The questions states: Ammonia has the formula $\ce{NH3}$. Household ammonia is a dilute aqueous solution of $\ce{NH3}$. Aqueous ammonia is a base that can neutralize acids.

If 18.2 mL of 0.800 M $\ce{HCl}$ solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentration of the ammonia? Express your answer with the appropriate units.

I used the values I was given; converted the $\ce{HCl}$ to moles; divided the moles of $\ce{HCl}$ by the liters of ammonia solution. However, I do not understand why this is correct.

$\ce{HCl}$ is an acid and $\ce{NH3}$ is a base. How does finding the molar mass of $\ce{HCl}$ help one calculate the molarity of $\ce{NH3}$?


1 Answer 1


Look at the chemical equation. One mole of $\ce{HCl}$ will neutralize one mole of $\ce{NH3}$. $\ce{HCl}$ is an acid so it loses a hydrogen ion and ammonia is a base so it gains a hydrogen ion.

$\ce{HCl + NH3 -> NH4Cl}$

So, you found the number of moles $\ce{HCl}$ added, and this equals the number of moles in the 5.00 mL sample of $\ce{NH3}$. Divide that by the volume and you get concentration, which is moles per liter. You don't need to find the molar mass of $\ce{HCl}$ and you didn't find it as described in your calculations above.

  • 1
    $\begingroup$ I guess I'm trying to have a deeper understanding of ratios when it comes to moles. Since the above reaction is 1:1 ratio I will assume that means that no matter if the elements were. let say, Chromium or Chlorine; these both would have the same amount of moles if they were part of a product of a reaction. For example If I was seeking the amount of moles of Chromium given that I had 100 grams of Chlorine in a reaction of CrCl. 100 grams of Cl equals 2.8206 moles Cl, thus since the ration is 1:1 Cr also equals 2.8206 moles. Is that correct? $\endgroup$
    – user137452
    Oct 30, 2014 at 2:10
  • $\begingroup$ @user137452 Yes. Your comment above is spot-on. If it's a balanced equation (i.e., the chemical equation has the lowest common integers like above), then there's a 1:1 ratio or 1:2 ratio or whatever is indicated in the equation. See more about limiting reagents which might help further your understanding. $\endgroup$ Oct 31, 2014 at 19:44

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