Question

A buffer solution is composed of 0.1 M acetic acid and 0.15 M sodium acetate. The change in pH of 1 L buffer solution upon addition of 50 mL of 1.0 M NaOH is ............... (rounded off to 2 decimals)

Hint:

To calculate the pH change in a buffer solution, use the Henderson-Hasselbalch equation before and after the addition of the strong base, considering the neutralization of the weak acid.

Answer 1

Step 1: Understanding the problem.
We are given a buffer solution composed of acetic acid (weak acid) and sodium acetate (conjugate base). The pH change is caused by the addition of NaOH, which will neutralize some of the acetic acid in the buffer.
Step 2: Using the Henderson-Hasselbalch equation.
The pH of a buffer solution is given by the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] Where:
- \( \text{pK}_a \) of acetic acid = 4.76
- \( [\text{A}^-] \) = concentration of acetate ions (conjugate base)
- \( [\text{HA}] \) = concentration of acetic acid
Step 3: Initial concentrations.
Initially, the concentrations are:
- \( [\text{HA}] = 0.1 \, \text{M} \)
- \( [\text{A}^-] = 0.15 \, \text{M} \)
Using the Henderson-Hasselbalch equation: \[ \text{pH} = 4.76 + \log \left( \frac{0.15}{0.1} \right) \] \[ \text{pH} = 4.76 + \log (1.5) = 4.76 + 0.176 = 4.936 \] Step 4: After addition of NaOH.
The moles of NaOH added are: \[ \text{moles of NaOH} = 0.05 \, \text{L} \times 1.0 \, \text{M} = 0.05 \, \text{moles} \] The moles of acetic acid before addition of NaOH are: \[ \text{moles of HA} = 1.0 \, \text{L} \times 0.1 \, \text{M} = 0.1 \, \text{moles} \] The NaOH will neutralize an equivalent amount of acetic acid: \[ \text{moles of HA remaining} = 0.1 - 0.05 = 0.05 \, \text{moles} \] The moles of acetate ions after neutralization will be: \[ \text{moles of A}^- = 0.15 + 0.05 = 0.2 \, \text{moles} \] Now, the new concentrations of \( \text{HA} \) and \( \text{A}^- \) are: \[ [\text{HA}] = \frac{0.05}{1.05} = 0.0476 \, \text{M}, \quad [\text{A}^-] = \frac{0.2}{1.05} = 0.1905 \, \text{M} \] Step 5: Re-calculating the pH.
Using the Henderson-Hasselbalch equation again: \[ \text{pH} = 4.76 + \log \left( \frac{0.1905}{0.0476} \right) \] \[ \text{pH} = 4.76 + \log (4.0) = 4.76 + 0.602 = 5.362 \] Step 6: Finding the change in pH.
The change in pH is: \[ \Delta \text{pH} = 5.362 - 4.936 = 0.426 \] Final Answer: \[ \boxed{0.43} \]