50 mL of 0.1 M NaOH is added to 75 mL of 0.1 M `NH_4Cl` to make a basic buffer. If `pK_a" of "NH_4^+` is 9.26, calculate pH.

Answer

To solve the problem of calculating the pH of a basic buffer created by mixing NaOH and NH4Cl, we can follow these steps:

### Step 1: Identify the components of the buffer
We have:
- 50 mL of 0.1 M NaOH
- 75 mL of 0.1 M NH4Cl

When NaOH is added to NH4Cl, it reacts to form NH4OH (the weak base) and NaCl. Thus, we have a buffer solution consisting of the weak base NH4OH and its salt NH4Cl.

### Step 2: Calculate the number of moles of each component
To find the number of moles, we use the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \]

For NaOH:
\[ \text{Moles of NaOH} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \]

For NH4Cl:
\[ \text{Moles of NH4Cl} = 0.1 \, \text{M} \times 0.075 \, \text{L} = 0.0075 \, \text{moles} \]

### Step 3: Calculate the pKb of NH4+
Given:
- pKa of NH4+ = 9.26
- pKb can be calculated using the relation:
\[ pK_a + pK_b = 14 \]
Thus,
\[ pK_b = 14 - 9.26 = 4.74 \]

### Step 4: Calculate the concentrations of NH4OH and NH4Cl
Next, we need to find the concentrations of NH4OH and NH4Cl in the final solution.

The total volume of the solution after mixing is:
\[ 50 \, \text{mL} + 75 \, \text{mL} = 125 \, \text{mL} = 0.125 \, \text{L} \]

Concentration of NH4Cl:
\[ [\text{NH4Cl}] = \frac{0.0075 \, \text{moles}}{0.125 \, \text{L}} = 0.06 \, \text{M} \]

Concentration of NH4OH:
Since NaOH reacts with NH4Cl to form NH4OH, the moles of NH4OH formed will be equal to the moles of NaOH added (0.005 moles). Therefore, the concentration of NH4OH is:
\[ [\text{NH4OH}] = \frac{0.005 \, \text{moles}}{0.125 \, \text{L}} = 0.04 \, \text{M} \]

### Step 5: Use the Henderson-Hasselbalch equation
For a basic buffer, the Henderson-Hasselbalch equation is given by:
\[ pOH = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Substituting the values:
\[ pOH = 4.74 + \log\left(\frac{0.06}{0.04}\right) \]
Calculating the log term:
\[ \log\left(\frac{0.06}{0.04}\right) = \log(1.5) \approx 0.176 \]
Thus,
\[ pOH = 4.74 + 0.176 = 4.916 \]

### Step 6: Calculate the pH
Finally, we can find the pH using the relation:
\[ pH = 14 - pOH = 14 - 4.916 = 9.084 \]

### Final Answer
The pH of the buffer solution is approximately **9.08**.

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