Can I get some help finding the molarity of HCl based on the data I gathered? When it reads 25.00/.03, I was using a volumetric flask with a differentiation of +/- 0.03mL

Question

Trial 1 Trial 2 Trial 3

Initial buret reading 0.31 0.40 18.76

final buret reading 28.43 28.59 46.71

Vol. of NaOH used 28.12 28.19 27.95

Molarity of NaOH 0.2512 0.2512 0.2512

Vol. of HCl used 20.00/.03 25.00/.03 25.00/.03

Molarity of HCl

Answer 1

The balanced equation for this neutralization is...

HCl + NaOH ==> NaCl + H2O

Volume of NaOH used = final buret reading - initial buret reading

I'll do the calculations for Trial 1, and you repeat them for Trials 2 and 3.

Volume NaOH = 28.43 ml - 0.31 ml = 28.12 ml

moles NaOH used = 28.12 ml x 1 L/1000 ml x 0.2512 mol/L = 0.007064 moles

moles HCl = 0.007064 mol NaOH x 1 mol HCl/mol NaOH = 0.007064 moles HCl

Molarity (M) of HCl = 0.007064 moles/20.00 ml x 1000 ml/L = 0.3532 M