"indicatior" - 1 õppematerjal

HCl lahus

from the results. Results Volumes of NaOH to neutralize the solution: 3 V NaOH 1=10,7 cm 3 V NaOH 2=10,8 cm 3 V NaOH 3 =10,8 cm Volume of acid in experiment: 3 3 V HCl =10 cm =0,01 dm Concentration of used solutions: C M , Hcl ≈ 0,1 M C M , NaOH =0,1004 M Volumes of HCl till the change of colour of the solution (controll solution no.2 + indicatior): 3 V HCl 1=4,8 cm 3 V HCl 2=4,6 cm 3 V HCl 3=4,6 cm Volume of controll solution in experiments: V solution no .2=10 cm3 =0,01 dm3 Actual molarity of controll solution: C M , solution no.2 ,actual =0,0108 M Calculations Reaction: HCl+ NaOH → NaCl+ H 2 O Calculations of part A: 1) Calculation of the molar concentration of HCl through the amount of NaOH solution needed for titration: