Question #74884
1 Answer
Explanation:
Ascorbic acid,
#"HC"_ 6"H"_ 7"O"_ (6(aq)) + "NaOH"_ ((aq)) -> "NaC"_ 6"H"_ 7"O"_ (6(aq)) + "H"_ 2"O"_((l))#
The acid and the base react in a
Use the molarity and volume of the sodium hydroxide solution to determine how many moles of base were needed for the reaction
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#
In your case, you will have
#n_(NaOH) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(14.50 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters")) = "0.001450 moles NaOH"#
This means that the ascorbic acid solution contained
To find how many grams of ascorbic acid would contain this many moles, use the acid's molar mass
#0.001450 color(red)(cancel(color(black)("moles HC"_6"H"_7"O"_6))) * "176.12 g"/(1color(red)(cancel(color(black)("moles HC"_6"H"_7"O"_6)))) = "0.255374 g"#
Since the total mass of the sample is said to be equal to
#"purity" = (0.255374 color(red)(cancel(color(black)("g"))))/(0.258color(red)(cancel(color(black)("g")))) xx 100 = "98.98%"#
Rounded to three sig figs, the answer should be
#"purity" = color(green)(|bar(ul(color(white)(a/a)"99.0%"color(white)(a/a)|)))#