Methylamine, CH3NH2, is a weak base. CH3NH2 (aq) + H2O (l) –> CH3NH3+ (aq) + OH- (aq) If the pH of a 0.150 M solution of the amine is 10.01, what is the value of pKb?
The Correct Answer and Explanation is:
To find the value of pKbpK_bpKb for methylamine (CH3NH2), we need to follow these steps:
Step 1: Write the equilibrium expression for the reaction.
The dissociation of methylamine in water can be represented as:CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)\text{CH}_3\text{NH}_2 (\text{aq}) + \text{H}_2\text{O} (\text{l}) \rightleftharpoons \text{CH}_3\text{NH}_3^+ (\text{aq}) + \text{OH}^- (\text{aq})CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)
The equilibrium expression for this reaction is:Kb=[CH3NH3+][OH−][CH3NH2]K_b = \frac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]}Kb=[CH3NH2][CH3NH3+][OH−]
Step 2: Calculate the concentration of OH−\text{OH}^-OH−.
The pH of the solution is given as 10.01. We can first convert pH to pOH:pOH=14−pH=14−10.01=3.99\text{pOH} = 14 – \text{pH} = 14 – 10.01 = 3.99pOH=14−pH=14−10.01=3.99
Now, we calculate the concentration of OH−\text{OH}^-OH− using the relationship:[OH−]=10−pOH=10−3.99≈1.0×10−4 M[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-3.99} \approx 1.0 \times 10^{-4} \, \text{M}[OH−]=10−pOH=10−3.99≈1.0×10−4M
Step 3: Use the ICE table to find the equilibrium concentrations.
Let’s assume that the initial concentration of CH3NH2 is 0.150 M and the change in concentration at equilibrium due to dissociation is xxx. From the balanced equation, the change in concentrations of CH3NH2, CH3NH3+, and OH- will be as follows:
| Substance | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| CH3NH2 | 0.150 | -x | 0.150 – x |
| CH3NH3+ | 0 | +x | x |
| OH- | 0 | +x | x |
From the pOH calculation, we know that at equilibrium, [OH−]=1.0×10−4 M[\text{OH}^-] = 1.0 \times 10^{-4} \, \text{M}[OH−]=1.0×10−4M, so x=1.0×10−4x = 1.0 \times 10^{-4}x=1.0×10−4.
Step 4: Solve for KbK_bKb.
Now we can substitute the equilibrium concentrations into the expression for KbK_bKb:Kb=(1.0×10−4)(1.0×10−4)0.150−1.0×10−4≈1.0×10−80.150K_b = \frac{(1.0 \times 10^{-4})(1.0 \times 10^{-4})}{0.150 – 1.0 \times 10^{-4}} \approx \frac{1.0 \times 10^{-8}}{0.150}Kb=0.150−1.0×10−4(1.0×10−4)(1.0×10−4)≈0.1501.0×10−8Kb≈6.67×10−8K_b \approx 6.67 \times 10^{-8}Kb≈6.67×10−8
Step 5: Calculate pKbpK_bpKb.
Finally, the pKbpK_bpKb is the negative logarithm of KbK_bKb:pKb=−logKb=−log(6.67×10−8)≈7.18pK_b = -\log K_b = -\log(6.67 \times 10^{-8}) \approx 7.18pKb=−logKb=−log(6.67×10−8)≈7.18
Conclusion:
The value of pKbpK_bpKb for methylamine is approximately 7.18.
