# Experiment 11: ACID-BASE TITRATIONSample Data Collection and Results PagesVolume of NaOH used in the Titration:Rough trialTrial 3Trial 1Trial 2Initial reading (mL)o.00mLZ1.0SmLO .0omo.00MLFinal reading (mL)43.15 mLZ1.50mL21.6om Lz1.45MLVolume dispensed (mL)21.55MLZI. L5mL21.5OML21.o0OmL21.6omLAverage volume of NaOH used (Trials 1-3)O. 2012 MMolarity of NaOHMOLESMOLARITY =Unknown Vinegar #_N ALITERSCalculation of moles HC2H3O2 in 5.00 mL vinegar:acetie acid =21. o om Ha0HMoles of

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With the attached data, how do o calculate the molarity of vinegar? help_outlineImage TranscriptioncloseExperiment 11: ACID-BASE TITRATION Sample Data Collection and Results Pages Volume of NaOH used in the Titration: Rough trial Trial 3 Trial 1 Trial 2 Initial reading (mL) o.00mL Z1.0SmL O .0om o.00ML Final reading (mL) 43.15 mL Z1.50mL 21.6om L z1.45ML Volume dispensed (mL) 21.55ML ZI. L5mL 21.5OML 21.o0OmL 21.6omL Average volume of NaOH used (Trials 1-3) O. 2012 M Molarity of NaOH MOLES MOLARITY = Unknown Vinegar #_N A LITERS Calculation of moles HC2H3O2 in 5.00 mL vinegar: acetie acid =21. o om Ha0H Moles of fullscreen
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Step 1

The number of moles for a given chemical species refers to the ratio of mass of the species to that of its molar mass. The formula is shown below:

Step 2

The molarity or the concentration can be defined as the ratio of number of moles of solute which are present per liter of the given solution. The SI units of molarity are moles/L. The mathematical formula as follows:

Step 3

The molarity as well as the volume of an unknown solution during the titr... help_outlineImage TranscriptioncloseMу-м,V, M, is the molarity of the stock solution V is the volume of the stock solution M, is the molarity of the unknown solution V is the volume of the unknown solution fullscreen

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