Problem 166 - Chemistry

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 \(\left(\mathrm{HBrO}, K_{\mathrm{a}}=\right.\) \(\left.2 \times 10^{-9}\right) .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, tell what has to be included to solve the problem correctly.

Step-by-Step Solution

Answer

The initial calculation of the pH of a $1.0 \times 10^{-6} M$ solution of hypobromous acid (HBrO) with a $K_a = 2 \times 10^{-9}$ gives a value based on the assumption that the change in [HBrO] is negligible. However, the mistake is not considering the contribution of hydronium ions from the auto-ionization of water in such a dilute weak acid solution. To solve the problem correctly, include the hydronium ions concentration from water ($1.0 \times 10^{-7} M$) in the \( K_a \) expression: \(2 \times 10^{-9} = \frac{(x + 1.0 \times 10^{-7})^2}{1.0 \times 10^{-6} - x}\). Solve for \(x\) (the concentration of \(H^+\) from HBrO) and then calculate the correct pH.

Step 1: Understanding the Problem

The student needs to calculate the pH of an aqueous solution of a weak acid with a given concentration using the acidity constant. Afterwards, the student has to find out what is wrong with the calculated answer and identify what is missing in the problem to solve it correctly.

Step 2: Initial Calculation

To calculate the pH of a weak acid, an expression is set up using the \(K_a\) constant for the dissociation reaction of the acid: \[ HBrO \rightarrow H^+ + BrO^- \] The expression can be set as: \(K_a = \frac{[H^+][BrO^-]}{[HBrO]}\) For weak acids, it is assumed that the [H+] and [BrO-] are x and the change in [HBrO] is negligible. Using the given concentration of 1.0x10^-6 M, we can simplify the equation to: \(2 \times 10^{-9} = \frac{x \times x }{1.0 \times 10^{-6}}\) Solve for x (which equals to [H+]) and then calculate the pH (pH=-log[H+]).

Step 3: Identify the Mistake and the Missing Element

The mistake in the initial calculation is the assumption that the change in [HBrO] is negligible and the concentration stays around 1.0x10^-6 M. However, for very dilute weak acids, this assumption may not hold. The missing element is the contribution of hydronium ions from the auto-ionization of water.

Step 4: Including the Auto-ionization of Water

In reality, water auto-ionizes to a small extent to hydrogen ions and hydroxide ions. Thus, the actual \(H^+\) concentration is the sum of the hydronium ion provided by the weak acid and by the water, and should be considered in the calculation.

Step 5: Appropriate Solution

To obtain a correct solution, perform the calculation as described in step 2 but with an amended \( K_a \) expression that accounts for the hydronium ions provided by the weak acid and the auto-ionization of water. The revised expression for \( K_a \) becomes \(2 \times 10^{-9} = \frac{(x + 1.0 x10^{-7} ) ^2}{1.0 x 10^{-6} - x}\). Here, \(x\) is the concentration of \(H^+\) from HBrO, and \(1.0 x 10^{-7} M\) is the concentration of \(H^+\) from water at room temperature. Solve for \(x\) to calculate the correct pH.

Key Concepts

Weak Acids

Weak acids are a fascinating group of substances characterized by their partial dissociation in water. This means they do not completely ionize, making them less efficient at donating protons compared to strong acids. Why is this important? Because it influences how we calculate their pH. In a weak acid solution such as hypobromous acid (

• Partial Ionization: Only a small fraction of acid molecules dissociate into ions.
• Equilibrium: A balance between the undissociated acid and the ions formed exists.

To solve a problem involving weak acids, we need to account for this equilibrium using the acidity constant (\(K_a\)). The formula typically becomes:\[K_a = \frac{[H^+][A^-]}{[HA]}\]Where \([H^+]\) is the concentration of hydrogen ions, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the weak acid. Since the dissociation is minimal, calculating exact pH requires us to make some assumptions, but sometimes these assumptions simplify the reality too much.

Auto-ionization of Water

Water, despite often being a neglected part of many calculations, can influence pH significantly when dealing with very dilute solutions or weak acids. Its auto-ionization can provide extra hydrogen ions (\(H^+\)) to the solution. This is crucial to consider, especially when the weak acid's concentration is similar to or lower than that of the water's auto-ionization, which is at about \(1.0 \times 10^{-7} M\) under normal conditions.

• What Happens? In pure water, a small amount of water molecules dissociate into water ions.
  • Formula: \(2 H_2O \leftrightarrow H_3O^+ + OH^-\)
  • This produces \(H^+onumber\) and \(OH^-\) ions, balancing out to maintain electrical neutrality.
• When to Consider: At low concentrations of the acid, ignoring this can cause inaccuracies in pH calculations.
• Why Important? It can significantly impact the ion concentration, leading to incorrect assumptions if not properly considered.

Acidity Constant

The acidity constant, often represented as \(K_a\), is a crucial concept in understanding the behavior of acids in a solution. It measures the strength of a weak acid by determining its tendency to donate protons. For weak acids, \(K_a\) helps to predict the acid's dissociation level, key to finding the pH of the solution.

• What is \(K_a\)? It quantifies the equilibrium between undissociated acids and their ions.
• The Formula:\[K_a = \frac{[H^+][A^-]}{[HA]}\]This equation helps to show how much of the acid stays undissociated versus how much breaks down into ions.
• Low \(K_a\) Value: Indicates a weak tendency for dissociation, thus a weak acid.
• High \(K_a\) Value: Suggests a stronger acid, more dissociation.

Understanding \(K_a\) allows for more accurate pH calculations and aids in pinpointing why certain assumptions (like ignoring water's auto-ionization) can lead to errors. This balance is crucial in determining how to approach solving exercises accurately.