The pKa value of NH4+ = NH3 + H+ is 9.25. How many mL of 0.1 M NH4Cl must one mix with how many mL 0.1 M NH3 to make a buffer whose pH = 9.0?
The correct answer and explanation is :
To solve this problem, we will use the Henderson-Hasselbalch equation for buffer solutions:
[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)
]
In this scenario:
- pKa = 9.25 (for the NH4+/NH3 system),
- pH = 9.0 (desired pH of the buffer),
- Base = NH3 (ammonia),
- Acid = NH4+ (ammonium ion from NH4Cl).
Step 1: Rearranging the Henderson-Hasselbalch equation
First, we plug the given values into the equation:
[
9.0 = 9.25 + \log \left( \frac{[\text{NH3}]}{[\text{NH4+}]} \right)
]
Next, solve for the ratio of ([\text{NH3}]) to ([\text{NH4+}]):
[
9.0 – 9.25 = \log \left( \frac{[\text{NH3}]}{[\text{NH4+}]} \right)
]
[
-0.25 = \log \left( \frac{[\text{NH3}]}{[\text{NH4+}]} \right)
]
[
\frac{[\text{NH3}]}{[\text{NH4+}]} = 10^{-0.25}
]
Using a calculator:
[
\frac{[\text{NH3}]}{[\text{NH4+}]} \approx 0.562
]
Step 2: Setting up the equation for molar amounts
Let ( x ) represent the volume of 0.1 M NH4Cl (which gives the amount of NH4+), and ( y ) represent the volume of 0.1 M NH3 (which gives the amount of NH3). The concentrations of NH4+ and NH3 are 0.1 M, so the number of moles of each is given by:
[
\text{moles of NH4+} = 0.1x
]
[
\text{moles of NH3} = 0.1y
]
Now, we use the ratio of the concentrations of NH3 to NH4+:
[
\frac{0.1y}{0.1x} = 0.562
]
Simplifying:
[
\frac{y}{x} = 0.562
]
Thus, we have:
[
y = 0.562x
]
Step 3: Total volume constraint
The total volume of the buffer solution is ( x + y ). Therefore, substitute ( y ) from the previous equation:
[
x + 0.562x = \text{Total Volume}
]
[
1.562x = \text{Total Volume}
]
Step 4: Solving for ( x ) and ( y )
Assuming we want to make 1 liter (1000 mL) of buffer solution:
[
1.562x = 1000 \text{ mL}
]
[
x \approx \frac{1000}{1.562} \approx 640.0 \text{ mL}
]
Now, substitute ( x ) into the equation ( y = 0.562x ):
[
y = 0.562 \times 640.0 \approx 359.4 \text{ mL}
]
Final Answer:
To prepare 1 liter of buffer with a pH of 9.0:
- Mix 640 mL of 0.1 M NH4Cl (providing NH4+),
- Mix 359.4 mL of 0.1 M NH3 (providing NH3).
Explanation:
This solution is based on the equilibrium between the weak acid (NH4+) and its conjugate base (NH3), where the pH depends on their ratio. By adjusting the volumes of NH4Cl and NH3, we can achieve the desired pH of 9.0 using the Henderson-Hasselbalch equation.