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

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

.sam. · Answer

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
$$[M_{NH_3}]=\frac{n(NH_3)}{V_{NH_3}}$$