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.