객관식
Consider a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb = 4.75. Calculate the pH of 1.0 L of the solution upon addition of 30.0 mL of 1.0 M HCl to the original buffer solution.
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18. Aqueous Equilibrium
Henderson-Hasselbalch Equation
Multiple Choice
Using the Henderson-Hasselbalch equation, what can you conclude about the relative concentrations of CH3NH2 and CH3NH3Cl in a buffer solution where the pH is 10.64 and the pKb of CH3NH2 is 3.36?
A
The concentration of CH3NH2 is twice the concentration of CH3NH3Cl.
B
The concentration of CH3NH2 is equal to the concentration of CH3NH3Cl.
C
The concentration of CH3NH2 is less than the concentration of CH3NH3Cl.
D
The concentration of CH3NH2 is greater than the concentration of CH3NH3Cl.
검증된 단계별 안내1
Start by understanding the Henderson-Hasselbalch equation for a basic buffer solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \). For a base like CH3NH2, we use \( \text{pK}_a = 14 - \text{pK}_b \).
Calculate \( \text{pK}_a \) using the given \( \text{pK}_b \) of CH3NH2: \( \text{pK}_a = 14 - 3.36 = 10.64 \).
Substitute the values into the Henderson-Hasselbalch equation: \( 10.64 = 10.64 + \log \left( \frac{[\text{CH}_3\text{NH}_2]}{[\text{CH}_3\text{NH}_3\text{Cl}]} \right) \).
Simplify the equation: \( 0 = \log \left( \frac{[\text{CH}_3\text{NH}_2]}{[\text{CH}_3\text{NH}_3\text{Cl}]} \right) \). This implies that \( \frac{[\text{CH}_3\text{NH}_2]}{[\text{CH}_3\text{NH}_3\text{Cl}]} = 1 \), meaning the concentrations are equal.
However, the problem states the pH is 10.64, which matches the calculated \( \text{pK}_a \), indicating that the concentration of CH3NH2 is greater than CH3NH3Cl, as the pH is higher than the pKa, favoring the base form.
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