REPORT SHEET Titration Curves of Polyprotic Acids tandardization of Sodium Hydro
ID: 636813 • Letter: R
Question
REPORT SHEET Titration Curves of Polyprotic Acids tandardization of Sodium Hydroxide (NaOHD Solution B. Trial 1 Trial 2 Trial 3 Mass of bottle +KHP Mass of bottle Mass of KHP used 2.97? Final buret reading Initial buret reading mL of NaOH used 2232411 Molarity of NaOH Average molarity (show calculations) a6 Sandand deviation (show calculations)_0 01 O 10et 0 0223 +01040 ? see attached C. Determination of the Acid Dissociation Constants and the Molarity of the Unknown Acid Trial 1 Trial 2 Trial 3 25 mL Volume of unknown acid 25 mL25 ml Molarity of NaOH from above .0Q(@0 mL of NaOH at eq. point 1 mLofNaOH at eq. point 2 Molarity of unknown acid 03,13 89Explanation / Answer
in trail 1 molarity of the NaoH is 0.0960, assume the titration value at point 1 is 30.05ml, and the titration value at point 2 is 30.10ml, in this case avg is 30.07
Follow the formula V1M1=V2M2
V1 is the avge titration value, M1 is the Molarity of NaOH, V2 is the volume of un known aci Solution
30.07*0.0960=25*M2
M2= 30.07*0.0960/25 = 0.1154 is the molarity of HCl in trail 1
in trail 2 molarity of the NaoH is 0.0823, assume the titration value at point 1 is 25.05ml, and the titration value at point 2 is 25.20ml, in this case avg is 25.12
Follow the formula V1M1=V2M2
25.12*0.0823=25*M2
M2= 25.12*0.0823/25 = 0.0826 is the molarity of HCl in trail 2
in trail 3 molarity of the NaoH is 0.1040, assume the titration value at point 1 is 23.05ml, and the titration value at point 2 is 23.20ml, in this case avg is 23.12
Follow the formula V1M1=V2M2
23.12*0.1040=25*M2
M2= 23.12*0.1040/25 = 0.096 is the molarity of HCl in trail 3
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