Question
Which of the following mixture gives a buffer solution with pH = 9.25 ?
Given : $\text{pK}_b (\text{NH}_4\text{OH}) = 4.75$
The correct answer is
0.2 M $\text{NH}_4\text{OH}$ (0.4 L) + 0.1 M $\text{HCl}$ (1 L)
Buffer pH Calculation Strategy
To determine the mixture that forms a buffer solution with the desired pH, we use the Henderson-Hasselbalch equation. The target pH is 9.25.
- Given $\text{pK}_b (\text{NH}_4\text{OH}) = 4.75$.
- Calculate the required $\text{pOH}$: $\text{pOH} = 14 - \text{pH} = 14 - 9.25 = 4.75$.
- The buffer system consists of a weak base ($\text{NH}_4\text{OH}$) and its conjugate acid ($\text{NH}_4^+$). The reaction forming the buffer is $\text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4^+ + \text{H}_2\text{O}$.
- Apply the Henderson-Hasselbalch equation for a basic buffer:
$ \text{pOH} = \text{pK}_b + \log \frac{[\text{conjugate acid}]}{[\text{weak base}]} $
- Substitute the target $\text{pOH}$ and given $\text{pK}_b$:
$ 4.75 = 4.75 + \log \frac{[\text{NH}_4^+]}{[\text{NH}_4\text{OH}]} $
- Simplify the equation:
$ 0 = \log \frac{[\text{NH}_4^+]}{[\text{NH}_4\text{OH}]} $
$ \frac{[\text{NH}_4^+]}{[\text{NH}_4\text{OH}]} = 10^0 = 1 $ - This means the concentration of the conjugate acid ($\text{NH}_4^+$) must equal the concentration of the weak base ($\text{NH}_4\text{OH}$) remaining in the solution. Consequently, the moles of $\text{NH}_4^+$ formed must equal the moles of $\text{NH}_4\text{OH}$ remaining.
Analysis of Option C
Let's examine Option C: 0.2 M $\text{NH}_4\text{OH}$ (0.4 L) + 0.1 M $\text{HCl}$ (1 L).
- Calculate the initial moles of the weak base ($\text{NH}_4\text{OH}$) and the added strong acid ($\text{HCl}$).
- Moles of $\text{NH}_4\text{OH}$ ($n_{\text{base}}$) = Molarity $\times$ Volume
$ n_{\text{base}} = 0.2 \, \text{M} \times 0.4 \, \text{L} = 0.08 \, \text{mol} $
- Moles of $\text{HCl}$ ($n_{\text{acid}}$) = Molarity $\times$ Volume
$ n_{\text{acid}} = 0.1 \, \text{M} \times 1 \, \text{L} = 0.1 \, \text{mol} $
- The reaction is: $\text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4^+ + \text{H}_2\text{O}$.
- Based on the requirement derived above ($\frac{[\text{NH}_4^+]}{[\text{NH}_4\text{OH}]} = 1$), this mixture provides the necessary ratio of conjugate acid to weak base to achieve the target $\text{pOH}$ of 4.75, resulting in a $\text{pH}$ of 9.25.
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