20mL of orange juice (citric acid), weighing 20.31g, was neutralized by 20.4mL, 0.0900M of NaOH. Given the following equation: C6H8O7 + 3NaOH --> Na3C6H5O7 + H2O. What is the percentage by mass of citric acid in the orange juice?

1 Answer
Nov 1, 2015

Answer:

#"0.579%"#

Explanation:

The idea here is that you need to find the moles of citric acid by using the molarity of the sodium hydroxide solution, then use the mass of the orange juice to find the percent concentration by mass of citric acid.

So, start with the balanced chemical equation for this neutralization reaction

#"C"_5"H"_8"O"_text(7(aq]) + color(red)(3)"NaOH"_text((aq]) -> "Na"_3"C"_6"H"_5"O"_text(7(aq]) + "H"_2"O"_text((l])#

Notice that your reaction needs #color(red)(3)# moles of sodium hydroxide for every mole of citric acid. This means that you can find the number of moles of citric acid by using the molarity of the sodium hydroxide solution and this moel ratio.

#color(blue)(c = n/V implies n = c * V)#

#n_"NaOH" = "0.090 M" * 20.4 * 10^(-3)"L" = "0.001836 moles NaOH"#

This means that the orange juice contained

#0.001836color(red)(cancel(color(black)("moles NaOH"))) * "1 mole citric acid"/(color(red)(3)color(red)(cancel(color(black)("moles NaOH")))) = "0.0006120 moles C"_3"H"_8"O"_7#

Now that you know how many moles of citric acid you had in the orange juice, use its molar mass to determine how many moles of citric acid would contain this many moles

#0.000612color(red)(cancel(color(black)("moles"))) * "192.124 g"/(1color(red)(cancel(color(black)("mole")))) = "0.11758 g C"_3"H"_8"O"_7#

Now, a solution's percent concentration by mass is defined as the ratio between the mass of the solute, which in your case is citric acid, and the total mass of the solution, multiplied by #100#.

#color(blue)("%m/m" = "mass of solute"/"mass of soluteion" xx 100)#

In your case, the percent concentration by mass of the orange juice will be

#"%m/m" = (0.11758color(red)(cancel(color(black)("g"))))/(20.31color(red)(cancel(color(black)("g")))) xx 100 = color(green)("0.579% C"_3"H"_8"O"_7)#