Chemistry
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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

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TatianaPadilla

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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