# Question 5cde3

May 25, 2016

$\text{0.1 N}$

#### Explanation:

In the context of an acid - base reaction, normality represents the number of equivalents of hydronium cations, ${\text{H"_3"O}}^{+}$, or hydroxide anions, ${\text{OH}}^{-}$, produced in solution by one mole of an acid or of a base present in one liter of solution.

In your case, a $\text{0.2 N}$ base will produce $0.2$ equivalents of hydroxide anions for every mole of base dissolved per liter of solution.

You can thus say that you have

15 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.2 equiv. OH"^(-))/(1color(red)(cancel(color(black)("L base solution")))) = "0.0030 equiv. OH"^(-)

You know that a neutralization reaction implies

${\text{H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O}}_{\left(l\right)}$

Since every equivalent of hydronium cations requires $1$ equivalent of hydroxide anions, you can say that the acid solution contained $0.0030$ equivalents of hydronium cations.

Therefore, the normality of the acid will be

"normality acid" = ("0.0030 equiv. H"_3"O"^(+))/(30 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"0.1 N"color(white)(a/a)|)))#

The answer is rounded to one sig fig.