Questions: Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that you have 135 mL of a buffer that is 0.100 M in both acetic acid (CH3COOH) and its conjugate base (CH3COO^-). Calculate the maximum volume of 0.500 M HCl that can be added to the buffer before its buffering capacity is lost.

Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that you have 135 mL of a buffer that is 0.100 M in both acetic acid (CH3COOH) and its conjugate base (CH3COO^-). Calculate the maximum volume of 0.500 M HCl that can be added to the buffer before its buffering capacity is lost.

Transcript text: Attempt 3 ${ }^{\circ}$ Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that you have 135 mL of a buffer that is 0.100 M in both acetic acid $\left(\mathrm{CH}_{3} \mathrm{COOH}\right)$ and its conjugate base $\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right)$. Calculate the maximum volume of 0.500 M HCl that can be added to the buffer before its buffering capacity is lost.

Solution

Solution Steps

Step 1: Understanding Buffer Capacity

Buffer capacity is the ability of a buffer solution to resist changes in pH upon the addition of an acid or base. The buffer in question consists of acetic acid ($\mathrm{CH}_3\mathrm{COOH}$) and its conjugate base ($\mathrm{CH}_3\mathrm{COO}^-$).

Step 2: Initial Moles of Buffer Components

Calculate the initial moles of acetic acid and acetate ion in the buffer solution: \[ \text{Volume of buffer} = 135 \, \text{mL} = 0.135 \, \text{L} \] \[ \text{Concentration of acetic acid} = 0.100 \, \text{M} \] \[ \text{Concentration of acetate ion} = 0.100 \, \text{M} \] \[ \text{Moles of acetic acid} = 0.135 \, \text{L} \times 0.100 \, \text{M} = 0.0135 \, \text{mol} \] \[ \text{Moles of acetate ion} = 0.135 \, \text{L} \times 0.100 \, \text{M} = 0.0135 \, \text{mol} \]

Step 3: Reaction with HCl

When HCl is added to the buffer, it reacts with the acetate ion ($\mathrm{CH}_3\mathrm{COO}^-$) to form acetic acid ($\mathrm{CH}_3\mathrm{COOH}$): \[ \mathrm{CH}_3\mathrm{COO}^- + \mathrm{HCl} \rightarrow \mathrm{CH}_3\mathrm{COOH} + \mathrm{Cl}^- \] The buffer capacity is lost when all the acetate ions are consumed.

Step 4: Maximum Moles of HCl

The maximum moles of HCl that can be added is equal to the initial moles of acetate ion: \[ \text{Moles of HCl} = 0.0135 \, \text{mol} \]

Step 5: Volume of 0.500 M HCl

Calculate the volume of 0.500 M HCl that corresponds to 0.0135 moles: \[ \text{Concentration of HCl} = 0.500 \, \text{M} \] \[ \text{Volume of HCl} = \frac{\text{Moles of HCl}}{\text{Concentration of HCl}} = \frac{0.0135 \, \text{mol}}{0.500 \, \text{M}} = 0.0270 \, \text{L} = 27.0 \, \text{mL} \]