Chapter 16: Problem 44
An aqueous solution of a strong base has a pH of 9.78 at \(25^{\circ} \mathrm{C}\). Calculate the concentration of the base if the base is (a) \(\mathrm{LiOH}\) and (b) \(\mathrm{Ba}(\mathrm{OH})_{2}\).
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hydroxide Ion Concentration
For example, if the pOH is 4.22, substituting this into the equation gives: \[ [\text{OH}^-] \approx 6.03 \times 10^{-5} \text{ M} \] This concentration reflects the strength of the base in the solution.
Remember, the hydroxide ion concentration is directly tied to the solution's basic nature.
Strong Base Properties
- For \(\text{LiOH}\), it dissociates into \(\text{Li}^+\) and \(\text{OH}^-\)
- For \(\text{Ba(OH)}_2\), it forms one \(\text{Ba}^{2+}\) ion and two \(\text{OH}^-\) ions
In basic solutions, strong bases raise the hydroxide ion concentration significantly, thus increasing the pH of the solution.
Acid-Base Equilibrium
- How acidic or basic a solution is
- The concentration of hydrogen ions \(\text{H}^+\) and hydroxide ions \(\text{OH}^-\)
This relationship remains constant at \(25^\circ \text{C}\), providing a reliable tool for calculations.
Ba(OH)₂ Dissociation
This means the hydroxide ion concentration is double the concentration of \(\text{Ba(OH)}_2\). With a \[ [\text{OH}^-] \approx 6.03 \times 10^{-5} \text{ M} \], we find: \[ [\text{Ba(OH)}_2] = \frac{6.03 \times 10^{-5}}{2} \approx 3.015 \times 10^{-5} \text{ M} \]
Understanding this dissociation is key to solving problems related to strong bases like \(\text{Ba(OH)}_2\). This knowledge ensures precise calculations.