Courtnee A. answered • 04/04/24
Chemistry Expert with Bachelor's Degree
To calculate the pH of the buffer solution, we first need to determine the concentration of acetic acid (\(CH_3COOH\)) and sodium acetate (\(CH_3COONa\)) in the buffer. Then we can use the Henderson-Hasselbalch equation to find the pH.
1. Calculate moles of acetic acid and sodium acetate:
Moles of acetic acid = volume (in L) * concentration (in mol/L)
\( \text{Moles of acetic acid} = 8.8 \, \text{mL} \times 0.0030 \, \text{mol/mL} = 0.0264 \, \text{mol} \)
Moles of sodium acetate = mass (in g) / molar mass (in g/mol)
\( \text{Moles of sodium acetate} = 3.5 \, \text{g} / 82.03 \, \text{g/mol} = 0.0427 \, \text{mol} \)
2. Calculate the total volume of the buffer solution:
Total volume = volume of acetic acid + volume of water
\( \text{Total volume} = 8.8 \, \text{mL} + 55.6 \, \text{mL} = 64.4 \, \text{mL} = 0.0644 \, \text{L} \)
3. Calculate the concentrations of acetic acid and sodium acetate in the buffer:
Concentration of acetic acid = moles / total volume
\( \text{Concentration of acetic acid} = 0.0264 \, \text{mol} / 0.0644 \, \text{L} = 0.410 \, \text{M} \)
Concentration of sodium acetate = moles / total volume
\( \text{Concentration of sodium acetate} = 0.0427 \, \text{mol} / 0.0644 \, \text{L} = 0.662 \, \text{M} \)
4. Use the Henderson-Hasselbalch equation to calculate the pH of the buffer:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
The pKa of acetic acid is approximately 4.76.
\[ \text{pH} = 4.76 + \log \left( \frac{0.662}{0.410} \right) \]
\[ \text{pH} = 4.76 + \log(1.6146) \]
\[ \text{pH} = 4.76 + 0.208 \]
\[ \text{pH} = 4.97 \]
Therefore, the pH of the buffer solution is approximately 4.97.
Next, let's calculate the pH of the new solution after mixing 32 mL of the buffer solution with 1.0 mL of 6.0 M HCl:
1. Calculate the moles of HCl added:
Moles of HCl = volume (in L) * concentration (in mol/L)
\( \text{Moles of HCl} = 0.0010 \, \text{L} \times 6.0 \, \text{mol/L} = 0.0060 \, \text{mol} \)
2. Determine the moles of acetic acid and sodium acetate remaining after neutralization:
Moles of acetic acid remaining = initial moles - moles of HCl added
\( \text{Moles of acetic acid remaining} = 0.0264 \, \text{mol} - 0.0060 \, \text{mol} = 0.0204 \, \text{mol} \)
Moles of sodium acetate remaining = initial moles
\( \text{Moles of sodium acetate remaining} = 0.0427 \, \text{mol} \)
3. Calculate the total volume of the new solution:
Total volume = volume of buffer + volume of HCl added
\( \text{Total volume} = 32 \, \text{mL} + 1.0 \, \text{mL} = 33 \, \text{mL} = 0.033 \, \text{L} \)
4. Calculate the concentrations of acetic acid and sodium acetate in the new solution:
Concentration of acetic acid = moles / total volume
\( \text{Concentration of acetic acid} = 0.0204 \, \text{mol} / 0.033 \, \text{L} = 0.618 \, \text{M} \)
Concentration of sodium acetate = moles / total volume
\( \text{Concentration of sodium acetate} = 0.0427 \, \text{mol} / 0.033 \, \text{L} = 1.294 \, \text{M} \)
5. Use the Henderson-Hasselbalch equation to calculate the pH of the new solution:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
\[ \text{pH} = 4.76 + \log \left( \frac{1.294}{0.618} \right) \]
\[ \text{pH} = 4.76 + \log(2.089) \]
\[ \text{pH} = 4.76 + 0.32 \]
\[ \text{pH} = 5.08 \]
Therefore, the pH of the new solution after mixing is approximately 5.08.
