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

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

Answer 2

Answer:

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

$=0.25M \times 45.0mL = 0.25 mol/L \times 0.045L = 0.01125 mol HCl$

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)