Question #5cde3

1 Answer
May 25, 2016

#"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.