# If 22.15mL of 0.100M sulfuric acid is required to neutralize 10.0mL of lithium hydroxide solution, what is the molar concentration of the base? How would I approach this problem?

## ${H}_{2} S {O}_{4}$ (aq) + 2LiOH(aq) $\rightarrow$ $L {i}_{2} S {O}_{4}$ + $2 {H}_{2} O$

Nov 29, 2015

$\text{0.443 M}$

#### Explanation:

Your strategy when it comes to neutralization reactions is to use the balanced chemical equation to help you determine how many moles of each reactant must be present in solution in order for complete neutralization to take place.

In your case, the balanced chemical equation looks like this

${\text{H"_2"SO"_text(4(aq]) + color(red)(2)"LiOH"_text((aq]) -> "Li"_2"SO"_text(4(aq]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between sulfuric acid and lithium hydroxide. This tells you that the reaction will always consume $\textcolor{red}{2}$ times more moles of lithium hydroxide than moles of sulfuric acid.

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

$\textcolor{b l u e}{\text{molarity" = "moles of solute"/"liters of solution}}$

If you determine the number of moles of sulfuric acid present in the initial solution, you can sue the aforementioned mole ratio to figure out how many moles of lithium hydroxide would be needed in order to make sure that all the moles of acid are neutralized.

So, the number of moles of sulfuric acid will be

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

$n = {\text{0.100 M" * 22.15 * 10^(-3)"L" = "0.002215 moles H"_2"SO}}_{4}$

This means that you'd need

0.002215 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(red)(2)" moles LiOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.00443 moles LiOH"

Now all you need to do is use the volume of the solution to figure out the solution's molarity

c = "0.00443 moles"/(10.0 * 10^(-3)"L") = color(green)("0.443 M")

So remember, chemical reactions are all about mole ratios. As far as solutions are concerned, start by focusing on moles, the use molarity and volume to help you find what you need.