# What volume of, in cm^3, of 0.200 mol of dm^(-3) NaOH is required to neutralize 25.0 cm^3 of 0.100 mol dm^3 of H2SO4?

## A. 12.5 B. 25.0 C. 50.0 D. 100 The answer is B. 25.0, why?

Feb 17, 2016

${\text{25.0 cm}}^{3}$

#### Explanation:

Start by writing the balanced chemical equation for this neutralization reaction

$\textcolor{p u r p \le}{2} {\text{NaOH"_text((aq]) + "H"_2"SO"_text(4(aq]) -> "Na"_2"SO"_text(4(aq]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

As you know, the stoichiometric coefficients of the reactants will tell you the mole ratio that exists between the two reactants.

In this case, you have a $\textcolor{p u r p \le}{2} : 1$ mole ratio between sodium hydroxide and sulfuric acid. This tells you that in order for a complete neutralization to occur, you need to have twice as many moles of sodium hydroxide than you do of sulfuric acid.

Now, the problem provides you with the molarity and volume of the sulfuric acid solution, which is an indirect way of providing you with the number of moles of sulfuric acid.

As you know, a solution's molarity tells you how many moles of solute you get per liter, or , like you have here, per ${\text{dm}}^{3}$ of solution.

$\textcolor{b l u e}{c = {n}_{\text{solute"/V_"solution}}}$

Plug in your values and find the number of moles of sulfuric acid - do not forget to convert the volume of the solution from ${\text{cm}}^{3}$ to ${\text{dm}}^{3}$

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

n_(H_2SO_4) = "0.100 mol" color(red)(cancel(color(black)("dm"^(-3)))) * 25.0 * 10^(-3)color(red)(cancel(color(black)("dm"^(3)))

${n}_{{H}_{2} S {O}_{4}} = {\text{0.00250 moles H"_2"SO}}_{4}$

The aforementioned mole ratio tells you that you'd need

0.00250 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(purple)(2)color(white)(a)"moles NaOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.00500 moles NaOH"

Now all you have to do is use the molarity of the sodium hydroxide solution to determine what volume would contain this many moles

$\textcolor{b l u e}{c = \frac{n}{V} \implies V = \frac{n}{c}}$

V_(NaOH) = (0.00500 color(red)(cancel(color(black)("moles"))))/(0.200color(red)(cancel(color(black)("moles")))/"dm"^3) = "0.0250 dm"^3

This is, of course, equivalent to

V_(NaOH) = 0.0250 * 10^3"cm"^3 = color(green)("25.0 cm"^3)