How many moles of `NH_(4)Cl` should be added to `200mL` solution of `0.18 M NH_(4)OH` to have a `pH` of `9.60. K_(b)` of `NH_(4)OH= 2 xx 10^(-5)`

Explanation

To solve the problem of how many moles of `NH₄Cl` should be added to `200 mL` of a `0.18 M NH₄OH` solution to achieve a `pH` of `9.60`, we can follow these steps:

### Step 1: Calculate the pOH
Given that the pH is `9.60`, we can find the pOH using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
So,
\[ \text{pOH} = 14 - \text{pH} = 14 - 9.60 = 4.40 \]

### Step 2: Calculate pKb
We are given the \( K_b \) of \( NH₄OH \) as \( 2 \times 10^{-5} \). We can calculate \( pK_b \) using the formula:
\[ pK_b = -\log(K_b) \]
Thus,
\[ pK_b = -\log(2 \times 10^{-5}) \]
Calculating this gives:
\[ pK_b \approx 4.70 \]

### Step 3: Use the Henderson-Hasselbalch equation
For a basic buffer, the relationship between pOH, pKb, and the concentrations of salt and base can be expressed as:
\[ \text{pOH} = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Substituting the known values:
\[ 4.40 = 4.70 + \log\left(\frac{[\text{Salt}]}{0.18}\right) \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ \log\left(\frac{[\text{Salt}]}{0.18}\right) = 4.40 - 4.70 = -0.30 \]
Taking the antilog:
\[ \frac{[\text{Salt}]}{0.18} = 10^{-0.30} \]
Calculating \( 10^{-0.30} \):
\[ 10^{-0.30} \approx 0.5012 \]
Thus,
\[ [\text{Salt}] = 0.5012 \times 0.18 \approx 0.0902 \, M \]

### Step 5: Calculate the moles of salt
To find the moles of salt (NH₄Cl) needed, we use the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \]
The volume is given as \( 200 \, mL = 0.200 \, L \):
\[ \text{Moles of Salt} = 0.0902 \, M \times 0.200 \, L = 0.01804 \, moles \]

### Final Answer
Therefore, the number of moles of \( NH₄Cl \) that should be added is approximately:
\[ \text{Moles of } NH₄Cl \approx 0.018 \, moles \]