Question

A buret 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?

Answer 1

Answer:

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:

$NaOH+HA--->NaA+H_{2}O$

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 = $\frac{mmole}{volumne(mL)}=\frac{0.33}{25}=0.0132M$