Question

If 45.0 mL of 0.25 M HCl is required to completely neutralize 25.0 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

Answer 1

The balanced chemical equation is

`1HCl+1NH_3>1NH_4 Cl`

The conversions are

Molarity of HCl and volume gives us moles HCl

Moles HCl to moles
`NH_3` (using mole ratio 1 : 1)

Moles
`NH_3` to Molarity
`NH_3`

Molarity HCl = (moles solute HCl) / (volume of solution in L)

Rearranging the formula

We get moles HCl = Molarity × volume

`=0.25M * 45.0mL = 0.25 mol/L * 0.045L = 0.01125 mol HCl`

moles HCl = moles
`NH_3` = 0.01125 mol
`NH_3`

Molarity
`NH_3` =(moles solute
`NH_3`) / (volume of solution in L)

`=\frac {(0.01125 mol )}{25.0mL} = \frac {(0.01125 mol )}{0.025L}`

=0.450 mol/L or M (Answer)

Answer 2

molar concentration = number of moles / volume

number of moles = molar concentration × volume

Number of moles of HCl = 0.25 × 45 = 11.25 mmoles

From the reaction:

11.25 mmoles of HCl will neutralize 11.25 mmoles of NH₃

Molar concentration of NH₃ solution = 11.25 / 25 = 0.45 M