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

(a) The balanced chemical equation for the neutralization reaction is:

CH3CH2NH2 + HI → CH3CH2NH3+ + I-

In this reaction, the nitrogen atom in ethylamine (CH3CH2NH2) is being protonated.

To calculate the equilibrium constant for the neutralization reaction, we can consider two separate reactions:
1. CH3CH2NH2 + H2O → CH3CH2NH3+ + OH-
2. HI + H2O → H3O+ + I-

The equilibrium constant for the net neutralization reaction can be calculated by multiplying the equilibrium constants for these two reactions:

Knet = Kb × Ka

Where Kb is the base dissociation constant for ethylamine and Ka is the acid dissociation constant for HI.

Given: pKb = 3.367

We can calculate Kb using the equation:

Kb = 10^(-pKb)

Kb = 10^(-3.367) = 4.066 × 10^(-4)

The value of Ka for HI is 1.0 × 10^9 (approximated as a strong acid).

Therefore, the equilibrium constant for the neutralization reaction is:

Knet = (4.066 × 10^(-4)) × (1.0 × 10^9) = 4.066 × 10^5

(b) i) Before the addition of any HI solution:
- The pH can be calculated using the pOH equation:
pOH = -log[OH-] = -log(Kw/Kb) = -log(1.0 × 10^(-14)/4.066 × 10^(-4)) ≈ 10.39
pH = 14 - pOH ≈ 3.61

- [CH3CH2NH2] = initial concentration = 0.105 mol L-1
- [CH3CH2NH3+] = 0 (no HI added yet)

ii) After the addition of 20.0 mL of HI solution:
- The volume of the solution is now 40.0 + 20.0 = 60.0 mL = 0.060 L
- In this case, half of the initial ethylamine will be neutralized by the added HI.

- [CH3CH2NH2] = initial concentration - moles neutralized by HI
= 0.105 - (0.060 L × 0.150 mol L-1/2) ≈ 0.0985 mol L-1

- [CH3CH2NH3+] = moles neutralized by HI
= 0.060 L × 0.150 mol L-1/2 ≈ 0.0045 mol L-1

iii) At the equivalence point:
- The volume of HI required to reach the equivalence point can be calculated using the equation:

molesCH3CH2NH2 = molesHI
(0.040 L × 0.105 mol L-1) = (V × 0.150 mol L-1)
V ≈ 0.028 L = 28.0 mL

At the equivalence point, the volume of HI added is 40.0 + 28.0 = 68.0 mL, and the CH3CH2NH2 is fully neutralized.

- [CH3CH2NH2] = 0 mol L-1
- [CH3CH2NH3+] = initial concentration + moles added by HI
= 0 + (0.068 L × 0.150 mol L-1) ≈ 0.0102 mol L-1

iv) After the addition of 60.0 mL of HI solution:
- The volume of the solution is now 40.0 + 60.0 = 100.0 mL = 0.100 L

- [CH3CH2NH2] = 0 mol L-1
- [CH3CH2NH3+] = initial concentration + moles added by HI
= 0 + (0.100 L × 0.150 mol L-1) ≈ 0.0150 mol L-1

Please note, the calculations are approximate and rounding off has been done for simplicity.

(a) The balanced chemical equation for the neutralization reaction between ethylamine (CH3CH2NH2) and hydroiodic acid (HI) is:

CH3CH2NH2 + HI → CH3CH2NH3+ + I-

In this reaction, the nitrogen atom in ethylamine is being protonated by the hydrogen ion from hydroiodic acid.

To calculate the equilibrium constant (Kb) for the neutralization reaction, we need to consider the two separate reactions involved:

1) CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH- (Kb1)
2) HI + H2O ⇌ H3O+ + I- (Kw)

The overall neutralization reaction can be written as the sum of these two reactions:

CH3CH2NH2 + HI ⇌ CH3CH2NH3+ + I-

By calculating the equilibrium constant for each reaction, we can determine the equilibrium constant for the overall neutralization reaction using the equation:

K overall = Kb1 / Kw

(b) To calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at different stages of the titration, we need to consider the stoichiometry and equilibrium relationships.

i) Before adding any HI solution:
At this stage, the concentration of [CH3CH2NH2] is 0.105 mol L-1 and [CH3CH2NH3+] is 0 mol L-1. The [OH-] concentration can be determined using Kb expression for ethylamine:

Kb1 = [CH3CH2NH3+][OH-] / [CH3CH2NH2]

Since [OH-] can be considered negligible compared to [CH3CH2NH2], we can assume Kb1 ≈ [CH3CH2NH3+][OH-].
Therefore, at this stage, [OH-] ≈ 0 and the solution is basic. To calculate the pH, we can use the equation pH = 14 - pOH.

ii) After adding 20.0 mL of HI solution:
The stoichiometry of the reaction shows that 1 mole of CH3CH2NH2 reacts with 1 mole of HI. Therefore, at this stage, we have used 20.0 mL / 40.0 mL = 0.5 of the initial moles of ethylamine. Thus, the remaining moles of ethylamine are 0.105 mol - 0.105 mol x 0.5 = 0.0525 mol.
The moles of CH3CH2NH3+ formed are also 0.0525 mol. We can calculate the concentration of [CH3CH2NH2] and [CH3CH2NH3+] using their moles and the new total volume of 40.0 mL + 20.0 mL = 60.0 mL.

iii) At the equivalence point:
At the equivalence point, the moles of CH3CH2NH2 and HI are stoichiometrically equal. Therefore, 0.105 mol of HI has reacted with 0.105 mol of CH3CH2NH2. The moles of CH3CH2NH3+ formed are also 0.105 mol. We can calculate the concentration of [CH3CH2NH2] and [CH3CH2NH3+] using their moles and the total volume of 40.0 mL + 40.0 mL = 80.0 mL.

iv) After adding 60.0 mL of HI solution:
By adding 60.0 mL of HI solution, we have used a total volume of 40.0 mL + 60.0 mL = 100.0 mL. The moles of HI used can be calculated from its concentration and volume. Subtracting the moles of HI used from the initial moles of CH3CH2NH2 will give the remaining moles of CH3CH2NH2. The moles of CH3CH2NH3+ formed can be obtained from the stoichiometry.

To calculate the pH at each stage, we can use the equation pH = -log[H+], where [H+] concentration is calculated from the [CH3CH2NH3+] concentration.

Note: To obtain precise numerical values, more specific data about volume measurements and concentrations would be required.