Question

Define conjugate acid-base pair. The hydroxyl ion concentration in aqueous solution of NaOH is 2 \( \times \) 10\(^{-4}\) mol dm\(^{-3}\). Calculate pH of the solution.

Hint:

Always remember the fundamental relationship \( \text{pH} + \text{pOH} = 14 \) for aqueous solutions at 25\(^\circ\)C. If you are given [H\(^+\)], find pH first. If you are given [OH\(^-\)], find pOH first, then convert to pH.

Answer 1

Part 1: Definition of Conjugate Acid-Base Pair A conjugate acid-base pair consists of two species that differ from each other by the presence of a single proton (H\(^+\)). When a Brønsted-Lowry acid donates a proton, the species that remains is its conjugate base. When a Brønsted-Lowry base accepts a proton, the species formed is its conjugate acid.
Example: In the reaction \( \text{HCl} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{Cl}^- \), HCl is the acid and Cl\(^-\) is its conjugate base. H\(_2\)O is the base and H\(_3\)O\(^+\) is its conjugate acid.

Part 2: Calculation of pH
Step 1: Given Information
Hydroxyl ion concentration, [OH\(^-\)] = 2 \( \times \) 10\(^{-4}\) mol dm\(^{-3}\).
Step 2: Calculate pOH
The pOH is the negative logarithm of the hydroxyl ion concentration. \[ \text{pOH} = -\log_{10}[\text{OH}^-] \] \[ \text{pOH} = -\log_{10}(2 \times 10^{-4}) \] \[ \text{pOH} = -(\log_{10}2 + \log_{10}10^{-4}) \] \[ \text{pOH} = -(0.3010 - 4) = 3.699 \]
Step 3: Calculate pH The sum of pH and pOH at 298 K (25\(^\circ\)C) is 14. \[ \text{pH} + \text{pOH} = 14 \] \[ \text{pH} = 14 - \text{pOH} \] \[ \text{pH} = 14 - 3.699 = 10.301 \] The pH of the solution is 10.301.