A titration of 15.0 cm3 of household ammonia, NH3, required 38.57 cm3 of 0.78M HCl. Calculate the molarity of the ammonia.

PLEASE SHOW ME HOW TO SOLVE THIS. Thank you very much!

Doesn't one mole of HCl neutralize one mole of NH3?

38.57 cm^3 = 0.03857 liters
There are (0.78)*(0.03857) = 0.0301 moles of HCl in the 0.78M titrating solution.

Assuming there are the same number of moles of NH3 in the 15.0 cm^3 of ammonia solution, calculate its molarity

Molarity = mols/liters.... so do .0301/.015.... and you get 2.01 M.

To solve this problem, we can use the concept of stoichiometry and the balanced chemical equation between ammonia (NH3) and hydrochloric acid (HCl):

NH3 + HCl -> NH4Cl

First, let's determine the number of moles of hydrochloric acid (HCl) used in the titration:

Moles of HCl = Volume of HCl (in L) x Molarity of HCl
= (38.57 cm3 / 1000) L x 0.78 mol/L
= 0.0300576 mol

According to the stoichiometry of the balanced equation, 1 mole of HCl reacts with 1 mole of NH3 to form 1 mole of NH4Cl. Therefore, the number of moles of ammonia (NH3) used in the titration is also 0.0300576 mol.

Next, let's calculate the molarity of ammonia (NH3):

Molarity of NH3 = Moles of NH3 / Volume of NH3 (in L)

It is given that the volume of ammonia is 15.0 cm3. To convert it to liters, divide by 1000:

Volume of NH3 (in L) = 15.0 cm3 / 1000
= 0.015 L

Now we can calculate the molarity of ammonia:

Molarity of NH3 = 0.0300576 mol / 0.015 L
= 2.00384 mol/L

Therefore, the molarity of ammonia (NH3) is approximately 2.00 mol/L.

Note: It is important to use the correct units and to convert them when necessary to ensure consistent and accurate calculations.