Asample of nitric acid has a mass of 8.2g. it is dissolved in 1l of water. a 25ml 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?

18.075 mL of NaOH solution was added to achieve neutralization

Explanation:

First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:

NaOH + HNO3 → NaNO3 + H2O

From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:

63.01 g of HNO3 1 mole

8.2 g of HNO3 x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3

So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.

1000 mL solution 0.13014 moles of HNO3

25 mL (aliquot) x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles

So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:

0.18 moles of NaOH 1000 mL Solution

0.0032535 moles x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL

As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.

idk what the question is about