Titration

TITRATION LABORATORY REPORT

The equivalence volume of the second titration was 15 .1 mL 0 .5 M NaOH

The pH at the equivalence point was 12 .39

The molarity of the HCl from the volumes of acid and base at the equivalence point is (Macid (Vacid (Mbase (Vbase (0 .5 M (15 .1 mL (Macid ---------------------- 0 .302 M HCl

25 mL 4 . Titration curve

The equivalence volume can be determined from looking at the titration curve 's sudden rise , which reflects the actual transition in the pH level from acidic

to basic

According to the titration curve , the equivalence volume is 15 mL

The equivalence volume is

M1V1 M2V2

M1 (25 mL (0 .5 M (15 mL

M1 0 .3 M HCl

The titration curve may be used even when the raw data is available because this provides the same prediction with regards to the acid-base reaction of solutions and the corresponding change in pH

The number of moles of NaOH in the solution gradually increases as it added by increments to the HCl solution . The initial molarity was 0 .001 M and by the equivalence point , there was 0 .015 M of NaOH in the solution

The molarity reported by the other individuals in the team was the same because the same settings were followed in the virtual chemistry laboratory exercise . If ever the values of the other members were different , this may be possibly due to the amount of NaOH or HCl that was loaded into the...