Question #b33d8

2 Answers
Jul 27, 2017


I can't give you a "correct" answer with the information given, but I can show you how to solve it, which I did.


This is a really annoying unit conversion problem before we can even solve it, hold on:

#(0.4mol)/(dm^3) * (dm^3)/(cm^3) * (cm^3)/(mL) * (10^3mL)/L = 400M#

#3 NaOH (aq) + C_6H_8O_7 (aq) → Na_3C_6H_5O_7 (aq) + 3 H_2O(l)#

You usually need the volume of one or the other solution in order to accurately solve this problem, but I will assume we are trying to neutralize #1 L# of citric acid.

#1 L * (0.005mol)/L * (3NaOH)/(C_6H_8O_7)* L/(400mol) approx 0.375mL#

This appears to be a bit unreasonable, so perhaps in your problem the amount of citric acid needing to be neutralized is larger.

Jul 28, 2017


The required volume is #"38 cm"^3#.


Step 1. Write the balanced equation for the reaction

#"H"_3"C"_6"H"_5"O"_7 + "3NaOH" → "Na"_3"C"_6"H"_5"O"_7 + 3"H"_2"O"#


#"H"_3"Cit" + "3NaOH" → "Na"_3"Cit" + 3"H"_2"O"#

Citric acid is a tribasic acid. Its titration curve will look something like the one below.

Titration curve

Step 2. Calculate the moles of #"NaOH"#

#"Moles of NaOH" = 0.005 color(red)(cancel(color(black)("mol H"_3"Cit"))) × "3 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_3"Cit")))) = "0.015 mol NaOH"#

Step 3. Calculate the volume of #"NaOH"#

#"Volume of NaOH" = 0.015 color(red)(cancel(color(black)("mol NaOH"))) × ("1000 cm"^3 "NaOH")/ (0.4 color(red)(cancel(color(black)("mol NaOH")))) ="38 cm"^3 "NaOH"#

Note: The answer can have only one significant figure, because that is all you gave for the concentration of the #"NaOH"# and for the moles of citric acid.

However, I calculated the answer to two significant figures.