# Question aa2ed

Oct 20, 2017

$\text{0.9 M}$

#### Explanation:

Start by writing a balanced chemical equation that describes this neutralization reaction.

${\text{H"_ 2"SO"_ (4(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"SO"_ (4(aq)) + "H"_ 2"O}}_{\left(l\right)}$

Now, notice that it takes $2$ moles of sodium hydroxide to neutralize $1$ mole of sulfuric acid. This means that in order for a complete neutralization to take place, the solution must contain twice as many moles of sodium hydroxide than of sulfuric acid.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{2 \times \text{moles of H"_2"SO"_4 = "moles of NaOH}}}}$

In your case, you know that a solution of sulfuric acid that has a molarity of $\text{3 M}$ and a volume of $\text{11.56 mL}$ was required to neutralize a solution of sodium hydroxide that has a volume of $\text{25.00 mL}$.

As you know, the number of moles of sulfuric acid needed can be calculated by using the molarity and the volume of the solution.

11.56 color(red)(cancel(color(black)("mL solution"))) * ("3 moles H"_2"SO"_4)/(10^3color(red)(cancel(color(black)("mL solution")))) = ( (11.56 * 3)/10^3)color(white)(.)"moles H"_2"SO"_4

If you take $x$ $\text{M}$ to be the molarity of the sodium hydroxide solution

$\text{molarity NaOH" = x color(white)(.)"M}$

you can say that the neutralization required

25.00 color(red)(cancel(color(black)("mL solution"))) * (xcolor(white)(.)"moles NaOH")/(10^3color(red)(cancel(color(black)("mL solution")))) = ((25.00 * x)/10^3)color(white)(.)"moles NaOH"

Now all you have to do is use the aforementioned $1 : 2$ mole ratio that exists between the two reactants to say that the number of moles of sodium hydroxide must be two times the number of moles of sulfuric acid.

2 " " xx " " overbrace((11.56 * 3)/color(red)(cancel(color(black)(10^3))))^(color(blue)("no. of moles of H"_2"SO"_4)) = overbrace((25.00 * x)/color(red)(cancel(color(black)(10^3))))^(color(blue)("no. of moles of NaOH"))#

Now just solve for $x$ to find

$2 \cdot 11.56 \cdot 3 = 25.00 \cdot x$

$x = \frac{2 \cdot 11.56 \cdot 3}{25.00} = 0.9$

This means that the molarity of the sodium hydroxide solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{molarity NaOH = 0.9 M}}}}$

The answer is rounded to one significant figure, the number of sig figs you have for the molarity of the sulfuric acid solution.