Chapter 16: Problem 30
Calculate the number of moles of \(\mathrm{KOH}\) in \(5.50 \mathrm{~mL}\) of a \(0.360 \mathrm{M} \mathrm{KOH}\) solution. What is the \(\mathrm{pOH}\) of the solution at \(25^{\circ} \mathrm{C} ?\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity
- \( M = \frac{n}{V} \)
To calculate the number of moles in a solution given molarity, you can rearrange this formula to solve for \( n \):
- \( n = M \times V \)
Concentration
- Higher concentration means more solute particles are packed into the same volume, making reactions faster due to increased particle interactions.
- In our context, we talk about potassium hydroxide (KOH) in water.
pOH
- \( \text{pOH} = -\log[\text{OH}^-] \)
As a strong base, potassium hydroxide fully dissociates in solution, meaning the hydroxide ion concentration directly equals the molarity of the KOH solution. Calculating pOH from a known concentration helps chemists understand how alkaline a solution is, which is crucial for processes that depend on a specific pH or pOH environment.
Potassium Hydroxide
- \( \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \)
KOH's complete dissociation is particularly significant because it directly influences the solution's properties.
- Increased hydroxide ion concentration leads to a decrease in pOH, effectively making the solution more basic.