DATA TABLE Concentration of NaOH LOOX10 Trial 1 Trial 2 Trial 3 Trial 4 1.2 2.3
ID: 587751 • Letter: D
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DATA TABLE Concentration of NaOH LOOX10 Trial 1 Trial 2 Trial 3 Trial 4 1.2 2.3 Initial buret reading Final buret reading Volume of NaOH added Moles of NaOH added Moles of HCI reacted 2. Molarity of diluted HC Molarity of undiluted HCI Average molarity of undiluted HC Precision CALCULATIONS 1. 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 HCI (liters) = mol L = M 5. Calculate the molarity of the undiluted acid: MV1 M2V2. 6. Calculate the precision of your data based on the Average Molarity of the Undiluted HCIExplanation / Answer
2. As given in the question, the molarity of acid can be calculated from the volume of base added and its concentration using the law of equivalence. However, as the volume of HCl used is absent, the number of moles of NaOH is first calculated from the volume of NaOH added as 0.1mol/Lx(27.5mL/1000mL) = 0.00275moles of NaOH.
3. Moles of acid = moles of base as per law of conservation of mass. This gives the moles of acid as 0.00275moles.
4. Since the precise volume of HCl used is not given, assuming the quantity to be V (in L), the molarity can be calculated as 0.00275/V. For instance, if V is 1, molarity = 0.00275M and if V = 0.5 (i.e. 500mL/1000mL), the molarity will be 0.0055M.
5. If the molarity of diluted HCl is found as 0.00275/V in Molarity, then the molarity of the undiluted HCl will be Mundiluted = (0.00275/VxVdiluted)/Vundiluted.
6. Without the actual value of undiluted HCl, only with the experimental value, the precision cannot be calculated.
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