Question

Construct the titration curve, for the titration of 50.0 mL of 0.125 M trimethylamine with 0.175 M HC1 (aq). (Use graph paper and show all your work for full CREDIT) Calculate the initial pH of the solution in the flask? Calculate the pH of the solution mixture after 15 mL of titrant is added? Calculate the pH at the equivalence volume? Calculate the pH at of the mixture after 45 mL of titrant is added? Calculate the pH of the mixture after 60 mL of titrant is added?

Answer 1

The amount of HCl needed for complete titration of tri methyl aniline is 50 X 0.125 = (V) X 0.175

V = 35.714 ml

1. Insert values into the K_b expression for acetic acid. The K_b for tri methyl amine is 6.3 x 10^-5.

6.3 x 10^-5 = [(x) (x)] / 0.125

x = 2.807 x 10^-3 M

pOH = 2.5517

pH = 11.4482

2.

1. Calculate moles of acid and base in solution before reaction:

(CH₃)₃N : 6.25 x 10^-3 mol

HCl : (0.0015 L) (0.175 mol/L) = 2.625 x 10^-3 mol

2) Determine amounts of base and tri methyl ammonium ion after reaction:

(CH₃)₃N : 6.25 x 10^-3 mol - 2.625 x 10^-3 mol = 3.625 x 10^-3 mol
(CH₃)₃NH⁺ : 2.625 x 10^-3 mol

3) Use Henderson-Hasselbalch equation to determine pOH of buffered solution:

pOH = pK_b + log (salt/base)

pOH = 4.2 + log (2.625 x 10^-3 / 3.625 x 10^-3)

pOH = 4.2 + log 0.724

pOH = 4.059

pH = 14 - 4.059 = 9.94

3.

The solution is now completely composed of a salt of a weak base. The pH of this solution will be acidic.

1) Calculate molarity of tri methyl ammonium ions:

6.25 x 10^-3 mol / 0.085714 L = 0.0729 M

2) Calculate the K_a of tri methyl ammonium

K_w = K_aK_b

1.00 x 10^-14 = (x) (6.25 x 10^-5 )

x = 1.587 x 10^-10

3) Calculate pH of the solution:

1.587 x 10^-10 = [(x) (x)] / (0.0729-x)

x = 0.34 x 10^-5 M

pH = 4.53147

4. After adding 35.714 ml of HCl the amount of tri methyl aniline the total amount is titrated and after that addition of HCl indicates the accumuation of H⁺ ions in the solution and the resultant solution will be acidic In nature.

The concentration of H⁺ ions is obtained by using (M_acidVacid - M_baseV_base) / (V_acid + V_base )

[H⁺] = ((0.175 X 45 ) - (50 X 0.125)) / (50 + 45) = 0.01710

pH = - log[H⁺] = 1.767

5. Addition of more amount of HCl results in addition of more amount of H⁺ ions accumulation and a greater increase in pH

[H⁺] = ((0.175 X 60) - (50 X 0.125)) / (50 + 60) = 0.038636

pH = - log[H⁺] = 1.413

The data used to draw the graph are volume of the amount of HCl added to tri methyl aniline

Volume of HCl added pH of the resulting solution

intially 0 ml 11.4482

5 ml 10.5884

10 ml 10.210

15 ml 9.94

20 ml 9.695

25 ml 9.432

30 ml 9.0798

35 ml 8.1098

35.714 ml 4.53147

40 ml 2.079

45 ml 1.767

60 ml 1.413

Titration of Trimethyl amine with HCl 13.0 12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 o 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 10 20 35 Volume of HCl added