Methylamine, CH3NH2, is a weak base that reacts according to the reaction
CH3NH2 + H2O <–> CH3NH3 + OH”
The value of the ionization constant, Kb, is 5.25 x 104. Methylamine reacts to form salts such as methylammonium nitrate, (CH3NH3)
(NO3).
b. Calculate the pH of a solution made by adding 0.0150 mole of solid methylammonium nitrate to 150.0 mL of a 0.125 molar solution of methylamine. Assume that no volume change occurs.
Answer:
Methylamine, CH3NH2, is a weak base that reacts according to the reaction
CH3NH2 + H2O <-> CH3NH3 + OH”
The value of the ionization constant, Kb, is 5.25 x 104. Methylamine reacts to form salts such as methylammonium nitrate, (CH3NH3″)
(NO3).
How many moles of either NaOH or HCI (state clearly which you choose) should be added to the solution in (b) to produce a solution with
a pH of 11.00? Assume that no volume change occurs.
The Correct Answer and Explanation is :
Part 1: Calculating the pH of the Solution
To calculate the pH of the solution made by adding methylammonium nitrate (CH₃NH₃NO₃) to the methylamine (CH₃NH₂) solution, we need to consider the reactions and equilibrium between methylamine and water, as well as the contribution of the methylammonium ion (CH₃NH₃⁺), which acts as a weak acid.
Step 1: Write the equilibrium equations and determine known quantities.
- Methylamine (CH₃NH₂) is a weak base that reacts with water: [
CH₃NH₂ + H₂O \rightleftharpoons CH₃NH₃⁺ + OH⁻
]
Kb for CH₃NH₂ = 5.25 x 10⁻⁴. - The methylammonium ion (CH₃NH₃⁺) from methylammonium nitrate reacts as a weak acid: [
CH₃NH₃⁺ + H₂O \rightleftharpoons CH₃NH₂ + H₃O⁺
]
The ionization constant Ka for CH₃NH₃⁺ can be found using the relationship between Ka and Kb: [
K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{5.25 \times 10^{-4}} = 1.90 \times 10^{-11}
]
Step 2: Set up the equilibrium for CH₃NH₃⁺ dissociation.
The concentration of CH₃NH₃⁺ is initially 0.0150 moles in 0.150 L, so:
[
\text{Concentration of CH₃NH₃⁺} = \frac{0.0150 \, \text{mol}}{0.150 \, \text{L}} = 0.100 \, \text{M}
]
Now, for the dissociation of CH₃NH₃⁺ in water:
[
\text{CH₃NH₃⁺} + H₂O \rightleftharpoons CH₃NH₂ + H₃O⁺
]
The change in concentration can be written as:
[
\begin{aligned}
\text{Initial concentration:} & \quad [CH₃NH₃⁺] = 0.100 \, \text{M} \quad \text{(CH₃NH₂ and H₃O⁺ are initially 0)} \
\text{Change in concentration:} & \quad -x \quad \text{(for CH₃NH₃⁺, +x for CH₃NH₂ and +x for H₃O⁺)} \
\text{Equilibrium concentrations:} & \quad [CH₃NH₃⁺] = 0.100 – x, \quad [CH₃NH₂] = x, \quad [H₃O⁺] = x
\end{aligned}
]
The equilibrium expression for Ka is:
[
K_a = \frac{[CH₃NH₂][H₃O⁺]}{[CH₃NH₃⁺]} = \frac{x^2}{0.100 – x}
]
Substitute the known value of Ka:
[
1.90 \times 10^{-11} = \frac{x^2}{0.100 – x}
]
Assume ( x ) is small compared to 0.100 M (because Ka is small), so ( 0.100 – x \approx 0.100 ):
[
1.90 \times 10^{-11} = \frac{x^2}{0.100}
]
Solving for ( x ):
[
x^2 = (1.90 \times 10^{-11}) \times 0.100 = 1.90 \times 10^{-12}
]
[
x = \sqrt{1.90 \times 10^{-12}} = 1.38 \times 10^{-6} \, \text{M}
]
Thus, the concentration of H₃O⁺ is ( 1.38 \times 10^{-6} ) M. The pH is:
[
\text{pH} = -\log[H₃O⁺] = -\log(1.38 \times 10^{-6}) = 5.86
]
Therefore, the pH of the solution is approximately 5.86.
Part 2: Adding NaOH or HCl to Achieve a pH of 11.00
To achieve a pH of 11.00, we need to adjust the concentration of ( H₃O⁺ ) to a level that corresponds to this pH. The target pH of 11.00 corresponds to a target ( [H₃O⁺] ) of:
[
\text{pH} = 11.00 \quad \Rightarrow \quad [H₃O⁺] = 10^{-11} \, \text{M}
]
Since the initial concentration of ( H₃O⁺ ) is ( 1.38 \times 10^{-6} \, \text{M} ), we need to reduce the ( [H₃O⁺] ) concentration. This can be done by adding NaOH (a strong base) to neutralize some of the ( H₃O⁺ ).
To calculate the moles of NaOH needed, we use the change in ( [H₃O⁺] ):
[
\Delta[H₃O⁺] = (1.38 \times 10^{-6} – 10^{-11}) \, \text{M} \approx 1.38 \times 10^{-6} \, \text{M}
]
Now, the moles of ( H₃O⁺ ) to be neutralized:
[
\text{Moles of } H₃O⁺ = 1.38 \times 10^{-6} \, \text{M} \times 0.150 \, \text{L} = 2.07 \times 10^{-7} \, \text{mol}
]
Since NaOH and H₃O⁺ react in a 1:1 ratio, the moles of NaOH required are the same:
[
\text{Moles of NaOH} = 2.07 \times 10^{-7} \, \text{mol}
]
Thus, to achieve a pH of 11.00, you should add approximately 2.07 × 10⁻⁷ moles of NaOH.