Question

Question #5cde3

Answer 1

`"0.1 N"`

Explanation:

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

In your case, a `"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

`"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_((l))`

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.