HCl lahus
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:
V HCl × C M , HCl=V NaOH × C M , NaOH
V NaOH × C M , NaOH
C M , HCl=
V HCl
0,0108 dm 3 × 0,1004 M
C M , HCl= =0, 1084 M
0,01 dm 3
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