# Question 3f2c9

Oct 14, 2017

${\text{2.94 mol L}}^{- 1}$

#### Explanation:

Start by writing the balanced chemical equation that describes this neutralization reaction

${\text{NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl}}_{\left(a q\right)}$

Notice that ammonia and hydrochloric acid neutralize each other in a $1 : 1$ mole ratio. This tells you that in order to have a complete neutralization, you need to mix equal number of moles of ammonia and of hydrochloric acid.

Use the molarity and the volume of the hydrochloric acid solution to find the number of moles of hydrochloric acid needed for your reaction.

18.4 color(red)(cancel(color(black)("mL solution"))) * overbrace("0.800 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 0.800 M")) = "0.01472 moles HCl"

You can thus say that this sample of hydrochloric acid will neutralize $0.01472$ moles of ammonia--remember, the two reactants neutralize each other in a $1 : 1$ mole ratio!

So, you know that $\text{5.00 mL}$ of ammonia solution contain $0.01472$ moles of ammonia. In order to find the molarity of the solution, you need to find the number of moles present in ${10}^{3} \text{mL" = "1 L}$ of this solution.

10^3 color(red)(cancel(color(black)("mL solution"))) * "0.01472 moles NH"_3/(5.00color(red)(cancel(color(black)("mL solution")))) = "2.944 moles NH"_3#

Since this represents the number of moles of ammonia present in exactly ${10}^{3} \text{mL" = "1 L}$ of this solution, you can say that the molarity of the solution will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 2.94 mol L}}^{- 1}}}}$

The answer is rounded to three sig figs.