Suppose the acid you were titrating was sulfuric acid, H2SO4, in place of acetic

Question

Suppose the acid you were titrating was sulfuric acid, H2SO4, in place of acetic acid. Using your data from part B(assume the volume of acid used was also 5.00 mL), calculate the molarity of the sulfuric acid.

Trial 1: Vol of vinegar: 5.00 mL, Inital Buret: 0.10 mL, Final Buret: 14.30mL, Vol of NaOH used: 14.20mL

Trial 2: Vol of vinegar: 5.00 mL, Initial Buret: 14.30 mL, Final Buret: 28.72mL, Vol of NaOH used: 28.20 mL

Trial 3: Vol of vingear: 5.00 mL, Initial Buret: 14.20mL, Final Buret: 28.20, Vol of NaOH used: 14.00mL

Explanation / Answer

To neutralize V volume of M molarity sulfuric acid , we need 2*V volume of M molarity NaOH.

Volume of acid = V = 5 ml

Trial 1: volume of NaOH used = 14.20 mL

Let the molarity of NaOH be M1.

moles of NaOH = 0.0142*M1

moles of sulfuric acid = 0.005*M

hence 0.005*M = 0.0142*M1

hence M= 0.258*M1 , where M1 is the molarity of the NaOH used in the 1st trial.

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