Question

I am looking for help with just trial one. I am not sure what equations to use or how to go about even answering these questions. I figure if I can get help with trial 1 I should be able to do trial 2 on my own.
( This week has been really bad, I lost my grandfather, so if you could "dumb" down the steps as much as possible I would appreciate it, my brain is just not working well. I have went and seen my instructor and that helped a little but nothing stuck)

82 CHEM 132L General Chemistry ll Week 2 Trial #1 Trial #2 mL of NaOH added PH pH mL ol NaOH added 7.00 00 .54 0.00 h. l, 00 13.05 1-92 15.155.3 22.00 | ↓1.4 L 10 4014 26.00 5 Print your titration curves foe both Trial-1 AND Trial lab report. If you did not record data abowe, please print the data table, too · 4,5 1.0T 52 ー Lab 11: Analysis of an Unknown Acid83 Trial 2 Trial # 1 olume of NaOH added to reach the equivalence point Volume of Na0H added to reach M the equivalence point pH at te the equirvalence point pK, of unknown acid Average pk, of unknown acid Possible identity of unknown weak acid Molarity of unknown acid [show cakculations) Average molarity of unknown acid

Answer 1

Hello, first, I'm sorry about your grandfather, I hope I can be of help, I'll explain:

The volume of NaOH added at the equivalence point, read in the graph, is the volume of NaOH that is read in the sharp jump of pH that you can see in your curve (you can take an average between the values ​​18 and 19 mL) , this is: 18.5 mL

The volume of NaOH at the midpoint of equivalence is calculated by dividing the previous volume by 2:

V 1/2 = 18.5 / 2 = 9.25 mL

The pH at the mean point of equivalence is read in the graph, at the volume of 9.25 mL, this pH is approximately 4.60.

The pKa of the acid, is equal to the pH at the midpoint of equivalence, this is 4.60.

The identity of the acid, you can suppose, comparing the obtained pKa with a table of proposed acids (the closest one).

The molarity of the acid is calculated with the volume of the acid titrated, with the volume of NaOH at the equivalence point and its concentration, with the equation:

[Acid] = [NaOH] * V NaOH / V Acid

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