Answer to Question #125160 in General Chemistry for mike

Question

A sample of nitric acid has a mass of 8.2g. It is dissolved in 1.00L of water. A 25.00mL aliquot of this acid is titrated with NaOH. The concentration of the NaOH is 0.18M. What titre volume was added to the aliquot to achieve neutralisation?

Answer 1

Nitric Acid ( HNO₃ ) Has Molar Mass = 1 + 14 + 3 * 16 = 63 g / mol ..

Moles of Nitric Acid = ( 8.2 / 63 ) mol . = 0.13 mol.

Volume of Solution = 1 liter .

So , Molarity Of Nitric Acid = ( 0.13 / 1 ) mol / liter ..

25 ml of 0.13 M Nitric Acid Solution Have Moles of Nitric Acid =(25 *0.13 ) /1000 Mole .......... 1.

25 ml of 0.13 M Nitric Acid Solution Titrated With 0.18 M NaOH Solution .

Let Volume Of NaOH Solution Is V Liter . .

So , Moles Of NaOH =( V * 0.18 ) Mole ........... 2.

1 Mole Of Nitric Acid Titrated ( Reacted ) With 1 Mole Of NaOH Solution .

So , Moles Of HNO₃ = Moles Of NaOH .

So , 1 = 2 .

( 25 *0.13 ) /1000 = V * 0.18 .

So , V = 0.01806 Liter . = 18.06 ml .