Chapter 15: Problem 127
A solution contains \(4.25 \mathrm{~g}\) of ammonia per \(250.0 \mathrm{~mL}\) of solution. Electrical conductivity measurements at \(25^{\circ} \mathrm{C}\) show that \(0.42 \%\) of the ammonia has reacted with water. Write the equation for this reaction and calculate the \(\mathrm{pH}\) of the solution.
Short Answer
Step by step solution
Write the Chemical Equation
Calculate Moles of Ammonia in Solution
Determine Moles of Ammonia That Reacted
Calculate Concentration of \(\text{OH}^-\) Ions
Calculate \(\text{pOH}\) and \(\text{pH}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ammonia Reaction with Water
- Ammonia accepts a proton (\(\text{H}^+\)) from a water molecule.
- This process produces ammonium ions (\(\text{NH}_4^+\)) and hydroxide ions (\(\text{OH}^-\)).
Chemical Equilibrium
The concept of equilibrium is significant because:
- The concentrations of ammonia, water, ammonium, and hydroxide ions stabilize over time.
- The equilibrium position depends on the extent of the reaction and the conditions such as temperature and concentration.
Electrical Conductivity
Here's why this is important:
- Conductivity measurements can help calculate how much ammonia reacts with water.
- By knowing the percentage of ammonia that forms ions, we can understand the solution's ion concentration more deeply.
Molarity and Moles Conversion
Here’s how it works in practice:
- Start with the mass of ammonia, \(4.25 \text{ g}\), and convert it to moles using its molar mass \(17.03 \text{ g/mol}\).
- Calculate the total moles: \[\text{moles of } \text{NH}_3 = \frac{4.25 \text{ g}}{17.03 \text{ g/mol}} \approx 0.25 \text{ mol} \]
- Determine the moles that reacted by accounting for the \(0.42\%\) conversion: \[ \text{moles of reacted } \text{NH}_3 = 0.25 \text{ mol} \times 0.0042 \approx 0.00105 \text{ mol}\]
- Finally, calculate the concentration of ions in the solution by dividing moles by volume in liters.