You are given 60 mL of 0.50 M phosphate buffer, pH = 6.83, to test. The starting composition of the buffer, both in terms of the concentration and the molar quantity of the two major phosphate species, is: %3D Concentration of HPO,: 0.304 M Molar quantity of HPO: 18.2 mmol Concentration of H,PO“: 0.196 M Molar quantity of H,PO = 11.8 mmol You add 1.7 mL of 1.00 M HCl to the buffer. Calculate the molar quantity of H,O* added as HCl, and the final molar quantity of HPO, and H,PO,¯ at equilibrium. а.

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### Buffer Capacity and pH Calculation

**Problem Statement:**
Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb" five times as much acid or base as a 0.1 M buffer for a given pH change. 

In this problem, you begin with a buffer of known pH and concentration and calculate the new pH after adding a specific quantity of acid or base. You have 60 mL of 0.50 M phosphate buffer with a pH of 6.83 to test. The starting composition of the buffer, expressed in terms of concentration and molar quantity of the two major phosphate species, is:

| Species            | Concentration (M) | Molar Quantity (mmol) |
|--------------------|-------------------|-----------------------|
| \( \text{HPO}_4^{2-} \) | 0.304             | 18.2                  |
| \( \text{H}_2\text{PO}_4^- \) | 0.196             | 11.8                  |

#### Tasks:

**a.** You add 1.7 mL of 1.00 M HCl to the buffer. Calculate:
- The molar quantity of \( \text{H}_3\text{O}^+ \) added as HCl.
- The final molar quantities of \( \text{HPO}_4^{2-} \) and \( \text{H}_2\text{PO}_4^- \) at equilibrium.

**b.** Determine:
- The new \( \text{HPO}_4^{2-}/\text{H}_2\text{PO}_4^- \) ratio.
- The new pH of the solution using the Henderson-Hasselbalch equation. Given \( \text{p}K_a \) of \( \text{H}_2\text{PO}_4^- \) is 6.64. Note: Use the molar ratio rather than the concentration ratio because both species are in the same volume.

**c.** Take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.
Transcribed Image Text:### Buffer Capacity and pH Calculation **Problem Statement:** Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb" five times as much acid or base as a 0.1 M buffer for a given pH change. In this problem, you begin with a buffer of known pH and concentration and calculate the new pH after adding a specific quantity of acid or base. You have 60 mL of 0.50 M phosphate buffer with a pH of 6.83 to test. The starting composition of the buffer, expressed in terms of concentration and molar quantity of the two major phosphate species, is: | Species | Concentration (M) | Molar Quantity (mmol) | |--------------------|-------------------|-----------------------| | \( \text{HPO}_4^{2-} \) | 0.304 | 18.2 | | \( \text{H}_2\text{PO}_4^- \) | 0.196 | 11.8 | #### Tasks: **a.** You add 1.7 mL of 1.00 M HCl to the buffer. Calculate: - The molar quantity of \( \text{H}_3\text{O}^+ \) added as HCl. - The final molar quantities of \( \text{HPO}_4^{2-} \) and \( \text{H}_2\text{PO}_4^- \) at equilibrium. **b.** Determine: - The new \( \text{HPO}_4^{2-}/\text{H}_2\text{PO}_4^- \) ratio. - The new pH of the solution using the Henderson-Hasselbalch equation. Given \( \text{p}K_a \) of \( \text{H}_2\text{PO}_4^- \) is 6.64. Note: Use the molar ratio rather than the concentration ratio because both species are in the same volume. **c.** Take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.
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