Ethylamine, CH3CH2NH2, is an organic base with pKb = 3.367 at 298 K. In an experiment, a 40.0 mL sample of 0.105 mol L-1 CH3CH2NH2 (aq) is titrated with 0.150 mol L-1 HI(aq) solution at 298 K. (1a) Write a balanced chemical equation for the neutralization reaction upon which this titration is based, and indicate clearly which atom is being protonated. Then, calculate the equilibrium constant for the neutralization reaction. (Hint: To calculate the equilibrium constant, you may find it helpful to represent the neutralization reaction as the sum of two separate reactions.) (1b) Calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at the following stages of the titration. i) before the addition of any HI solution. ii) after the addition of 20.0 mL of HI solution. iii) at the equivalence point. iv) after the addition of 60.0 mL of HI solution
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To answer the questions, we need to understand the neutralization reaction and the equilibrium constant. Let's break it down step by step.
1a) The neutralization reaction between ethylamine (CH3CH2NH2) and hydrogen iodide (HI) can be represented by the following balanced chemical equation:
CH3CH2NH2 + HI -> CH3CH2NH3+ + I-
In this reaction, the ethylamine molecule is being protonated, meaning it gains a hydrogen ion (H+) to form the ethylammonium cation (CH3CH2NH3+).
To calculate the equilibrium constant for this neutralization reaction, we can break it down into two separate reactions:
1) CH3CH2NH2 + H2O -> CH3CH2NH3+ + OH-
2) HI + H2O -> H3O+ + I-
The equilibrium constant (Kw) for the ionization of water is 1.0 x 10^-14 at 298 K.
Therefore, we can use the concentrations of hydroxide ion (OH-) and hydronium ion (H3O+) to calculate the equilibrium constant for each individual reaction.
K1 = [CH3CH2NH3+][OH-] / [CH3CH2NH2] (from reaction 1)
K2 = [H3O+][I-] / [HI] (from reaction 2)
Finally, the equilibrium constant for the neutralization reaction is the product of K1 and K2:
K = K1 * K2
1b) Let's calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at different stages of the titration:
i) Before the addition of any HI solution:
- The initial volume of ethylamine solution is 40.0 mL, which corresponds to 0.040 L.
- [CH3CH2NH2] = 0.105 mol L^-1 * 0.040 L = 0.0042 mol
- [CH3CH2NH3+] = 0 mol (no HI added yet)
- Since ethylamine is a weak base, it undergoes hydrolysis in water, resulting in the formation of OH-. Therefore, the pH will be slightly basic, but we need more information to calculate the exact pH.
ii) After the addition of 20.0 mL of HI solution:
- The total volume of the solution is now 40.0 mL + 20.0 mL = 60.0 mL = 0.060 L.
- The moles of HI added can be calculated as [HI] = 0.150 mol L^-1 * 0.020 L = 0.003 mol.
- The moles of ethylamine remaining can be calculated as 0.0042 mol - 0.003 mol = 0.0012 mol.
- The moles of ethylammonium formed can be calculated as 0.003 mol.
- [CH3CH2NH2] = 0.0012 mol / 0.060 L = 0.02 mol L^-1
- [CH3CH2NH3+] = 0.003 mol / 0.060 L = 0.05 mol L^-1
- Note: The pH can be calculated using the Henderson-Hasselbalch equation, but we need additional information, such as the pKa of ethylamine or the pH of the HI solution.
iii) At the equivalence point:
- The equivalence point is reached when the moles of HI added are equal to the moles of ethylamine initially present.
- The total volume of the solution is 40.0 mL + 40.0 mL = 80.0 mL = 0.080 L.
- The moles of HI added can be calculated as [HI] = 0.150 mol L^-1 * 0.040 L = 0.006 mol.
- The moles of ethylamine initially present are 0.105 mol L^-1 * 0.040 L = 0.0042 mol.
- Since HI is a strong acid/base, it fully dissociates into H3O+ and I- ions.
- Therefore, all the moles of ethylamine react with the moles of HI added, resulting in the formation of moles of ethylammonium.
- [CH3CH2NH2] = 0 mol (all used up)
- [CH3CH2NH3+] = 0.006 mol / 0.080 L = 0.075 mol L^-1
- The pH at the equivalence point will depend on the pKa of ethylamine or the pH of the HI solution.
iv) After the addition of 60.0 mL of HI solution:
- The total volume of the solution is now 40.0 mL + 60.0 mL = 100.0 mL = 0.100 L.
- The moles of HI added can be calculated as [HI] = 0.150 mol L^-1 * 0.060 L = 0.009 mol.
- The moles of ethylamine remaining can be calculated as 0.0042 mol - 0.009 mol = -0.0048 mol (negative because all ethylamine is used up).
- The moles of ethylammonium formed can be calculated as 0.009 mol.
- [CH3CH2NH2] = 0 mol (all used up)
- [CH3CH2NH3+] = 0.009 mol / 0.100 L = 0.09 mol L^-1
- The pH will depend on the pKa of ethylamine or the pH of the HI solution.
Note: To calculate the exact pH values, we need additional information, such as the pKa of ethylamine or the pH of the HI solution.