K_a$, $K_b$, and the ion product of water ($K_w$):

$K_w = K_a imes K_b$

At 25°C, $K_w$ is approximately $1.0 imes 10^{-14}$. Rearranging the equation to solve for $K_b$:

K_b = rac{K_w}{K_a}

Substituting the values for $K_w$ and $K_a$:

K_b = rac{1.0 imes 10^{-14}}{1.29 imes 10^{-5}}

Calculating this gives us the $K_b$ value:

$K_b imes 7.75 imes 10^{-10}$

This $K_b$ value is essential for the next step, which involves setting up an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of the species involved in the hydrolysis reaction. The small value of $K_b$ indicates that the propionate ion is a weak base, and the hydrolysis reaction will not proceed to completion. This means that only a small fraction of the propionate ions will react with water to form propionic acid and hydroxide ions. The accurate calculation of $K_b$ is a critical step in determining the pH of the sodium propionate solution, as it provides the foundation for understanding the equilibrium concentrations of hydroxide ions, which directly influence the pH.

Setting up the ICE Table

The process of calculating pH for a 0.68 M sodium propionate solution requires setting up an ICE (Initial, Change, Equilibrium) table. This table is a crucial tool for organizing the concentrations of reactants and products in an equilibrium reaction, allowing us to determine the equilibrium concentrations of the species involved. For the hydrolysis of the propionate ion ($C_2H_5CO_2^-$) in water, the equilibrium reaction is:

$C_2H_5CO_2^-(aq) + H_2O(l) ightleftharpoons HC_2H_5CO_2(aq) + OH^-(aq)$

The ICE table is structured with three rows representing the Initial concentrations, the Change in concentrations, and the Equilibrium concentrations. The first column represents the propionate ion, the second column represents propionic acid ($HC_2H_5CO_2$), and the third column represents hydroxide ions ($OH^-$). Water is a pure liquid and its concentration does not change significantly, so it is not included in the ICE table.

Here's how we set up the ICE table:

• Initial (I): At the beginning, we have a 0.68 M solution of sodium propionate, which dissociates completely into 0.68 M propionate ions. The initial concentrations of propionic acid and hydroxide ions are approximately 0 M, as the hydrolysis reaction has not yet occurred.
• Change (C): As the reaction proceeds towards equilibrium, some of the propionate ions will react with water, forming propionic acid and hydroxide ions. We represent this change with '-x' for the propionate ion and '+x' for both propionic acid and hydroxide ions, where 'x' is the change in concentration.
• Equilibrium (E): The equilibrium concentrations are the sum of the initial concentrations and the changes in concentrations. For propionate ions, it is 0.68 - x, and for both propionic acid and hydroxide ions, it is x.

The ICE table can be summarized as follows:

Species	$C_2H_5CO_2^-$	$HC_2H_5CO_2$	$OH^-$
Initial (I)	0.68	0	0
Change (C)	-x	+x	+x
Equilibrium (E)	0.68 - x	x	x

This ICE table is a critical tool for determining the equilibrium concentrations of hydroxide ions, which are essential for calculating the pH of the solution. By using the $K_b$ value calculated earlier and the equilibrium concentrations from the ICE table, we can set up an equation to solve for 'x', the equilibrium concentration of hydroxide ions. The small value of $K_b$ often allows us to make an approximation that simplifies the calculation, making it easier to solve for 'x'. This step is crucial in the overall process of determining the pH of the sodium propionate solution.

Calculating the Hydroxide Ion Concentration ([OH-])

The next crucial step in pH calculation for the 0.68 M sodium propionate solution involves using the ICE table and the base dissociation constant ($K_b$) to determine the hydroxide ion concentration ($[OH^-]$). From the ICE table, we established the equilibrium concentrations of the species involved in the hydrolysis reaction. The equilibrium expression for the reaction:

$C_2H_5CO_2^-(aq) + H_2O(l) ightleftharpoons HC_2H_5CO_2(aq) + OH^-(aq)$

can be written in terms of $K_b$ as follows:

K_b = rac{[HC_2H_5CO_2][OH^-]}{[C_2H_5CO_2^-]}

Substituting the equilibrium concentrations from the ICE table, we get:

K_b = rac{x imes x}{0.68 - x} = rac{x^2}{0.68 - x}

We previously calculated $K_b$ to be approximately $7.75 imes 10^{-10}$. Now, we can set up the equation:

7.75 imes 10^{-10} = rac{x^2}{0.68 - x}

Since $K_b$ is very small, we can assume that 'x' is much smaller than 0.68, and thus, the term (0.68 - x) can be approximated as 0.68. This simplification makes the calculation much easier:

7.75 imes 10^{-10} imes rac{x^2}{0.68}

Now, we can solve for 'x':

$x^2 = 7.75 imes 10^{-10} imes 0.68$

$x^2 = 5.27 imes 10^{-10}$

Taking the square root of both sides:

x = imes rac{5.27 imes 10^{-10}}

$x imes 2.29 imes 10^{-5} M$

This value of 'x' represents the equilibrium concentration of hydroxide ions ($[OH^-]$) in the solution. It is important to verify the assumption we made earlier that 'x' is much smaller than 0.68. In this case, $2.29 imes 10^{-5}$ is indeed much smaller than 0.68, so our approximation is valid. Now that we have the hydroxide ion concentration, we can proceed to calculate the $pOH$ and subsequently the pH of the solution. This step is a critical link in the chain of calculations required to determine the final pH value.

Calculating the pOH and pH

The final step in calculating the pH of the 0.68 M sodium propionate solution involves using the calculated hydroxide ion concentration ($[OH^-]$) to first find the $pOH$ and then determine the pH. We previously calculated the hydroxide ion concentration to be approximately $2.29 imes 10^{-5} M$. The $pOH$ is defined as the negative logarithm (base 10) of the hydroxide ion concentration:

$pOH = -log[OH^-]$

Substituting the value of $[OH^-]$:

$pOH = -log(2.29 imes 10^{-5})$

Calculating this gives us:

$pOH imes 4.64$

Now that we have the $pOH$, we can calculate the pH using the relationship between pH, $pOH$, and the ion product of water ($K_w$):

$pH + pOH = 14$

This relationship is derived from the fact that the sum of pH and $pOH$ is equal to the negative logarithm of $K_w$, which is 14 at 25°C. Rearranging the equation to solve for pH:

$pH = 14 - pOH$

Substituting the calculated $pOH$ value:

$pH = 14 - 4.64$

$pH imes 9.36$

The problem asks for the answer to be rounded to one decimal place. Therefore, the pH of the 0.68 M sodium propionate solution at 25°C is approximately 9.4.

This final calculation provides the answer to the problem and demonstrates the step-by-step process of determining the pH of a solution containing the salt of a weak acid. The pH value of 9.4 indicates that the solution is basic, which is expected since sodium propionate is the salt of a weak acid and a strong base. The hydrolysis of the propionate ion generates hydroxide ions, leading to this basic pH. This detailed calculation not only solves the specific problem but also illustrates the general method for calculating the pH of such solutions, providing a valuable understanding of acid-base chemistry and equilibrium principles.

Conclusion

In conclusion, we have successfully calculated the pH of a 0.68 M solution of sodium propionate ($NaC_2H_5CO_2$) at 25°C. This calculation involved several key steps, each building upon the previous one to arrive at the final answer. We began by understanding the chemistry of the system, recognizing that sodium propionate is the salt of a weak acid, propionic acid ($HC_2H_5CO_2$), and that the propionate ion ($C_2H_5CO_2^-$) undergoes hydrolysis in water, producing hydroxide ions and making the solution basic. The given $pK_a$ of propionic acid (4.89) was crucial in determining the acid dissociation constant ($K_a$) and subsequently the base dissociation constant ($K_b$) for the propionate ion. This $K_b$ value is a quantitative measure of the strength of the propionate ion as a base and is essential for understanding the extent of the hydrolysis reaction.

We then used the calculated $K_b$ value to set up an ICE (Initial, Change, Equilibrium) table, which allowed us to organize and track the concentrations of the species involved in the hydrolysis reaction. The ICE table provided a clear picture of the equilibrium concentrations of propionate ions, propionic acid, and hydroxide ions. By substituting these equilibrium concentrations into the $K_b$ expression, we were able to solve for the hydroxide ion concentration ($[OH^-]$). A key simplification was made by assuming that the change in concentration ('x') was much smaller than the initial concentration of the propionate ion, which allowed us to avoid solving a quadratic equation. This assumption was validated by the small value of $K_b$.

Once the hydroxide ion concentration was determined, we calculated the $pOH$ using the formula $pOH = -log[OH^-]$. The $pOH$ value was then used to find the pH using the relationship $pH + pOH = 14$. The final calculated pH of the solution was approximately 9.4, rounded to one decimal place as requested. This pH value confirms that the solution is indeed basic, as expected for a solution of the salt of a weak acid. This detailed calculation demonstrates the importance of understanding acid-base equilibria and the application of equilibrium constants in determining the pH of solutions. The step-by-step approach used in this article provides a clear and comprehensive method for tackling similar pH calculation problems in chemistry and related fields. The ability to accurately calculate pH is fundamental in many scientific disciplines, making this a valuable skill for students and professionals alike.