Chemistry Forum

Calculate the pH at the beginning, at the midpoint, and at the equivalence point for the titration also choose an indicator:
a) of 50 mL of 0.1M HCl with 0.1 M NaOH
b) of 50 mL of 0.1 M CF₃COOH with 0.1 M NaOH
c) 50 mL of 0.1 M triethylamine with 0.1 M HCl

I am working on these questions.

My answer to a) : At the beginning, pH of NaOH will be very large, but when we start dropping 50 mL of HCl, pH will start decreasing. At some time, time will come when 0.005 mol of HCl will be equal to moles of NaOH i-e 0.005 mol. This point is called equivalence point. At this point, the pH of the solution will be 7.

Please be aware that M is a concentration and not the same like moles. You mix it up.
The solution is 0,1 M equal to 0,1 mol/ l. The moles in 50 ml is 0.005 mol and not 0.005 M.

Thanks for pointing out my mistake. I corrected my mistake in my post. What would be indicator for the reaction in a) Will it be Phenol red?

Indicator pH must be around +- 1 pH around equivalent point. Strong acid and strong Base means pH 6 to 8. Phenol red would be fine 6.4 - 8.2
Bromine thymolblue, neutral red,rosol acid, m-nitrophenol also possible.

Answer to a) At the beginning, pH of the solution, before adding any amount of NaOH, is 1(pH=-log(0.1 )=1). After adding 0.0025 moles of NaOH to 50.0 mL of HCl, at the mid-point, the pH will be 1.48(pH=-log(0.0025/0.0525)=1.32). At the equivalence point pH will be 7.

Is this answer correct?

Is the titration in question b) strong acid strong base or weak acid strong base titration?

Is the titration in question c), weak base strong acid titration?

a is correct
b no. Check properties of Trifluorine acetic acid.
c yes

My answer to b)
At the beginning, pH of the solution, before adding any amount of NaOH is 1.15(pH=-log(0.0707)=1.15). After adding 0.0025 mol of NaOH, to the 50 mL of CF₃COOH, at the mid point, the pH of the solution is 1.48(pH=-log(0.0025/0.075)=1.48). At the equivalence point, pH will be 7

Is this answer correct?

a) is now wrong, why you changed to 0,00525
You have 50 ml + 25 ml = 75 ml = 0,075 l
b) is wrong too

Thanks, you are correct. I wrongly computed the new volume after adding NaOH. I corrected my last post . You may check it now. Is my answer to b) is correct now?

Now its correct. What is your opinion to trifluorine acetic acid.

Trifluoroacetic acid is a strong organic acid with pKa =0

My answer to c) At the beginning, the pH of 50 mL C₆H₁₅N is 7.49. At the midpoint, pH of 0.075 mL mixture of HCl and C₆H₁₅N will be log(0.025/0.075)=1.48 At the equivalence point, pH will be less than 7.
Note: Kb of C₆H₁₅N= 5.3e-4

Is this answer correct?

No, its a weak base so you have to calculate with HENDERSON Hasselbalch equation.

So, my corrected answer to c)

At the beginning, the pH of 50 mL solution of C₆H₁₅N(triethyamine) is 10.72
At the mid-point, pH of 0.075 mL mixture of C₆H₁₅N and HCl is 10.72

At the equivalence point, pH of 0.100 mL solution of 0.1M C₆H₁₅N and 0.1 M HCl is 6.01.

Show your calculation. At midpoint its wrong. HCl is not the point, which acid we are dealing with.

Yes. My pH computation at the midpoint for C₆H₁₅N + HCl reaction was wrong. Now, after working some more computations, at the mid-point, my answer for pH of 0.075mL mixture of HCl and C₆H₁₅N is 6.1.

Is this answer correct?