Calculate the pH of a 0.52 M NH3 (Kb = 1.8×10^-5) solution.
a. 5.03
b. 2.51
c. 8.97
d. 0.52
e. 11.49
The correct answer and explanation is :
To calculate the pH of a 0.52 M NH₃ (ammonia) solution, we need to understand that ammonia is a weak base that partially ionizes in water. The dissociation of ammonia in water is represented as:
[
\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-
]
We are given the base dissociation constant (Kb) of ammonia, which is ( 1.8 \times 10^{-5} ), and the concentration of ammonia, which is 0.52 M. Using this information, we can set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of the species at equilibrium.
Step 1: Write the equilibrium expression
The equilibrium expression for the dissociation of ammonia is:
[
\text{Kb} = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}
]
Substituting the known values:
[
1.8 \times 10^{-5} = \frac{[ \text{NH}_4^+ ][ \text{OH}^- ]}{[ \text{NH}_3 ]}
]
Step 2: Set up the ICE table
Let ( x ) represent the concentration of ( \text{OH}^- ) produced at equilibrium. Therefore, the concentration of ( \text{NH}_4^+ ) will also be ( x ), and the concentration of ( \text{NH}_3 ) will decrease by ( x ).
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| NH₃ | 0.52 | -x | 0.52 – x |
| NH₄⁺ | 0 | +x | x |
| OH⁻ | 0 | +x | x |
Step 3: Solve the equation
Substitute the equilibrium concentrations into the Kb expression:
[
1.8 \times 10^{-5} = \frac{x \cdot x}{0.52 – x}
]
Assume that ( x ) is small compared to 0.52, so we approximate ( 0.52 – x \approx 0.52 ). This simplifies the equation to:
[
1.8 \times 10^{-5} = \frac{x^2}{0.52}
]
Solving for ( x ):
[
x^2 = (1.8 \times 10^{-5}) \times 0.52
]
[
x^2 = 9.36 \times 10^{-6}
]
[
x = \sqrt{9.36 \times 10^{-6}} = 3.06 \times 10^{-3} \, \text{M}
]
Step 4: Calculate the pOH and pH
Now that we know the concentration of OH⁻ (( x = 3.06 \times 10^{-3} ) M), we can calculate the pOH:
[
\text{pOH} = -\log[\text{OH}^-] = -\log(3.06 \times 10^{-3}) = 2.51
]
Since pH and pOH are related by the equation:
[
\text{pH} + \text{pOH} = 14
]
Substitute the value for pOH:
[
\text{pH} = 14 – 2.51 = 11.49
]
Final Answer:
The pH of the 0.52 M NH₃ solution is 11.49.
Thus, the correct answer is e. 11.49.