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
After I went to bed last night and before I went to sleep, I worked the Keq in my head BUT you know how difficult that can be without pen and paper; therefore PLEASE double check this part to make sure I didn't slip up somewhere with something extra or something missing. If I call ethylamine, BNH2, then
if pKb = 3.367 the Kb = 4.3E-4
I followed the hint in the problem. The first equation is the ionization of the BNH2 in water.
BNH2 + HOH ==> BNH3^+ + OH^-
The second equation is the titration of the OH^- with H^+. It is
H^+ + OH^- ==> H2O
Now add those equations together and set up the Keq expression.
Now if you look at the expression you will (I think) a (BNH3^+)(OH^-)/(BNH2) which of course is Kb.
Check those off and look further. You should find a (H^+)(OH^-) in the denominator which means Kw. Further you should see (H2O) in the numerator and (H2O) in the denominator) so they cancel. If I wasn't too sleepy all that ends up with Keq = Kb/Kw but check it out.
For the titration itself, do this.
1. Determine the volume HCl to reach the equivalence point, approximately 28 mL.
2. For the beginning (zero mL HCl), set up an ICE chart for
.........BNH2 + HOH ===> BNH3^+ + OH^-
I........0.105............0........0
C..........-x..............x.......x
E........0.105-x............x......x
Set up the Kb expression substitute the equilibrium values and solve for x = (OH^-); convert that to pH.
3. For all points between zero and the eq. point, use the Henderson-Hasselbalch equation.
4. For the eq point, use the hydrolysis of the salt. If you have 0.004 mols of the salt produced (that's approximate remember), then mols/L will give you about 0.004 mols/(0.040+0.028) = about 0.06M
........BNH3^+ + HOH ==> H3O^+ + BNH2
I.......0.06M.............0........0
C.......-x................x........x
E......0.06-x.............x........x
Ka for BNH3^+ = Kw/Kb for BNH2) = (x)(x)/0.06-x and solve for x = (H3O^+) and convert to pH.
5. For all point after the eq point, that is simply the excess H^+ that has been added. Don't forget to take into account the dilution of the excess HCl.
If you have questions post your work and explain what you don't understand.
To answer these questions, we need to understand the neutralization reaction and how titrations work. Let's break it down step by step:
1a) The balanced chemical equation for the neutralization reaction between ethylamine (CH3CH2NH2) and hydroiodic acid (HI) can be written as follows:
CH3CH2NH2 + HI → CH3CH2NH3+ + I-
In this reaction, the hydrogen atom (H) from HI is protonating the nitrogen atom (N) in ethylamine, leading to the formation of the ethylammonium ion (CH3CH2NH3+).
To calculate the equilibrium constant (K) for this neutralization reaction, we can consider it as the sum of two separate reactions: the dissociation of ethylamine and the protonation of ethylamine.
The reactions are as follows:
1. CH3CH2NH2 → CH3CH2NH- + H+
2. HI → H+ + I-
The equilibrium constant for reaction 1 can be represented as Ka = [CH3CH2NH-][H+]/[CH3CH2NH2]. Since ethylamine is a weak base, we can assume that [H+] in the presence of ethylamine is negligible. Thus, Ka = [CH3CH2NH-]/[CH3CH2NH2].
Similarly, the equilibrium constant for reaction 2 (dissociation of HI) can be represented as Kb = [H+][I-]/[HI]. In this case, the concentration of [H+] is equal to the concentration of [I-] formed during the neutralization reaction.
The equilibrium constant (K) for the overall reaction can be calculated using the relationship K = Ka * Kb.
1b) Now let's solve the pH and concentration calculations at different stages of the titration:
i) Before the addition of any HI solution:
Since no HI has been added yet, the concentration of CH3CH2NH3+ is 0, and the concentration of CH3CH2NH2 is 0.105 mol L-1. The pH can be calculated using the pKb value: pOH = pKb + log([CH3CH2NH3+]/[CH3CH2NH2]). Convert pOH to pH using the relationship pH = 14 - pOH.
ii) After the addition of 20.0 mL of HI solution:
At this stage, we need to calculate how much HI has reacted with CH3CH2NH2. Since the HI concentration is 0.150 mol L-1 and the volume added is 20.0 mL, we can calculate the moles of HI added. From the stoichiometry of the balanced equation, we know that 1 mole of HI reacts with 1 mole of CH3CH2NH2. Thus, the concentration of CH3CH2NH2 will decrease and the concentration of CH3CH2NH3+ will increase. Use the moles of HI reacted to calculate the new concentrations of CH3CH2NH2 and CH3CH2NH3+.
iii) At the equivalence point:
The equivalence point is reached when the moles of acid added equal the moles of base initially present. Calculate the moles of HI added at the equivalence point and use it to determine the concentration of CH3CH2NH2 and CH3CH2NH3+.
iv) After the addition of 60.0 mL of HI solution:
Using similar calculations as in the previous steps, calculate the moles of HI reacted and the concentrations of CH3CH2NH2 and CH3CH2NH3+.
Remember to convert volumes to liters when calculating concentrations.
By following these steps, you'll be able to answer the questions regarding the neutralization reaction and the titration of ethylamine with HI.