The pH of 0.01 M ammonium sulphate `[K_b NH_3]=2xx10^(-5)`

Options

A. 5.5
B. 6.8
C. 8.6
D. 7

Explanation

To find the pH of a 0.01 M solution of ammonium sulfate, we can follow these steps:

### Step 1: Identify the components of ammonium sulfate
Ammonium sulfate \((NH_4)_2SO_4\) is a salt formed from the weak base ammonia \((NH_3)\) and the strong acid sulfuric acid \((H_2SO_4)\). In solution, it dissociates into ammonium ions \((NH_4^+)\) and sulfate ions \((SO_4^{2-})\).

### Step 2: Determine the relevant equilibrium
The ammonium ion \((NH_4^+)\) can donate a proton to water, establishing an equilibrium:
\[ NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+ \]
This reaction shows that ammonium ions act as a weak acid.

### Step 3: Calculate the \(K_a\) for \(NH_4^+\)
Using the relationship between \(K_a\) and \(K_b\):
\[ K_w = K_a \cdot K_b \]
where \(K_w = 1.0 \times 10^{-14}\) at 25°C and \(K_b\) for ammonia is given as \(2.0 \times 10^{-5}\).

First, calculate \(K_a\):
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-5}} = 5.0 \times 10^{-10} \]

### Step 4: Calculate \(pK_a\)
Now, calculate \(pK_a\):
\[ pK_a = -\log(K_a) = -\log(5.0 \times 10^{-10}) \approx 9.3 \]

### Step 5: Use the formula for pH
For a salt of a weak base and strong acid, the pH can be calculated using the formula:
\[ pH = 7 - \frac{1}{2} pK_a - \frac{1}{2} \log C \]
where \(C\) is the concentration of the salt (0.01 M).

### Step 6: Substitute values into the pH formula
Substituting the values:
\[ pH = 7 - \frac{1}{2} \times 9.3 - \frac{1}{2} \log(0.01) \]
Calculating \(\log(0.01)\):
\[ \log(0.01) = -2 \]
Now substituting this back:
\[ pH = 7 - \frac{1}{2} \times 9.3 - \frac{1}{2} \times (-2) \]
\[ pH = 7 - 4.65 + 1 = 7 - 3.65 = 5.35 \]

### Step 7: Final pH value
Thus, the pH of the 0.01 M ammonium sulfate solution is approximately **5.35**.

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