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Chapter 14: Problem 166

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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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+]).
03

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.
04

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.
05

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.

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