Part B How many grams of dry NH,CI need to be added to 2.00 L of a 0.100 M solution of ammonia, NH3. to prepare a buffer solution that has a pH of 8.85? Kb for ammonia is 1.8 x 10 Express your answer with the appropriate units. ▸ View Available Hint(s)

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## Base/Acid Ratios in Buffers

### Henderson-Hasselbalch Equation

**pH and pK Definitions:**
- Just as **pH** is the negative logarithm of \([H_3O^+]\), **pK\(_a\)** is the negative logarithm of \(K_a\).

\[
pK_a = -\log K_a
\]

- The Henderson-Hasselbalch equation is used to calculate the **pH** of buffer solutions:

\[
\text{pH} = \text{pK}_a + \log \left(\frac{[\text{base}]}{[\text{acid}]}\right)
\]

- Notice that the **pH** of a buffer has a value close to the **pK\(_a\)** of the acid, differing only by the logarithm of the concentration ratio \([\text{base}]/[\text{acid}]\).
- The Henderson-Hasselbalch equation in terms of **pOH** and **pK\(_b\)** is similar:

\[
\text{pOH} = \text{pK}_b + \log \left(\frac{[\text{acid}]}{[\text{base}]}\right)
\]

### Diagram Explanation
There is an image showing three boxes, each representing a different pH value scenario, with corresponding buffer component information:

1. **pH = 3.74**
   - Indicates that \([acetic \text{ } acid]\) is ten times greater than \([acetate]\).

2. **pH = 4.74**
   - Indicates that \([acetate] = [acetic \text{ } acid]\).

3. **pH = 5.74**
   - Indicates that \([acetate]\) is ten times greater than \([acetic \text{ } acid]\).

### Part B

**Problem Statement:**
- How many grams of dry \(NH_4Cl\) need to be added to 2.00 L of a 0.100 M solution of ammonia, \(NH_3\), to prepare a buffer solution that has a **pH** of 8.85? \(K_b\) for ammonia is \(1.8 \times 10^{-5}\).

- Express your answer with the appropriate units.

\( \
Transcribed Image Text:## Base/Acid Ratios in Buffers ### Henderson-Hasselbalch Equation **pH and pK Definitions:** - Just as **pH** is the negative logarithm of \([H_3O^+]\), **pK\(_a\)** is the negative logarithm of \(K_a\). \[ pK_a = -\log K_a \] - The Henderson-Hasselbalch equation is used to calculate the **pH** of buffer solutions: \[ \text{pH} = \text{pK}_a + \log \left(\frac{[\text{base}]}{[\text{acid}]}\right) \] - Notice that the **pH** of a buffer has a value close to the **pK\(_a\)** of the acid, differing only by the logarithm of the concentration ratio \([\text{base}]/[\text{acid}]\). - The Henderson-Hasselbalch equation in terms of **pOH** and **pK\(_b\)** is similar: \[ \text{pOH} = \text{pK}_b + \log \left(\frac{[\text{acid}]}{[\text{base}]}\right) \] ### Diagram Explanation There is an image showing three boxes, each representing a different pH value scenario, with corresponding buffer component information: 1. **pH = 3.74** - Indicates that \([acetic \text{ } acid]\) is ten times greater than \([acetate]\). 2. **pH = 4.74** - Indicates that \([acetate] = [acetic \text{ } acid]\). 3. **pH = 5.74** - Indicates that \([acetate]\) is ten times greater than \([acetic \text{ } acid]\). ### Part B **Problem Statement:** - How many grams of dry \(NH_4Cl\) need to be added to 2.00 L of a 0.100 M solution of ammonia, \(NH_3\), to prepare a buffer solution that has a **pH** of 8.85? \(K_b\) for ammonia is \(1.8 \times 10^{-5}\). - Express your answer with the appropriate units. \( \
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