Question

A 25.00-mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125-mL Erlenmeyer flask and 2 drops of phenolphthalein are added. The above sodium hydroxide solution (the titrant) is used to titrate the nitric acid solution (the analyte). If 12.75 mL of the titrant is dispensed from a burette in causing a color change of the phenolphthalein, what is the molar concentration of the nitric acid solution? (Show all steps for calculating the answer.)

Answer: 0.0611 M HNO3​

Answer 1

0.0611M of HNO3

Step-by-step explanation:

The concentration of the NaOH solution must be 0.1198M

The reaction of NaOH with HNO3 is:

NaOH + HNO3 → NaNO3 + H2O

1 mole of NaOH reacts per mole of HNO3.

That means the moles of NaOH used in the titration are equal to moles of HNO3.

Moles HNO3:

12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.

In 25.00mL = 0.025L -The volume of the aliquot-:

0.00153 moles HNO3 / 0.025L =

0.0611M of HNO3