How to Determine pH and pOH in a 0.250 M HF Solution Given Ka 3.5 x 10^-4

In this article, we will learn how to determine the pH and pOH of a 0.250 M hydrofluoric acid (HF) solution given its acid dissociation constant (Ka) is 3.5 x 10^-4. This process involves several steps, including writing the dissociation equation, setting up the Ka expression, defining initial concentrations, and solving for the equilibrium concentrations. Let's break down the procedure step by step.

1. Introduction to the Problem

HF is a weak acid that partially dissociates in water according to the following equilibrium:

Heterogeneous Equilibrium: HF ? H F-

We are given the Ka for HF as 3.5 x 10-4 and the initial concentration of HF as 0.250 M.

2. Step-by-Step Solution

Step 1: Write the Dissociation Equation

The dissociation can be written as:

HF ? H F-

Step 2: Set Up the Expression for Ka

The Ka expression is:

Ka frac{[H^ ][F^-]}{[HF]}

Step 3: Define Initial Concentrations

Let's define the initial concentration of HF:

[HF]_0 0.250 M

Let 'x' represent the amount of HF that dissociates. At equilibrium:

t[H^ ] x
t[F^-] x
t[HF] 0.250 - x

Step 4: Substitute into the Ka Expression

Substitute the concentrations into the Ka expression:

Ka frac{x cdot x}{0.250 - x} frac{x^2}{0.250 - x}

Step 5: Approximate x

Since Ka is small, we can assume x to be much smaller than 0.250 M. Thus:

Ka approx frac{x^2}{0.250}

Substitute the value of Ka:

Step 6: Solve for x

Rearrange the equation to solve for x:

x^2 8.75 times 10^{-5} x sqrt{8.75 times 10^{-5}} approx 9.35 times 10^{-3}

Step 7: Calculate Equilibrium Concentrations

Now we have:

t[F^-] x approx 9.35 times 10^{-3} M
t[HF] 0.250 - x approx 0.250 - 9.35 times 10^{-3} approx 0.24065 M

Step 8: Calculate pH and pOH

To find pH:

pH -log[H^ ] -log(9.35 times 10^{-3}) approx 2.03

To find pOH:

pH pOH 14
tpOH 14 - pH approx 14 - 2.03 approx 11.97

3. Summary of Results

t t ttConcentration ttValue (M) t t tt[HF] tt0.24065 t t tt[F^-] tt9.35 times 10^{-3} t t tt[H_3O^ ] tt9.35 times 10^{-3} t t ttpH tt2.03 t t ttpOH tt11.97 t t

These calculations provide the concentrations and pH/pOH of the solution at equilibrium.

4. Additional Insights and Considerations

Understanding how to solve problems involving weak acids such as HF is crucial in various fields, including environmental science, chemistry, and industrial processes. By applying these principles, we can gain insights into the behavior of weak acids in aqueous solutions and how to manage them effectively. Furthermore, this knowledge can be used in developing safer and more efficient chemical reactions and solutions.