If 18.8mL of 0.800M HCL solution are needed to neutralize 5.00mL of a household ammonia solution, what is the molar concentration of the ammonia?

How do I solve this?
Thank you.

To solve this question, you need to use the concept of stoichiometry and the balanced chemical equation of the reaction between HCl and ammonia (NH3). Here are the steps to solve this problem:

1. Write the balanced chemical equation:
HCl + NH3 → NH4Cl

2. Determine the ratio of moles between the reactants (HCl and NH3) based on the balanced equation. In this case, the ratio is 1:1, meaning one mole of HCl reacts with one mole of NH3.

3. Calculate the moles of HCl used:
moles of HCl = volume of HCl solution (in L) × molar concentration of HCl

Given: volume of HCl solution = 18.8 mL = 0.0188 L
Molar concentration of HCl = 0.800 M

moles of HCl = 0.0188 L × 0.800 M = 0.01504 moles

4. Since the ratio between HCl and NH3 is 1:1, you know that the moles of NH3 used will also be 0.01504 moles.

5. Calculate the molar concentration of NH3:
Molar concentration of NH3 = moles of NH3 / volume of NH3 solution (in L)

Given: volume of NH3 solution = 5.00 mL = 0.005 L

Molar concentration of NH3 = 0.01504 moles / 0.005 L = 3.008 M

Therefore, the molar concentration of the household ammonia solution is 3.008 M.

tgy

mL x M = mL x M