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

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