Titration and pH computation
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- Dhamnekar Winod
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Titration and pH computation
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.
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.
Last edited by Dhamnekar Winod on Tue Apr 27, 2021 8:33 am, edited 1 time in total.
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- ChenBeier
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Re: Titration and pH computation
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.
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.
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Re: Titration and pH computation
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?
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Re: Titration and pH computation
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.
Bromine thymolblue, neutral red,rosol acid, m-nitrophenol also possible.
- Dhamnekar Winod
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Re: Titration and pH computation
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?
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?
Last edited by Dhamnekar Winod on Wed Apr 28, 2021 6:58 am, edited 1 time in total.
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Re: Titration and pH computation
a is correct
b no. Check properties of Trifluorine acetic acid.
c yes
b no. Check properties of Trifluorine acetic acid.
c yes
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Re: Titration and pH computation
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?
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?
Last edited by Dhamnekar Winod on Wed Apr 28, 2021 8:53 am, edited 3 times in total.
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Re: Titration and pH computation
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
You have 50 ml + 25 ml = 75 ml = 0,075 l
b) is wrong too
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Re: Titration and pH computation
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?
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Re: Titration and pH computation
Now its correct. What is your opinion to trifluorine acetic acid.
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Re: Titration and pH computation
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?
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?
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Re: Titration and pH computation
No, its a weak base so you have to calculate with HENDERSON Hasselbalch equation.
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Re: Titration and pH computation
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.
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.
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Re: Titration and pH computation
Show your calculation. At midpoint its wrong. HCl is not the point, which acid we are dealing with.
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Re: Titration and pH computation
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?
Is this answer 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]