To calculate the pH of a solution prepared by dissolving 0.10 moles of NH₄Cl in 0.500 L of 0.40 M NH₃, we'll follow these steps:
Step 1: Determine the concentration of NH₄⁺ and NH₃ in the solution
1. Calculate the concentration of NH₄⁺:
The moles of NH₄Cl dissolved are 0.10 moles, and since the volume of the solution is 0.500 L, the concentration of NH₄⁺ can be calculated as follows:
[tex]\[\text{Concentration of NH₄⁺} = \frac{\text{moles of NH₄Cl}}{\text{volume of solution}} = \frac{0.10 \text{ moles}}{0.500 \text{ L}} = 0.20 \text{ M}\][/tex]
2. Concentration of NH₃:
The concentration of NH₃ remains 0.40 M as given.
Step 2: Calculate the pKa value from the given Ka value of NH₄⁺
The relationship between the pKa and Ka is:
[tex]\[ \text{pKa} = -\log(K_a) \][/tex]
Given:
[tex]\[ K_a = 5.56 \times 10^{-10} \][/tex]
We can determine:
[tex]\[ \text{pKa} = -\log(5.56 \times 10^{-10}) \approx 9.254925208417943 \][/tex]
Step 3: Use the Henderson-Hasselbalch equation to find the pH
The Henderson-Hasselbalch equation for a buffer solution is:
[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \][/tex]
In this case:
- The base is NH₃ with concentration 0.40 M
- The acid is NH₄⁺ with concentration 0.20 M
Substitute the values:
[tex]\[ \text{pH} = 9.254925208417943 + \log \left( \frac{0.40}{0.20} \right) \][/tex]
First, calculate the ratio:
[tex]\[ \frac{0.40}{0.20} = 2 \][/tex]
Now calculate the logarithm:
[tex]\[ \log(2) \approx 0.3010 \][/tex]
Finally, add the logarithm to the pKa:
[tex]\[ \text{pH} = 9.254925208417943 + 0.3010 = 9.555955204081924 \][/tex]
So, the pH of the solution is approximately 9.556.
Summary:
1. The concentration of NH₄⁺ in the solution is 0.20 M.
2. The pKa value of NH₄⁺ is approximately 9.255.
3. Using the Henderson-Hasselbalch equation, the pH of the solution is approximately 9.556.
These results are consistent with the numerical solution.
