An aqueous solution of methylamine CH3NH2 has a pH of 10.56 . How many grams of methylamine are there in 100.0mL of the solution? Round your answer to 2 significant digits.
The correct answer and explanation is:
To find the amount of methylamine (CH₃NH₂) in the solution, we must first determine the concentration of methylamine and then convert it to mass.
Step 1: Determine the pOH and pKb
Given that the pH of the solution is 10.56, we can find the pOH of the solution by using the relationship: pOH=14−pHpOH = 14 – pH pOH=14−10.56=3.44pOH = 14 – 10.56 = 3.44
Next, we use the pOH to calculate the concentration of hydroxide ions (OH⁻): [OH−]=10−pOH=10−3.44=3.60×10−4 M[OH⁻] = 10^{-pOH} = 10^{-3.44} = 3.60 \times 10^{-4} \, \text{M}
Step 2: Set up the equilibrium expression for methylamine
Methylamine (CH₃NH₂) is a weak base and reacts with water to form hydroxide ions and methylammonium ions (CH₃NH₃⁺): CH₃NH₂+H₂O⇌CH₃NH₃++OH−\text{CH₃NH₂} + \text{H₂O} \rightleftharpoons \text{CH₃NH₃}⁺ + \text{OH}⁻
The base dissociation constant (Kₓ) for methylamine is: Kb=[CH3NH3+][OH−][CH3NH2]K_b = \frac{[CH₃NH₃⁺][OH⁻]}{[CH₃NH₂]}
Step 3: Find the concentration of methylamine
From the information, we know that the concentration of OH⁻ is 3.60×10−4 M3.60 \times 10^{-4} \, \text{M}. This corresponds to the amount of CH₃NH₃⁺ produced at equilibrium, and since methylamine reacts in a 1:1 ratio with OH⁻, the concentration of CH₃NH₂ at equilibrium is the same: [CH3NH2]=3.60×10−4 M[CH₃NH₂] = 3.60 \times 10^{-4} \, \text{M}
Step 4: Calculate the mass of methylamine
To find the mass of methylamine, we use the formula: mass=molarity×molar mass×volume\text{mass} = \text{molarity} \times \text{molar mass} \times \text{volume}
The molar mass of CH₃NH₂ is: Molar mass of CH₃NH₂=12.01 (C)+3×1.008 (H)+14.01 (N)=31.06 g/mol\text{Molar mass of CH₃NH₂} = 12.01 \, (\text{C}) + 3 \times 1.008 \, (\text{H}) + 14.01 \, (\text{N}) = 31.06 \, \text{g/mol}
Now, calculating the mass in 100.0 mL (0.100 L) of solution: mass=(3.60×10−4 mol/L)×(31.06 g/mol)×(0.100 L)=1.12×10−3 g\text{mass} = (3.60 \times 10^{-4} \, \text{mol/L}) \times (31.06 \, \text{g/mol}) \times (0.100 \, \text{L}) = 1.12 \times 10^{-3} \, \text{g}
Thus, the mass of methylamine in 100.0 mL of the solution is 1.1 mg (rounded to 2 significant digits).