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. (a) 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.) (b) Calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at the following stages of the titration. i) before te 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

Question

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. (a) 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.) (b) Calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at the following stages of the titration. i) before te 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

Answer 1

(a) The neutralization reaction can be represented as follows:

CH3CH2NH2 (aq) + HI (aq) → CH3CH2NH3+ (aq) + I- (aq)

In this reaction, the proton (H+) is being added to the ethylamine molecule, resulting in the formation of ethylammonium ion (CH3CH2NH3+).

To calculate the equilibrium constant (K) for this neutralization reaction, we can break it down into two separate reactions:

1. Dissociation of ethylamine:

CH3CH2NH2 (aq) ⇌ CH3CH2NH2 (aq) + H2O (l)

2. Protonation of ethylamine:

CH3CH2NH2 (aq) + H+ (aq) ⇌ CH3CH2NH3+ (aq)

The equilibrium constant for the neutralization reaction (K) can be expressed as the product of the equilibrium constants for these two reactions:

K = [CH3CH2NH3+] / ([CH3CH2NH2] x [H+])

(b) To calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at different stages of the titration:

i) Before the addition of any HI solution:
At this stage, the concentration of ethylamine remains at its initial value of 0.105 mol L-1. The concentration of ethylammonium ion is zero, and the pH can be calculated using the pKb value.

ii) After the addition of 20.0 mL of HI solution:
The number of moles of HI added can be calculated using the volume and concentration of the HI solution. From the balanced chemical equation, we can determine that the stoichiometric ratio between HI and CH3CH2NH2 is 1:1. Use the number of moles of HI to determine the moles of ethylamine that reacted. Then, calculate the new concentration of ethylamine and the concentration of ethylammonium ion. The pH can be determined using the pKb value and the concentration of ethylammonium ion.

iii) At the equivalence point:
The equivalence point is reached when the moles of HI added are equal to the moles of ethylamine present initially. Use the volume and concentration of HI solution to calculate the moles of HI added. Then, calculate the concentration of ethylamine and ethylammonium ion at the equivalence point. The pH can be determined using the pKb value and the concentration of ethylammonium ion.

iv) After the addition of 60.0 mL of HI solution:
Repeat the steps in (ii) and (iii) to calculate the concentration of ethylamine, ethylammonium ion, and the pH at this stage.