An ammonia solution is 9.9% ammonia by mass and has a density of 0.99 g/mL. Calculate the pH of the solution. `K_b (NH_4OH) = 1.7 xx 10^(-5)`

Answer

To calculate the pH of the ammonia solution, we will follow these steps:

### Step 1: Calculate the mass of ammonia in the solution
Given that the ammonia solution is 9.9% by mass, in a 100 g solution, the mass of ammonia (NH₃) is:
\[ \text{Mass of NH}_3 = 9.9 \, \text{g} \]

### Step 2: Calculate the number of moles of ammonia
The molar mass of ammonia (NH₃) is approximately 17 g/mol. Therefore, the number of moles of ammonia is:
\[ \text{Moles of NH}_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{9.9 \, \text{g}}{17 \, \text{g/mol}} \approx 0.581 \, \text{mol} \]

### Step 3: Calculate the volume of the solution
Using the density of the solution (0.99 g/mL), we can find the volume of 100 g of the solution:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \, \text{g}}{0.99 \, \text{g/mL}} \approx 101.01 \, \text{mL} = 0.10101 \, \text{L} \]

### Step 4: Calculate the concentration of ammonia
The concentration of ammonia in the solution is given by:
\[ \text{Concentration of NH}_3 = \frac{\text{Moles of NH}_3}{\text{Volume of solution in L}} = \frac{0.581 \, \text{mol}}{0.10101 \, \text{L}} \approx 5.74 \, \text{mol/L} \]

### Step 5: Use the \(K_b\) expression to find \([\text{OH}^-]\)
The dissociation of ammonia can be represented as:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]
Using the \(K_b\) expression:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \approx \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \approx \frac{C\alpha^2}{C} = C\alpha^2 \]
Where \(C\) is the concentration of ammonia and \(\alpha\) is the degree of dissociation.

Rearranging gives:
\[ \alpha = \sqrt{\frac{K_b}{C}} \]

Substituting \(K_b = 1.7 \times 10^{-5}\) and \(C = 5.74 \, \text{mol/L}\):
\[ \alpha = \sqrt{\frac{1.7 \times 10^{-5}}{5.74}} \approx 0.0053 \]

### Step 6: Calculate \([\text{OH}^-]\)
Now we can find the concentration of hydroxide ions:
\[ [\text{OH}^-] = C\alpha = 5.74 \times 0.0053 \approx 0.0304 \, \text{mol/L} \]

### Step 7: Calculate the \(pOH\)
\[ pOH = -\log[\text{OH}^-] = -\log(0.0304) \approx 1.52 \]

### Step 8: Calculate the \(pH\)
Using the relation \(pH + pOH = 14\):
\[ pH = 14 - pOH = 14 - 1.52 \approx 12.48 \]

### Final Answer
The pH of the ammonia solution is approximately **12.48**.

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