An Erlenmeyer flask containing 20.0 mL of sulfuric acid of an unknown concentration was titrated with exactly 15.0 mL of 0.25 M NaOH solution to the second equivalence point. What was the concentration of sulfuric acid in the flask?

0.09375M

Explanation:

There are two methods of going about this, either we use dilution formula (easiest and fastest) or we solve through molarity.

Using dilution formula,

C2 (H2SO4) = ?

C1 (NaOH) = 0.25M

V2 (H2SO4) = 20cm³

V1 (NaOH) = 15cm³

However we can solve using molarity method

Equation of reaction =

2NaOH + H2SO4 》 Na2SO4 + 2H2O

O.25M of NaOH = 1000cm³

X moles = 15cm³

X = (0.25 * 15) / 1000

X = 0.00375 moles is present in 15cm³ of NaOH

From equation of reaction,

2 moles of NaOH requires 1 mole of H2SO4

Therefore

0.00375 / 2 = 0.001875 moles is present in H2SO4

From the reaction,

0.00187 moles of H2SO4 = 20 cm³

X moles = 1000cm³

X = (0.00187*1000) / 20 = 0.09375M

practical process described pest part of scientific investigation.

because practical knowledge is powerful then therotical knowledge.