To calculate the pH of a 0.33 M ammonia (NH₃) solution, we first need to recognize that ammonia is a weak base that partially ionizes in water according to the following equilibrium:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]
The equilibrium constant for this reaction is known as the base dissociation constant (\(K_b\)). At 25°C, the \(K_b\) for ammonia is approximately \(1.8 \times 10^{-5}\).
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Set up the expression for \(K_b\): \[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
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Define the changes in concentration: Let's assume \(x\) is the concentration of \(\text{OH}^-\) (and therefore \(\text{NH}_4^+\)) formed at equilibrium. The initial concentration of \(\text{NH}_3\) is 0.33 M, and at equilibrium, the concentrations will be:
- \([\text{NH}_3] = 0.33 - x\) (approximately 0.33 M since \(x\) is small)
- \([\text{NH}_4^+] = x\)
- \([\text{OH}^-] = x\)
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Substituting into the \(K_b\) expression: \[ 1.8 \times 10^{-5} = \frac{x \cdot x}{0.33 - x} \approx \frac{x^2}{0.33} \]
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Solving for \(x\): \[ x^2 = (1.8 \times 10^{-5}) \times 0.33 \] \[ x^2 = 5.94 \times 10^{-6} \] \[ x = \sqrt{5.94 \times 10^{-6}} \approx 2.44 \times 10^{-3} \text{ M} \]
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Now, calculate \([\text{OH}^-]\): \[ [\text{OH}^-] = x \approx 2.44 \times 10^{-3} \text{ M} \]
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Calculate the pOH: \[ \text{pOH} = -\log_{10}([\text{OH}^-]) = -\log_{10}(2.44 \times 10^{-3}) \approx 2.61 \]
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Finally, calculate the pH: \[ \text{pH} + \text{pOH} = 14 \] \[ \text{pH} = 14 - \text{pOH} = 14 - 2.61 \approx 11.39 \]
Thus, the pH of a 0.33 M ammonia solution is approximately 11.39.
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