At the end of the experiment you titrate the solution with 0.507 M HCl and it takes 38.30 mL to neutralize the ammonia. What is the equilibrium molarity of ammonia after the 2nd addition of ammonia

Respuesta :

Answer: check explanation

Explanation:

The balanced equation of reaction is given below;

HCl + NH3 -----------------------> NH4Cl.

We are given the volume in milliliters, let us convert them into Litres;

= 38.30 × 10^-3 Litres.

Here, we have an incomplete question (one parameter is missing- the volume of ammonia,NH3). Therefore, we assume that the volume of Ammonia, NH3 is 10mL(10× 10^-3 Litres).

Step one: we need to calculate the number of moles of HCl.

Number of moles of HCl= molarity × volume.

Number of moles of HCl= 0.507 M × 38.30× 10^-3 L.

Number of moles of HCl= 0.0194181 moles.

From the equation of reaction above, we have that one mole of ammonia is reacting with one mole of Hydrogen Chloride, HCl. Hence, the number of moles of ammonia is equal to the number of moles of Hydrogen Chloride,HCl.

Step two: calculate the molarity of Ammonia, NH3.

The molarity of ammonia= number of moles of ammonia/ volume of Ammonia,NH3.

Molarity of Ammonia= 0.0194181/10× 10^-3 moles NH3.

Molarity of Ammonia= 0.00000194181.

Molarity of Ammonia = 1.94181× 10^-6 M.