An initial 25.00 ml solution of nitric acid (hno3) was diluted to 250.00 ml, and15.00 ml of this diluted solution was titrated with a 0.2244 m solution of sodium hydroxide. if 21.33 ml of the sodium hydroxide solution were required to neutralize the acid, what is the nitric acid concentration of the initial 25.00 ml solution?

HNO3 concentration in the initial 25.00 mL solution is 3.19 M

Explanation:

We are working backward:

First we will determine the [HNO3] in 15 mL that reacted with Sodium hydroxide (NaOH).

mole of nitric acid = mole of NaOH

[HNO3] * 15 mL = 0.2244 M * 21.33 mL

[HNO3] = 0.319 M

The above is the [HNO3] in 250 mL solution that was transferred as 15 mL to react with NaOH.

Second, we can determine the [HNO3] in 25 mL because the 250 mL solution was made from the 25 mL.

[HNO3] * 25 mL = 0.319 M * 250 mL

[HNO3] = 3.19 M which is the concentration in the initial 25 mL solution.  

answer: h+ its less than oh

explanation:

a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.