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

Hcl + nh3 - > nh4cl (balanced eqn)

no. of mol of hcl = vol. (L) x molarity = 0.045 * 0.25 = 0.01125mol

ratio of hcl:nh3 after balancing eqn = 1:1

no. of mol of nh3 that is completely neutralised by hcl = 0.01125 * 1 = 0.01125mol

therefore, concentration of nh3 = mol / total volume (L) = 0.01125mol / 0.025L = 0.45M