Methylamine, CH3NH2, is a weak base that hydrolyzes water according to the equation shown below. The value of ionization constant, Kb, is 5.2 x 10-4 (a) Calculate the concentration of OH in a 0.225M methylamine solution. (b) Calculate the pH.
Methylamine, CH3NH2, is a weak base that hydrolyzes water according to the equation shown below. The value of ionization constant, Kb, is 5.2 x 10-4
(a) Calculate the concentration of OH in a 0.225M methylamine solution.
(b) Calculate the pH.
The Correct Answer and Explanation is :
Let’s break this down step by step.
Given Data:
- The concentration of methylamine ([CH_3NH_2]) = 0.225 M
- Ionization constant ((K_b)) = (5.2 \times 10^{-4})
- Methylamine hydrolyzes water according to the equation:
[
CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-
]
(a) Calculation of OH⁻ Concentration
To solve for the concentration of OH⁻, we will first set up an ICE (Initial, Change, Equilibrium) table.
Let’s assume that the initial concentration of (CH_3NH_2) is 0.225 M and that none of the (CH_3NH_3^+) and OH⁻ have been formed at the start, so their initial concentrations are 0.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| (CH_3NH_2) | 0.225 | -x | 0.225 – x |
| (CH_3NH_3^+) | 0 | +x | x |
| (OH^-) | 0 | +x | x |
The expression for (K_b) is:
[
K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}
]
Substituting the equilibrium concentrations into this equation:
[
5.2 \times 10^{-4} = \frac{x \cdot x}{0.225 – x}
]
Since (K_b) is small, we can assume that (x) is much smaller than 0.225 M, so we approximate (0.225 – x \approx 0.225).
Thus, the equation simplifies to:
[
5.2 \times 10^{-4} = \frac{x^2}{0.225}
]
Solving for (x):
[
x^2 = (5.2 \times 10^{-4}) \times 0.225 = 1.17 \times 10^{-4}
]
[
x = \sqrt{1.17 \times 10^{-4}} \approx 0.0108 \, \text{M}
]
So, the concentration of OH⁻ is approximately (0.0108 \, \text{M}).
(b) Calculation of pH
To calculate the pH, we need to find the pOH first using the concentration of OH⁻. The relationship between pOH and OH⁻ concentration is given by:
[
pOH = -\log[OH^-]
]
Substitute the value of ( [OH^-] = 0.0108 \, \text{M} ):
[
pOH = -\log(0.0108) \approx 1.97
]
Finally, we can calculate the pH using the relationship between pH and pOH:
[
pH + pOH = 14
]
[
pH = 14 – pOH = 14 – 1.97 = 12.03
]
Final Answers:
- (a) The concentration of OH⁻ in the solution is approximately (0.0108 \, \text{M}).
- (b) The pH of the solution is approximately 12.03.
Explanation:
The weak base methylamine ((CH_3NH_2)) reacts with water to produce hydroxide ions (OH⁻) and methylammonium ions ((CH_3NH_3^+)) through hydrolysis. The concentration of OH⁻ is determined by applying the ionization constant (K_b) in an equilibrium expression. Since (K_b) is small, we assume that the change in concentration is negligible compared to the initial concentration, simplifying the calculations. The concentration of OH⁻ is then used to determine the pOH, and from there, the pH is calculated. The pH value indicates a basic solution, which is consistent with the behavior of methylamine as a weak base.