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
To answer this question, we first need to write the balanced chemical equation for the neutralization reaction and determine which atom is being protonated.
(a) Balanced chemical equation for the neutralization reaction:
CH3CH2NH2 (aq) + HI (aq) -> CH3CH2NH3+ (aq) + I- (aq)
In this reaction, the nitrogen atom in ethylamine (CH3CH2NH2) is being protonated.
Next, let's calculate the equilibrium constant for the neutralization reaction by considering it as the sum of two separate reactions:
1. CH3CH2NH2 (aq) + H2O (l) -> CH3CH2NH3+ (aq) + OH- (aq)
This reaction involves the base ethylamine accepting a proton from water to form the conjugate acid.
2. HCl (aq) + OH- (aq) -> H2O (l) + I- (aq)
This reaction involves the acid HCl donating a proton to hydroxide ion to form water.
The overall reaction can be obtained by adding the two reactions:
CH3CH2NH2 (aq) + HCl (aq) -> CH3CH2NH3+ (aq) + I- (aq)
The equilibrium constant (K) for this reaction can be calculated as the product of the equilibrium constants (K1 and K2) for the individual reactions:
K = K1 * K2
The equilibrium constant expression for reaction 1 (K1) is given by:
K1 = [CH3CH2NH3+] / [CH3CH2NH2] * [OH-]
The equilibrium constant expression for reaction 2 (K2) is given by:
K2 = [H2O] / [HCl] * [I-] / [OH-]
By substituting the values of K1 and K2 into the equation for K, we can calculate K.
(b) To calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at different stages of the titration, we need to consider the stoichiometry and the concept of acid-base reactions.
i) Before the addition of any HI solution:
- The solution contains only ethylamine (CH3CH2NH2).
- The pH can be calculated from the pKb value and the concentration of ethylamine using the Henderson-Hasselbalch equation:
pH = pKb + log([CH3CH2NH3+]/[CH3CH2NH2])
- Since no HI solution has been added, the concentrations of [CH3CH2NH3+] and [CH3CH2NH2] are the same.
ii) After the addition of 20.0 mL of HI solution:
- Calculate the moles of HI added:
Moles of HI = concentration of HI × volume of HI solution (in L)
- The reaction between HI and ethylamine is 1:1, so the moles of HI consumed are equal to the moles of ethylamine.
- Calculate the remaining moles of ethylamine and use it to obtain the updated concentrations of [CH3CH2NH3+] and [CH3CH2NH2].
iii) At the equivalence point:
- The equivalence point occurs when the moles of acid (HI) added are stoichiometrically equal to the moles of base (ethylamine) present initially.
- Use this information to calculate the concentration of ethylamine, [CH3CH2NH2], and [CH3CH2NH3+] at the equivalence point.
iv) After the addition of 60.0 mL of HI solution:
- Repeat the calculations mentioned in step (ii) to determine the updated concentrations of [CH3CH2NH3+] and [CH3CH2NH2].
By following these steps and performing the necessary calculations, you will be able to determine the pH, [CH3CH2NH2], and [CH3CH2NH3+] at each stage of the titration.