# Question 49df4

May 2, 2016

No, you cannot have a complete neutralization in this case.

#### Explanation:

You start by writing the balanced chemical equation that describes this neutralization reaction.

Sulfuric acid, ${\text{H"_2"SO}}_{4}$, a strong acid, will react with ammonia, ${\text{NH}}_{3}$, a weak base, to form aqueous ammonium sulfate, ("NH"_4)_2"SO"_4, which will exist as ions in solution

${\text{H"_ 2"SO"_ (4(aq)) + color(red)(2)"NH"_ (3(aq)) -> 2"NH"_ (4(aq))^(+) + "SO}}_{4 \left(a q\right)}^{2 -}$

As you can see, the two reactants react in a $1 : \textcolor{red}{2}$ mole ratio, which tells you that a complete neutralization will require twice as many moles of ammonia as moles of sulfuric acid.

Now, you don't have to do any calculation here to say that sample of ammonia given to you will not be enough for a complete neutralization.

As you know, a solution's molarity tells you how many moles of solute you get per liter of solution.

Notice that the two solutions have the same volume and the same molarity, which can only mean that they must contain the same number of moles of solute.

Since you're mixing equal number of moles of sulfuric acid and ammonia, the reaction will stop before all the acid is neutralized.

That happens because ammonia will act as a limiting reagent, i.e. it will be completely consumed by the reaction before all the moles of sulfuric acid get the chance to react.

Now, what would you need in order to completely neutralize the sulfuric acid?

Since you must provide twice as many moles of ammonia to the reaction, you can either

• use the same volume, but double the molarity
• use the same molarity, but double the volume

The sulfuric acid solution contains

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

n_(H_2SO_4) = "3.0 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{{H}_{2} S {O}_{4}} = {\text{0.30 moles H"_2"SO}}_{4}$

This means that the ammonia solution must contain

0.30 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(red)(2)color(white)(a)"moles NH"_3)/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.60 moles NH"_3#

which you can get by using

$\text{100 mL}$ of a $\text{6.0-M}$ solution

or

$\text{200 mL}$ of a $\text{3.0-M}$ solution.