If 35.5 ml of 0.23 m hcl is required to completely neutralize 20.0 ml of nh3, what is the concentration of the nh3 solution? show all of the work needed to solve this problem. hcl + nh3 yields nh4cl

NH3 is neutralised by the equation:

HCL + NH3 -> NH4CL

In this equation there is a one to one relationship in terms of the number of moles of each reactant. I.e. To neutralise 1 mole of NH3 we require 1 mole of HCL.

To calculate the concentration of NH3 required, we must first calculate the number of moles of HCL used.

volume HCL = 35.5mL = 0.0355 litres 
concentration HCL = 0.23M = 0.23 mole/litre 
Note that the term "M" for concentration simply means moles/litre
number moles = concentration x volume
number moles HCL = 0.0355 x 0.23 = 0.008165 moles HCL
based on the equation, we know the number of moles of NH3 must be the same

So, 
moles NH3, n = 0.008165
volume NH3, v = 20.0mL = 0.020 litres
n = c x v
c = n / v
c = 0.008165 / 0.020
=0.41
i.e. the concentration of NH3 would be 0.41 moles/litre or 0.41M

This intuitively makes sense because there is less volume of NH3 required to be neturalised, in a one-to-one mole relationship. So the concentration of NH3 would need to be higher than that of HCL.