Chemistry Forum

5.9 g of trimethyl amine [ (CH₃)₃N ] is added to enough water to make 1.0 L of solution. 50.0 mL of this solution is placed in an Erlenmeyer flask and is then titrated with 0.100 M HCl.

a.Calculate the pH of the solution in the Erlenmeyer flask before the titration is started.

b.Calculate the pH of the solution at the midpoint of the titration.

c.Calculate the pH of the solution at the equivalence point.

d.What indicator would you choose for the titration?

My answers:

a) Molar mass of [(CH₃)₃ N] = 59.112 g/ gmol. This means 5.9 g of (CH₃)₃N is 0.1 M. This means 50 ml of [(CH₃)₃N] have 0.005 mol.
\(K_b\) of (CH₃)₃N = 6.4e-5 . After adding H₂O, to [(CH₃)₃N], the reaction is

[(CH₃)₃N] + H₂O <=> [(CH₃)₃NH+] + OH- This means \(0.000064= \frac{x*x}{(0.1-x)} \Rightarrow x= 0.00253 M, pOH=2.6 , pH= 14.0 - 2.6 = 11.4 \)

b) The reaction is \(Me_3N(aq) + HCl(aq) \rightarrow Me_3NH^+(aq) + Cl^-(aq) \) There is 1:1 molar ratio between Trimethylamine and Hydrochloric acid. So 0.005 mol of acid is required for this reaction. How to proceed further? I know some next steps which i shall give in my next reply.

a) is ok.
b) it is asked for the midpoint, what means half base is titrated. Use HH equation.
c) means all base is converted to ammonium salt. Use Kw and kb.
d) indicator range should between -1 target pH +1

Using Handerson-Hasselbalch Equation, my answer to b) is \( pH= 11.4 + \log{\frac{1}{1}}=11.4.\) Is this correct?

No pH 11.4 you have already in beginning with no acid. Your answer from a. After adding acid the pH must drop. Use right pks value or calculate pOH first and use pKb. In logarithmen add the concentration of base and salt.

Using Handerson-Hasselbalch equation pH of the solution at the mid-point of titration is =\( 9.81 + \log{\frac{0.0025}{0.0025}}=9.81\). Is this correct?

Yes its correct.

My answer to part c) of this question.

This solution is now completely composed of salt of strong acid. The pH of this solution will be acidic.
1) Calculate the molarity of trimethylammonium cation. 0.05 mol/0.050 ml = 0.1 M.

2) Calculate the \( K_a\) of the trimethylamine cation \(K_w= K_aK_b \) \(\Rightarrow\) 1e-14=0.000064*x \(\Rightarrow x= 1.5625e-10\)

3) Calculate the pH of the solution
\(1.5625e-10=\frac{ x*x}{0.1} \Rightarrow x=3.9528e-6, \Rightarrow pH=5.403\)

Is this pH of the solution at equivalence point correct?

Is the above-mentioned answer correct?

pH = ½ (pKa + log cB + 14)

Where you get the 100 ml from.

Thanks for pointing out my mistake. I have made necessary corrections to my answer to c.

Is the final answer correct now?

It is correct

I want to know answers to the following questions?
1) What is the volume of HCl required for this reaction \([(CH_3)_3N] + HCl \rightarrow [(CH_3)_3NH^+] + Cl^- \)

My answer is 500 ml.

2) What is the final volume of \([(CH_3)_3NH^+]\) after reaction?
my answer is 550 ml.

Are these above answers correct?

No. You have only 50 ml 0,1 M Amin what requires the same amount of 0,1 M HCl. So you have to use 50 ml. Final volume would be 100 ml.

This reflects back to the other question regarding pH. The right concentrations have to be considered.

You have mentioned \(pH= \frac12 (pK_a + \log{(cB)} +14) \) How to use this formula here to get pH of 5.403 of the solution at equivalence point.

There is a mistake in.
The ka is 6,4*10^-5 what guides to 4,19 pka. We want to calculate the corresponding acid the Trimethylammonium means have to use negativ value

pH = ½ (pKa -log cB + 14) = (-4,19 +1+14)/2 = 5,4