DATA TABLE Concentration of NaOH Trial 1 Trial 2 Trial 3 Trial 4 1.2 2.3 34.3 30

Question

DATA TABLE Concentration of NaOH Trial 1 Trial 2 Trial 3 Trial 4 1.2 2.3 34.3 30.2 Initial buret readin Final buret reading Volume of NaOH added Moles of NaOH addecd Moles of HCI reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HC Precision CALCULATIONS I. One mole of HCl neutralizes one mole of NaOH. So that moles of acid = moles of base. Molarity, M, is composed of the quotient of the moles divided by the liters: M = mol/L 2. Use this to calculate moles of base from the known molarity of the NaOH: Volume (liters) x moles/liters = moles OH. 3. Remember: moles OH-= mol HCl 4. Calculate molarity of the acid: Moles HCl / volume HCl (liters) = mol/ M 5. Calculate the molarity of the undiluted acid: MAG = McV2. 6. Calculate the precision of your data based on the Average Molarity of the Undiluted HCI

Explanation / Answer

Concentration of NaoH is required to solve this problem. Please do provide and repost it again

Lets assume conc of NaOH is x mol/litre

So now For trial 1,

Moles Of NaOH added= 27.1x

Moles of HCl added= 27.1x as moles of NaOH=moles of HCl

Molarity of diluted HCl= 27.1x (M) assumed volume of the acid= 1 litre

now molarity of undiluted HCl can be calculated from V1S2=V2S2 formula.

So now For trial 2,

Moles Of NaOH added= 27.9x

Moles of HCl added= 27.9x as moles of NaOH=moles of HCl

Molarity of diluted HCl= 27.9x (M) assumed volume of the acid= 1 litre

now molarity of undiluted HCl can be calculated from V1S2=V2S2 formula.

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