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 NH2 solution? Show all of the work needed to solve this problem.
HCl + NH3 yields NH4Cl

Answer 1

We have
\[\LARGE HCl + NH_3 \rightarrow NH_4Cl \]
First calculate the moles of HCl, using
\[\Large n(HCl)=[M_{HCl}](V_{HCl})\]
Since the mole conversion is 1:1,
\[n(HCl)=n(NH_3)\]
Using
\[n(NH_3)=[M_{NH_3}](V_{NH_3})\]
Concentration of \(NH_3\) will be
\[[M_{NH_3}]=\frac{n(NH_3)}{V_{NH_3}}\]