Question

Exactly 15.0 mL of an unknown acid was measured in a 25 ml graduated cylinder and poured into a 250 ml flask. The graduated cylinder was then rinsed with 25 mL of distilled water. The rinse was added to the 250 ml flask. An additional 75 mL of water was added to the flask along with 4-6 drops of the phenolphthalein indicator. The solution was then titrated with 0.288 M NaOH. The initial burette reading was 1.63 mL. After titration, the final burette reading was 41.17 ml. Calculate the molarity of the unknown acid. (Hint: The acid base ratio is 1:1) HA + NaOH → NaA + H2O

Answer 1

Answer:

Molarity of the unknown acid is 0,76M

Explanation:

In an acid base reaction:

HA + NaOH → NaA + H₂O

The moles of NaOH that you spend in the titration will be the same of the HA. Thus:

Volume spend ×molarity=moles of NaOH

(0,04117L-0,00163L)×0,288M = 0,0114 moles NaOH ≡ moles of HA

The sample of the unknown acid has a volume of 15,0mL=0,0150L. Thus, molarity of the unknown acid is:

$\frac{0,0114moles}{0,0150L}$ = 0,76M

I hope it helps!