Question

10.0 ml of concentrated hno3 (15.5 m ) is added to 100.0 ml of water. how many moles of hno3 dissolve in the 100.0 ml of water?

Answer 1

154.99 moles of HNO3 get dissolved in 100ml of water when 10.0 ml of conc. HNO3 (15.5M) is added to it.

How does HNO3 dissolve in water ?

Since nitric acid is a moderately potent acid, we assume that the water solvent undergoes complete protonolysis in accordance with the equation:

$HNO_{3}(aq) + H_{2}O(l)$  ⇄ $H_{3}O^{+} + NO_{3}^{-}$

To the right is where the equilibrium is.

The aqueous solution is now stoichiometric for hydronium ion and nitrate ion, regardless of the initial nitric acid concentration.

Given:

V1 (Volume of nitric acid)  = 10.0ml

M1 (Molarity of nitric acid) = 15.5M

V (Volume of Water)  = 100.0ml

n (no. of moles of HNO3 dissolved)  = ?

Formula applied:

1) For dilution : M1V1 = M2V2

where : V2 = total volume of the final solution

             M2 = molarity of the final solution

2) Molarity  = ( no. of moles (mol) ) / ( total volume of the solution (l) )

Universal constants used :

Molar mass of HNO3 = 63.01g

Solution:

$M1*V1 = M2*V2 = > M2 = \frac{M1*V1}{V2} \\\\= > M2 = \frac{15.5*10}{110} mol ml^{-1} \\\\= > M2 = 1.409 mol ml^{-1}$

Now, calculate the no. of moles dissolved in the solution :

$M2 = \frac{n}{V} \\\\= > 1.409 mol ml^{-1} = \frac{n}{110} ml \\\\= > n = 1.409 * 110 mol\\\\= > n = 154.99 mol$

Learn more about nitric acid   :

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