# Question ebf5e

May 15, 2017

See below.

#### Explanation:

If you titrate with sodium hydroxide solution, the reaction is:

${C}_{3} {H}_{5} O {\left(C O O H\right)}_{3} \left(a q\right) + 3 N a O H \left(a q\right) + N {a}_{3} {C}_{3} {H}_{5} O {\left(C O O\right)}_{3} \left(a q\right) + 3 {H}_{2} O \left(i\right)$

Just as an example, lets say you had 100 ml of orange juice, which weighed 120 g. You titrated this against 0.1 M NaOH solution, and measured volume of titre as 20 ml.

First, work out how many moles of NaOH you used.

$M = \frac{n}{V}$ where M is molarity, n is number of moles and V is volume in litres.

Therefore: $n = V . M = \left(\frac{20}{1000}\right)$ x $0.1 =$0.002 moles

The equation tells you that 1 mole of citric acid reacts with 3 moles of NaOH, so you must have had (0.002/3) = 0.00067 moles of citric acid in your orange juice.

The molar mass of citric acid is 192.12 g, so 0.00067 moles is 192.12 x 0.00067 = 0.1287 g of citric acid.

The % of citric acid was therefore: (0.1287/120).100 = 0.1073%#

Please note that, with the exception of the reaction equation and the molar mass, I just made these numbers up to illustrate the procedures - I can't recall how much citric acid is really in orange juice!