Aburet is filled with 0.1517 m naoh a 25.0 ml portion of an unknown acid and two drops of indicator are added to an erlenmeyer flask, and the titration experiment is carried out. if the initial buret reading was 0.55 ml, and the buret reading at the end point was 22.50 ml, what is the molarity of the unknown acid?

molarity of acid =0.0132 M

Explanation:

We are considering that the unknown acid is monoprotic. Let the acid is HA.

The reaction between NaOH and acid will be:

Thus one mole of acid will react with one mole of base.

The moles of base reacted = molarity of NaOH X volume of NaOH

The volume of NaOH used = Final burette reading - Initial reading

Volume of NaOH used = 22.50-0.55= 21.95 mL

Moles of NaOH = 0.1517X21.95=3.33 mmole

The moles of acid reacted = 3.33 mmole

The molarity of acid will be =

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o