Acid-Base Titrations

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Dhamnekar Winod
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Acid-Base Titrations

Post by Dhamnekar Winod »

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.
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Re: Acid-Base Titrations

Post by ChenBeier »

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
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Re: Acid-Base Titrations

Post by Dhamnekar Winod »

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

Post by ChenBeier »

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.
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Re: Acid-Base Titrations

Post by Dhamnekar Winod »

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?
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Re: Acid-Base Titrations

Post by ChenBeier »

Yes its correct.
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Re: Acid-Base Titrations

Post by Dhamnekar Winod »

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?
Last edited by Dhamnekar Winod on Fri Mar 19, 2021 2:42 am, edited 1 time in total.
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Re: Acid-Base Titrations

Post by Dhamnekar Winod »

Is the above-mentioned answer correct?
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Re: Acid-Base Titrations

Post by ChenBeier »

pH = ½ (pKa + log cB + 14)

Where you get the 100 ml from.
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Re: Acid-Base Titrations

Post by Dhamnekar Winod »

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

Is the final answer correct now?
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Re: Acid-Base Titrations

Post by ChenBeier »

It is correct
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Re: Acid-Base Titrations

Post by Dhamnekar Winod »

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?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Acid-Base Titrations

Post by ChenBeier »

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.
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Re: Acid-Base Titrations

Post by Dhamnekar Winod »

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.
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Re: Acid-Base Titrations

Post by ChenBeier »

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
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