Question

Citric acid is a weak triprotic acid (abbreviated as H₃C for this problem, where “C” is analogous to “A” in our usual acid-base notation) A solution was prepared by dissolving 0.100 moles of sodium citrate (Na₃C) in 1.00L of water.
a) List all of the species present in solution
b) Write the charge balance for this solution
c) Write two mass balance equations for this solution

Answer 1

Citric Acid is a triprotic acid which mean it has a total of three acidic Hydrogens which can be released when it is treated with any base. To form Sodium Citrate, all the three acidic hydrogens have to be removed, i.e., we need excess quantity of base.

(a) All the species present in the solution of Sodium Citrate and water will be : Sodium ions (Na⁺), Citrate ions (C₆H₅O₇^-) and protons (H⁺).

(b) Charge balance equation for the given condition is:

Na₃C₆H₅O₇ + 3H₂O H₃C₆H₅O₇ + 3NaOH

(c) Mass balance equation:

Molecular weight of citric acid = 192 grams

Mass of citric acid = moles x molecular weight = 0.100 x 192 = 19.2 grams citric acid

Volume of water = 1 L or 1 Kg = 1000 grams

Total : Total mass of the resultant solution = 1019.2 grams.

Individual : 19.2 ; 1000 grams.