Role of hydrochloric acid and sodium hydroxide in isolation of citric acid from lemon juice

Question

I'm trying to extract citric acid from squeezed lemons. After studying the subject, I found a lab manual [1, p. 68]:

ISOLATION OF CITRIC ACID

Reagents: Lemon juice (ca. $\pu{100 mL}$ – 3 lemons)
$\ce{CaCl2}~\pu{5 g}$
$10\,\%$ $\ce{NaOH}$
$\pu{2 M}~\ce{H2SO4}$
$\pu{2 M}~\ce{HCl}$
$\pu{2 M}~\ce{NaOH}$

Glassware:
Magnetic stirrer
beakers $\pu{250 mL}$ (3)
Measuring cylinders (2)
filtering system with Büchner funnel
flask $\pu{100 mL}$
Pipettes (2)
Pasteur’s pipettes
beaker $\pu{50 mL}$
glass rod

Place a beaker (v. $\pu{250 mL})$ on a stirrer and pour into it $\pu{100 mL}$ of fresh squeezed lemon juice (weight it!). Add dropwise slowly and carefully $10\,\%$ aqueous $\ce{NaOH}$ basifying the mixture up to $\mathrm{pH} = 8.$ You will recognize this moment by changing of the colour from light yellow to light orange. Filter the obtained mixture on a Büchner funnel. (Caution! Change from time to time the filter papers. The pores of filter paper become blocked, so it is necessary to replace the filter paper with a new one. Repeat it as often as necessary. A clog paper can cause an explosion of the filter flask!).

Move the obtained transparent layer to the beaker, place it on a stirrer and add $\pu{50 mL}$ of $10\,\%$ aqueous $\ce{CaCl2}.$ Stir for 15 minutes, then heat it to boiling and filter calcium citrate $(\ce{Ca3C12H10O14})$ from the hot mixture on a Büchner funnel. Wash the obtained product on funnel with hot water. Dissolve the obtained crude product in small amount of $\pu{2 M}~\ce{HCl}~(\pu{5 mL}).$ Then neutralize the solution with $\pu{2 M}~\ce{NaOH}$ to approximately $\mathrm{pH} = 7.5$ and boil the mixture. Separate the sediment on Büchner funnel and dry on the air. Weight the product and calculate the yield relative to the amount of the fresh lemon juice.

Does anybody know the reason for adding hydrochloric acid and sodium hydroxide in the last paragraph?

This looks very useless to me, since in this case the alkaline and acid neutralize each other.

Reference

1. Przybył, A. K.; Kurek, J. Laboratory Of Organic Chemistry: Natural Products And Pharmaceuticals; Adam Mickiewicz University: Poznań, 2013. (PDF)

Answer 1

The $\ce{HCl}$ is added to dissolve calcium citrate by protonating the citrate to form (neutral) citric acid:

$$\ce{Ca3(C6H5O7)2(s) + 6HCl(aq) -> 3 Ca^{2+} (aq) + 6 Cl- (aq) + 2 C6H8O7 (aq)}$$

To recrystallize, you have to form citrate again. The $\mathrm{p}K_\mathrm{a}$ values of citrate range from 3.1 to 6.4, so at pH = 7.5 citric acid deprotonates fully to form citrate again:

$$ \ce{ C6H8O7 (aq) <=> C6H5O7^3- + 3H+(aq)}$$

On the other hand, you don't want the pH too basic because then calcium hydroxide will precipitate. In the presence of sufficiently high concentration of calcium ions, calcium citrate will precipitate again:

$$\ce{ 3 Ca^{2+} (aq) + 2 C6H5O7^3- <=> Ca3(C6H5O7)2(s) }$$

Each step in crystallization will have less contaminants, with some loss of the desired product. So it might look like you are going around in circles and losing product. However, the purity of the product will increase with every recrystallization step.

Answer 2

$100$ g lemon juice contains $6.08$ g citric acid, and $0.3$ g malic acid $\ce{(CHOHCOOH)2}$, according to K. Diem, Scientific Tables, CIBA-Geigy SA, Basel, 1972, p. 508

The first precipitation produces an impure precipitate of calcium citrate containing about $5$% calcium malate $\ce{CaC4H4O6}$. When redissolved in $\ce{HCl}$, neutralized and reprecipitated, the malate ions are eliminated. The second precipitate contains pure calcium citrate.