Question #5cde3
1 Answer
Explanation:
In the context of an acid - base reaction, normality represents the number of equivalents of hydronium cations,
In your case, a
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
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.