Calcium tartrate and treatment of wine waste-waters - ChemPRIME

Calcium tartrate and treatment of wine waste-waters

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Tartaric acid separated from grape juice
Tartaric acid crystals

Tartaric acid (TA) is a byproduct of wine production. This organic acid and its salts are used in foods such as fruit jellies, preserves, jams, baked goods, and confections. TA is hardly metabolized and degraded by yeast and spoilage bacteria providing microbiological stability to foods that contain it. In addition to its use as antimicrobial and acidulant, TA and its salts are used as emulsifiers, leavening, and anticacking agents.[1] TA is also used in the beverage industry and has non-food uses in textile coloring, galvanizing, and mirror production.

The increasing popularity in wine consumption in recent years has resulted in the increase of waste from wine making practices.[2] One liter of white wine generates the same amount of water pollution as 3 people in one day.[3] Waste-waters from wine production contain biodegradable compounds and fruit suspended solids; [4] their treatment is of great importance because their high pollutant activity. Moreover, waste treatment is of economic interest because the organic compounds present in waste from the wine making process can have value as additives, ingredients, and substrates in the food and pharmaceutical industries.

Waste derived from wine making

 Type of Waste Name Derived from Treatment[3] Liquid/Sludge Vinasse Vintage   process Anaerobic digestion, ozonation, thermopilic   anaerobic digestion, aerobic biodegradation,   sequencing batch reactor, electrodyalisis, and   wet oxidation Solid waste Grape   marc Skin,   stalks,   and seeds Combustion, solid-state fermentation,   incineration, composting, and pyrolysis. Vine   shoots Vine   pruning Lees Grapes   and yeast

The table above shows sources of waste in wine making as well as methods to treat them. A method devised by Rivas, et al. (2006) to treat distilled lees involves the reaction of tartaric acid with calcium-ions to form calcium tartrate. This salt of limited solubility is then redissolved with HCL to obtain TA. Upon removal of TA, the distilled lees can be used as nutrients for lactic acid bacteria (Lactobacilllus pentosus) for production of lactic acid.

In the method described above, addition of calcium ions prompts the precipitation of calcium tartrate. Once tartaric acid is removed from the solution as tartrate, it will partially dissociate into calcium and tartrate ions establishing the following equilibrium with the solid salt:

$\text{CaC}_{4}\text{H}_{4}\text{O}_{6}({s})\rightleftharpoons \text{Ca}^{2+}({aq}) + \text{C}_{4}\text{H}_{4}\text{O}_{6}^{2-}({aq})$

with the corresponding equilibrium constant

${K}_{sp} = [\text{Ca}^{2+}][\text{C}_{4}\text{H}_{4}\text{O}_{6}^{2-}]=\text{7.7} \times \text{10}^{-7} \text{mol}^{2} \text{dm}^{-6}\,$

If we designate as x the concentration of each of the ions, the concentration of calcium-ions at equilibrium is

\begin{align} x^{2}=\text{7.7} \times \text{10}^{-7} \text{mol}^{2} \text{dm}^{-6}\end{align}

From which

$x=\sqrt{\text{7.7}\times \text{10}^{-7}\text{ mol}^{2} \text{dm}^{-6} }= \text{8.77} \times \text{10}^{-4} \text{mol dm}^{-3}$

The concentration of calcium-ions at equilibrium is 8.77 x 10–4 mol dm–3. If now we increase the concentration of tartrate-ions, the equilibrium will shift to the left according to the Le Chatelier’s principle. More calcium tartrate will precipitate, decreasing the concentration of calcium ions. The decrease in concentration obtained in this way is often referred to as the common-ion effect.

Similarly, if an excess of calcium-ions is added to the solution, the concentration of tartrate-ion will decrease. Since in this process we are concerned about removing as much tartaric acid in the form of tartrate as possible, addition of calcium ions in excess will minimize dissociation of calcium tartrate increasing its the yield.

The solubility product can be used to calculate how much the calcium-ion concentration is decreased by the common-ion effect. Suppose we mix 10 cm3 of a saturated solution of calcium tartrate with 10 cm3 of concentrated sodium tartrate (4 M C4H6O6). Because of the twofold dilution, the concentration of tartrate will be 2 mol dm–3. Feeding this value into equation 7 from the solubility product section, we then have the result

or              $\text{7.7} \times \text{10}^{-7} \text{mol}^{2} \text{dm}^{-6}=[\text{Ca}^{2+}](\text{2 mol dm}^{-3})$

so that      $[\text{ Ca}^{2+}]=\frac{\text{7.7 }\times \text{ 10}^{-7}\text{ mol}^{2}\text{ dm}^{-6}}{\text{2 mol dm}^{-3}}=\text{3.85 }\times \text{ 10}^{-7}\text{ mol}\text{ dm}^{-3}$

We have thus lowered the calcium-ion concentration from an initial value of 8.77 x 10–4 mol dm–3 ) to a final value of 3.85 × 10–7 mol dm–3, a decrease of about a factor of 2000!.

 This option will not work correctly. Unfortunately, your browser does not support inline frames. Calcium tartrate

Because it tells us about the conditions under which equilibrium is attained, the solubility product can also tell us about those cases in which equilibrium is not attained. If extremely dilute solutions of Na2C4H6O6 and CaCl2 are mixed, for instance, it may be that the concentrations of calcium ions and chloride ions in the resultant mixture are both too low for a precipitate to form. In such a case we would find that the product Q, called the ion product and defined by

$Q=\text{(}c_{\text{Ca}^{2+}}\text{)(}c_{\text{C}_{4}\text{H}_{4}\text{O}_{6}^{2-}})$      (1)

In this case Q has a value below the solubility product, 7.7 × 10–7 mol2 dm–6. In order for equilibrium between the ions and a precipitate to be established, either the calcium-ion concentration or the tartrate-ion concentration or both must be increased until the value of Q is exactly equal to the value of the solubility product. The opposite situation, in which Q is larger than Ksp, corresponds to concentrations which are too large for the solution to be at equilibrium. When this is the case, precipitation occurs, lowering the concentration of both the lead and chloride ions, until Q is exactly equal to the solubility product.

To determine in the general case whether a precipitate will form, we set up an ion-product expression Q which has the same form as the solubility product, except that the stoichiometric concentrations rather than the equilibrium concentrations are used. Then if

$\text{Q}>{K}_{sp}\,$      precipitation occurs

while if      $\text{Q}<{K}_{sp}\,$      no precipitation occurs

EXAMPLE 1 Decide whether CaC2O4, calcium oxalate, will precipitate or not when (a)100 cm3 of 0.02 M CaCl2 and 100 cm3 of 0.02 M Na2C2O4 are mixed, and also when (b) 100 cm3 of 0.0001 M CaCl2 and 1000 cm3 of 0.0001 M Na2C2O4 are mixed. Ksp = 2.32 × 10–9 mol2 dm–6.

Solution

a) After mixing, the concentration of each species is halved. We thus have

${c}_{\text{Ca}^{2+}}=\text{0.01 mol dm}^{-3}={c}_{\text{C}_{2}\text{O}_{4}^{2-}}$

so that the ion-product Q is given by

${Q}={c}_{\text{Ca}^{2+}} \times {c}_{\text{C}_{2}\text{O}_{4}^{2-}} = \text{0.01 mol dm}^{-3} \times \text{ 0.01 mol dm}^{-3}$

or      ${Q}= \text{10}^{-4} \text{ mol}^{2}\text{dm}^{-6}\,$

Since Q is larger than Ksp (2.32 × 10–9 mol2 dm–6), precipitation will occur.

b) In the second case the concentration of each ion becomes

${c}_{\text{Ca}^{2+}}=\frac{\text{0.0001 mol dm}^{-3} \times \text{100 cm}^{3}}{\text{1100 cm}^{3}} = \text{9.09} \times \text{10}^{-6} \text{mol dm}^{-3}$

and      ${c}_{\text{C}_{2}\text{O}_{4}^{2-}}=\frac{\text{0.0001 mol dm}^{-3} \times \text{1000 cm}^{3}}{\text{1100 cm}^{3}} = \text{9.09} \times \text{10}^{-5} \text{mol dm}^{-3}$

thus      \begin{align}{Q}&={c}_{\text{Ca}^{2+}}\times {c}_{\text{C}_{2}\text{O}_{4}^{2-}}\\ \text{ }&=(\text{9.09} \times \text{10}^{-6}\text{mol dm}^{-3}) (\text{9.09} \times \text{10}^{-5} \text{mol dm}^{-3})\\ \text{ }&=\text{8.26} \times \text{10}^{-10}\text{mol}^{2} \text{dm}^{-6}\end{align}

Since Q is now less than Ksp, no precipitation will occur.

EXAMPLE 2 Calculate the mass of CaC2O4 precipitated when 100 cm3 of 0.0200 M CaCl2 and 100 cm3 of 0.0200 M Na2C2O4 are mixed together.

Solution In in part a of the previous example we determined that precipitation does actually occur. In order to find how much calcium oxalate is precipitated, we must concentrate on the amount of each species. Since 100 cm3 of 0.02 M CaCl2 was used, we have

$n_{\text{Ca}^{2+}}=\text{0.0200 }\frac{\text{mmol}}{\text{cm}^{3}}\times \text{ 100 cm}^{3}=\text{2.00 mmol}$

similarly      $n_{\text{C}_{2}\text{O}_{4}^{2-}}=\text{0.0200 }\frac{\text{mmol}}{\text{cm}^{3}} \times \text{ 100 cm}^{3}=\text{2.00 mmol}$

If we designate the amount of CaC2O4 that precipitates as x mmoles, we can set up the following table

 Species Ca2+ (aq) C2O42– (aq) Initial amount (mmol) $\text{2.00}\,$ $\text{2.00}\,$ Amount reacted (mmol) $-{x}\,$ $-{x}\,$ Equilibrium   amount (mmol) $(\text{2}-{x})\,$ $(\text{2}-{x})\,$ Equilibrium   concentration (mmol cm–3) $\frac{\text{2}-{x}}{\text{200}}$ $\frac{\text{2}-{x}}{\text{200}}$

Thus

${K}_{sp} = [\text{Ca}^{2+}][\text{C}_{2}\text{O}_{4}^{2-}]\,$

or      $\text{2.32 }\times \text{ 10}^{-9}\text{ mol}^{2}\text{ dm}^{-6}=\left( \frac{\text{2}-x}{\text{200}}\text{ mol dm}^{-3} \right)\left( \frac{\text{2}-x}{\text{200}}\text{ mol dm}^{-3} \right)$

Rearranging,

$\text{200}^{2}\times\text{2.32}\times\text{10}^{-9}=\text{9.28}\times\text{10}^{-3}=(\text{2}-{x})^{2}$

or              $\text{2}-{x}=\sqrt{\text{9.28}\times \text{10}^{-3}}=\text{0.096}$

so that      ${x}=\text{2}-\text{0.096}=\text{1.904}\,$

Since 1.904 mmol of CaC2O4 precipitated, its mass is

\begin{align}m_{\text{CaC}_{2}\text{O}_{4}}&=\text{1.904 mmol }\times \text{ 146.11 }\frac{\text{mg}}{\text{mmol}}\\ \text{ }&=\text{278.2 mg}=\text{0.278 g}\end{align}

Note: Since the solubility product of CaC2O4 is very small, about 95% of the calcium oxalate originally formed precipitates.

References

1. Food Additives, 2nd ed. 2002, Branen, A., Davidson, M.P., Salminen, S. and Thorngate III, J.H.
2. Pollard, K. 2009. Waste management in the Wine Industry. http://www.brocku.ca/tren/courses/tren3p14/Pollard_Kasey_4120903_TREN3P14%20Waste%20Management%20Paper.pdf
3. Arvanitoyannis, I. S., Ladas, D., and Mavromatis, A. 2006. Wine waste treatment methodology. Int. J. Food Sci. Tech. 41:1117-1151.
4. Rivas, B., Torrado, A., Moldes, A. B., and Dominguez, J. M. 2006. Tartaric acid recovery from distilled lees and use of the residual solid as an economic nutrient for Lactobacillus. J. Agric. Food Chem. 54:7904-7911.