Ask Question
3 October, 07:37

If 35.5 mL of 0.23 M HCl is required to completely neutralize 20.0 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

+3
Answers (1)
  1. 3 October, 11:17
    0
    NH3 is neutralised by the equation:

    HCL + NH3 - > NH4CL

    In this equation there is a one to one relationship in terms of the number of moles of each reactant. I. e. To neutralise 1 mole of NH3 we require 1 mole of HCL.

    To calculate the concentration of NH3 required, we must first calculate the number of moles of HCL used.

    volume HCL = 35.5mL = 0.0355 litres

    concentration HCL = 0.23M = 0.23 mole/litre

    Note that the term "M" for concentration simply means moles/litre

    number moles = concentration x volume

    number moles HCL = 0.0355 x 0.23 = 0.008165 moles HCL

    based on the equation, we know the number of moles of NH3 must be the same

    So,

    moles NH3, n = 0.008165

    volume NH3, v = 20.0mL = 0.020 litres

    n = c x v

    c = n / v

    c = 0.008165 / 0.020

    =0.41

    i. e. the concentration of NH3 would be 0.41 moles/litre or 0.41M

    This intuitively makes sense because there is less volume of NH3 required to be neturalised, in a one-to-one mole relationship. So the concentration of NH3 would need to be higher than that of HCL.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “If 35.5 mL of 0.23 M HCl is required to completely neutralize 20.0 mL of NH3, what is the concentration of the NH3 solution? Show all of ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers