0.006 50 M solution of ethanolamine, H_{2}NCH_{2}CH_{2}*OH has a pH of 10.64 at 25 deg * C Calculate the K_{b} of ethanolamine. What concentra- tion of undissociated ethanolamine remains at equilibrium?
The Correct Answer and Explanation is:
To solve for the Kb of ethanolamine and the concentration of undissociated ethanolamine at equilibrium, we need to break down the process step by step.
Step 1: Write the reaction for the dissociation of ethanolamine
Ethanolamine, a weak base, dissociates in water as follows:H2NC2H4OH+H2O⇌H2NC2H4OH2++OH−\text{H}_2\text{NC}_2\text{H}_4\text{OH} + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{NC}_2\text{H}_4\text{OH}_2^+ + \text{OH}^-H2NC2H4OH+H2O⇌H2NC2H4OH2++OH−
Here, ethanolamine reacts with water to produce ethanolamine protonated (ethanolammonium ion) and hydroxide ions. The equilibrium constant for this reaction is Kb.
Step 2: Use the given pH to find [OH⁻]
The pH is given as 10.64, so we can calculate the pOH using the relation:pOH=14−pH\text{pOH} = 14 – \text{pH}pOH=14−pHpOH=14−10.64=3.36\text{pOH} = 14 – 10.64 = 3.36pOH=14−10.64=3.36
Now, calculate the concentration of hydroxide ions ([OH⁻]):[OH−]=10−pOH=10−3.36≈4.37×10−4 M[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-3.36} \approx 4.37 \times 10^{-4} \, \text{M}[OH−]=10−pOH=10−3.36≈4.37×10−4M
Step 3: Set up an ICE table
We now set up the ICE (Initial, Change, Equilibrium) table for the dissociation of ethanolamine. Let x be the concentration of ethanolamine that dissociates:
| Component | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| Ethanolamine (H₂NC₂H₄OH) | 0.00650 | -x | 0.00650 – x |
| Ethanolamine ion (H₂NC₂H₄OH₂⁺) | 0 | +x | x |
| Hydroxide ion (OH⁻) | 0 | +x | x |
From the ICE table, we know that at equilibrium, [OH⁻] = x. Thus, x = 4.37 × 10⁻⁴ M.
Step 4: Solve for Kb
The expression for the base dissociation constant Kb is:Kb=[H2NC2H4OH2+][OH−][H2NC2H4OH]K_b = \frac{[\text{H}_2\text{NC}_2\text{H}_4\text{OH}_2^+][\text{OH}^-]}{[\text{H}_2\text{NC}_2\text{H}_4\text{OH}]}Kb=[H2NC2H4OH][H2NC2H4OH2+][OH−]
Substitute the known values into this equation:Kb=x×x0.00650−xK_b = \frac{x \times x}{0.00650 – x}Kb=0.00650−xx×xKb=(4.37×10−4)20.00650−4.37×10−4≈1.91×10−70.00611≈3.13×10−5K_b = \frac{(4.37 \times 10^{-4})^2}{0.00650 – 4.37 \times 10^{-4}} \approx \frac{1.91 \times 10^{-7}}{0.00611} \approx 3.13 \times 10^{-5}Kb=0.00650−4.37×10−4(4.37×10−4)2≈0.006111.91×10−7≈3.13×10−5
Thus, the Kb of ethanolamine is approximately 3.13 × 10⁻⁵.
Step 5: Find the concentration of undissociated ethanolamine at equilibrium
The concentration of undissociated ethanolamine at equilibrium is:[H2NC2H4OH]=0.00650−x=0.00650−4.37×10−4≈0.00606 M[\text{H}_2\text{NC}_2\text{H}_4\text{OH}] = 0.00650 – x = 0.00650 – 4.37 \times 10^{-4} \approx 0.00606 \, \text{M}[H2NC2H4OH]=0.00650−x=0.00650−4.37×10−4≈0.00606M
Final Answer:
- The Kb of ethanolamine is 3.13 × 10⁻⁵.
- The concentration of undissociated ethanolamine at equilibrium is approximately 0.00606 M.
