Question

If 17.8 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentration of the ammonia?

Answer 1

First write a balanced equation of HCl and NH3 as follows:

HCl(aq) + NH3(l) -----> NH4Cl(aq)

Now calculate the moles of HCl as follows:

17.8 mL of 0.800 M HCl or 0.0178 L

0.0178 L * 0.800 mol /HCl= 0.01424 mol HCl

The reaction show HCl and NH3 reacts in 1:1. Thus the number of moles of NH3 also 0.01424 moles.

Now, molarity = number of moles / volume in L

5.00 Ml = 0.005 L

= 0.01424 moles / 0.005 L

= 2.848 M

The molar concentration of the ammonia is 2.848 M