Methylamine, CH3NH2, is a weak base. CH3NH2(aq) + H2O(l) → CH3NH3+(aq) + OH-(aq) If the pH of a 0.065 M solution is 11.70, what is the value of Kb? Show the steps please.
The Correct Answer and Explanation is:
To find the value of the base dissociation constant (Kb) for methylamine (CH3NH2), we can use the given data and apply the equilibrium principles.
Step 1: Write the equilibrium expression
The equilibrium reaction for methylamine is:
CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)
The equilibrium constant expression for Kb is:Kb=[CH3NH3+][OH−][CH3NH2]K_b = \frac{[CH3NH3^+][OH^-]}{[CH3NH2]}Kb=[CH3NH2][CH3NH3+][OH−]
Step 2: Use the pH to find [OH-]
Given that the pH of the solution is 11.70, we can first calculate the pOH using the formula:pOH=14−pH=14−11.70=2.30\text{pOH} = 14 – \text{pH} = 14 – 11.70 = 2.30pOH=14−pH=14−11.70=2.30
Then, we can calculate the concentration of OH- ions using the relationship between pOH and [OH-]:[OH−]=10−pOH=10−2.30=5.012×10−3 M[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2.30} = 5.012 \times 10^{-3} \text{ M}[OH−]=10−pOH=10−2.30=5.012×10−3 M
Step 3: Set up an ICE table
Next, we will use an ICE (Initial, Change, Equilibrium) table to calculate the concentrations of all species at equilibrium. Initially, before any dissociation:
- [CH3NH2] = 0.065 M (the concentration of methylamine)
- [CH3NH3+] = 0 M
- [OH-] = 0 M
At equilibrium:
- [CH3NH2] = 0.065 – x (where x is the change in concentration due to dissociation)
- [CH3NH3+] = x
- [OH-] = x
From the pOH calculation, we know that at equilibrium, [OH-] = 5.012×10−3 M5.012 \times 10^{-3} \text{ M}5.012×10−3 M, so:x=5.012×10−3 Mx = 5.012 \times 10^{-3} \text{ M}x=5.012×10−3 M
Step 4: Substitute into the Kb expression
Now, we can substitute these equilibrium concentrations into the Kb expression:Kb=(5.012×10−3)(5.012×10−3)0.065−5.012×10−3K_b = \frac{(5.012 \times 10^{-3})(5.012 \times 10^{-3})}{0.065 – 5.012 \times 10^{-3}}Kb=0.065−5.012×10−3(5.012×10−3)(5.012×10−3)
Simplify the expression:Kb=(5.012×10−3)20.065−5.012×10−3=2.512×10−50.05999K_b = \frac{(5.012 \times 10^{-3})^2}{0.065 – 5.012 \times 10^{-3}} = \frac{2.512 \times 10^{-5}}{0.05999}Kb=0.065−5.012×10−3(5.012×10−3)2=0.059992.512×10−5Kb≈4.18×10−4K_b \approx 4.18 \times 10^{-4}Kb≈4.18×10−4
Final Answer:
The value of KbK_bKb for methylamine is approximately 4.18 × 10⁻⁴.
Explanation:
In this calculation, we first used the pH to find the concentration of OH- ions in the solution. Then, we applied the equilibrium principles by setting up an ICE table to find the concentration of CH3NH2, CH3NH3+, and OH- at equilibrium. Using the known equilibrium concentrations, we calculated Kb, which reflects the base’s strength. The higher the Kb value, the stronger the base.
