Problem 174 - Chemistry An Atoms First Approach

Question

Making use of the assumptions we ordinarily make in calculating the pH of an aqueous solution of a weak acid, calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-6}-M\) solution of hypobromous acid (HBrO, \(K_{\mathrm{a}}=2 \times 10^{-9} \mathrm{J} .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

Step-by-Step Solution

Answer

The issue with the calculation is that we assumed that the initial concentrations of [H^+] and [BrO^-] are 0 M, ignoring the contribution of water's autoionization to [H^+]. Since the acid concentration is very low ( \(1.0 \times 10^{-6} M\)), it's necessary to consider the ionization of water and its contribution to [H^+] and [OH^-]. To correctly solve the problem, we should use the \(K_w\) (Ion product of water), to account for the contribution from water's autoionization when calculating the initial concentrations of each species and the pH of the solution.

Step 1: Write the acid dissociation equation

First, we write the equation representing the dissociation of hypobromous acid (HBrO) in aqueous solution: \(HBrO \rightleftharpoons H^+ + BrO^-\)

Step 2: Setup ICE table

Next, we set up an ICE (Initial, Change, Equilibrium) table to identify the concentrations of each species involved in the reaction. Initial Concentrations: [ HBrO ] = \(1.0 \times 10^{-6} M\) [ H^+ ] = 0 M [ BrO^- ] = 0 M Change: [ HBrO ] = \(-x M\) [ H^+ ] = \(+x M\) [ BrO^- ] = \(+x M\) Equilibrium Concentrations: [ HBrO ] = \((1.0 \times 10^{-6} - x) M\) [ H^+ ] = \(x M\) [ BrO^- ] = \(x M\)

Step 3: Write the \(K_a\) expression

Write the expression for \(K_a\) based on the reaction: \(K_a = \dfrac{[H^+][BrO^-]}{[HBrO]} = 2 \times 10^{-9}\)

Step 4: Substitute equilibrium concentrations into the \(K_a\) expression

Substitute the equilibrium concentrations from the ICE table into the \(K_a\) expression: \(2 \times 10^{-9} = \dfrac{x^2}{(1.0 \times 10^{-6} - x)}\)

Step 5: Identify the problem

As we found the \(K_a\) expression, we can notice that this assumption we made in the ICE table ([H^+] and [BrO^-] starting at 0 M) is not accurate because we ignored the contribution of water's autoionization to [H^+] when the acid (HBrO) concentration is at a very low value ( \(1.0 \times 10^{-6} M\)). In other words, we should also consider how water ionizes to form H^+ and OH^- in this case. To properly approach the problem, we need to include the ionization of water and its contribution of [H^+] and [OH^-] in the solution when calculating the initial concentrations of each species using the \(K_w\) (Ion product of water) for the correct calculation of the pH.

Key Concepts

Acid Dissociation Constant (Ka)

The acid dissociation constant (\(K_a\)) is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the chemical reaction in which an acid donates a proton to water:

\[ HA \rightleftharpoons H^+ + A^- \]
In this expression, HA represents the acid, and A^- represents the conjugate base. The larger the value of the acid dissociation constant, the stronger the acid because it indicates a greater tendency of the acid to lose its proton. In the context of a weak acid like hypobromous acid (HBrO), the value of its \(K_a\), which is \(2 \times 10^{-9}\), hints at it being a relatively weak acid as it does not dissociate significantly in solution. Knowing the \(K_a\) value helps us to calculate the pH of the solution by understanding the degree of acid ionization.

ICE Table Method

The ICE table method is an organized way to keep track of initial concentrations, changes in concentrations, and equilibrium concentrations in a chemical reaction. ICE stands for Initial, Change, and Equilibrium. This method is really helpful for visualizing the shifts in species concentrations as a reaction proceeds. Initially, we write down the concentrations that we know before any reaction occurs. Then, we consider the changes that must occur to reach equilibrium, often represented by the variable \(x\). Lastly, we calculate the equilibrium concentrations by combining the initial concentrations with the changes.

Using the ICE table allows us to create a clear picture of how the concentration of hypobromous acid decreases while the concentrations of hydronium (\(H^+\)) and the bromite ion (\(BrO^-\)) increase. After filling in the table, we can apply the acid dissociation constant to solve for \(x\) and thus determine the pH of the solution.

Autoionization of Water

Autoionization of water is a critical concept in understanding acid-base chemistry as it refers to water's ability to act as both an acid and a base, which explains the small concentration of hydrogen (\(H^+\)) and hydroxide (\(OH^-\)) ions always present in pure water. The reaction is depicted as:

\[ 2H_2O \rightleftharpoons H^+ + OH^- \]
The equilibrium constant for this reaction is the ion product of water (\(K_w\)) and it is equal to \(1.0 \times 10^{-14}\) at 25°C. When we address very weak acids or their very dilute solutions, the autoionization of water and its contribution to the hydronium ion concentration cannot be ignored, as it might significantly affect the pH calculation. Due to this factor, adjustments in the initial concentrations must be considered in the ICE table for accurate results.

Hypobromous Acid pH Problem

Calculating the pH of hypobromous acid (HBrO) solution involves using the given acid dissociation constant (\(K_a\)) and the knowledge from the ICE table method, considering the influence of water's autoionization. For very dilute solutions of weak acids, such as the given \(1.0 \times 10^{-6} M\) HBrO, we can't simply assume that the hydronium ion concentration comes solely from the acid dissociation since the autoionization of water is comparable to the acid's contribution at this dilution. By taking into account the water's contribution to the hydronium ion concentration, a more accurate pH can be calculated, highlighting the necessity to adjust our initial assumptions and thus fully address the weak acid pH problem. This consideration rectifies the common mistake of overlooking the significant effect that the autoionization of water has on the pH in such dilute acid solutions.