Avolume of 25.0 ml of nitric acid, hno3, is titrated with .12 m naoh. to completely neutralize the acid 10.0 ml of naoh must be added. what is the molarity of nitric acid is

0.048M

Explanation:

Volume of acid = 25mL = 0.025L

Concentration of base NaOH = 0.12M

Volume of base NaOH = 10mL= 0.01L

Unknown:

concentration or molarity of the nitric acid = ?

Solution

We first write the balanced reaction equation:

           HNO₃ + NaOH → NaNO₃ + H₂O

In this problem, the focus is on the reacting acid and base which are on the left hand side.

Now, to find the molarity of the nitric acid that would be completely neutralized by the base, we adopt the mole concept.

First, we work from the known to the unknow. We can find the number of moles of the base given because its parameters are complete. The number of moles is given by the expression below:

        Number of moles = concentration x volume

To find the number of moles of the base, we put the parameters into the expression above:

Number of moles of NaOH = concentration of NaOH x volume of NaOH

                                              = 0.01 x0.12

                                              = 0.0012mol

From the number of moles of the base, we can find the unknown.

  In the reaction equation:

      1 mole of base neutralizes 1 mole of the acid

      0.0012mole of the base will neutralize 0.0012mole of the acid

Therefore, 0.0012mole of HNO₃ was neutralized.

Now we can calculate the molarity of the acid using the expression below:

     concentration of acid =

     concentration of acid = = 0.048M

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

false

because it does not have the potential to be entirely used up