# It takes 10 mL of 0.15 M KOH solution to neutralize 7.5 mL of HNO_3 solution. What is the molarity of HNO_3?

Jan 21, 2016

$\text{0.2 M}$

#### Explanation:

Start by writing the balanced chemical equation for this neutralization reaction

${\text{KOH"_text((aq]) + "HNO"_text(3(aq]) -> "KNO"_text(3(aq]) + "H"_2"O}}_{\textrm{\left(a q\right]}}$

Notice that you have $1 : 1$ mole ratios between potassium hydroxide, $\text{KOH}$, and nitric acid, ${\text{HNO}}_{3}$. This tells you that the reaction will always consume equal numbers of moles of the two reactants.

Now, molarity is defined as number of moles of solute per liters of solution.

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

Basically, all you have to do here is use the molarity and volume of the potassium hydroxide solution to determine how many moles of this strong base were used to neutralize the strong acid.

According to the aforementioned mole ratio, the number of moles of acid will be equal to the number of moles of base.

$c = \frac{n}{V} \implies n = c \cdot V$

${n}_{K O H} = \text{0.15 M" * 10 * 10^(-3)"L" = "0.0015 moles KOH}$

This of course implies that you have

0.0015 color(red)(cancel(color(black)("moles KOH"))) * "1 mole HNO"_3/(1color(red)(cancel(color(black)("mole KOH")))) = "0.0015 moles HNO"_3

This means that the nitric acid solution had a molarity of

$c = \frac{n}{V}$

c_(HNO_3) = "0.0015 moles"/(7.5 * 10^(-3)"L") = color(green)("0.2 M")

The answer is rounded to one sig fig, the number of sig figs you have for the volume of the potassium hydroxide solution.