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Chapter 20: Problem 52

The label of a certain brand of mayonnaise lists EDTA as a food preservative. How does EDTA prevent the spoilage of mayonnaise?

Short Answer

Expert verified
EDTA, or Ethylenediaminetetraacetic acid, works as a preservative by exploiting its chelating properties where it binds to metal ions and subsequently inhibits the growth of microbes and oxidation of fats in mayonnaise, ultimately lowering the rate of its spoilage.

Step by step solution

01

- Understanding of EDTA

EDTA, or Ethylenediaminetetraacetic acid, is a known compound widely used as a preservative. It's known for its strong chelating properties, which means it's able to bind with metal ions.
02

- Function of EDTA

EDTA inhibits enzymes by binding to metal ions that are essential for the enzyme's activity. This function of EDTA is often exploited in the food industry to extend the shelf-life of food products.
03

- Application to mayonnaise

In the context of mayonnaise, which is an emulsion of oil, egg yolk and acidic liquid, such as vinegar or lemon juice, spoilage can be caused by the growth of microbes or by the oxidation of fats. EDTA, due to its ability to bind metal ions, prevent certain enzyme activities and hence inhibit the growth of these microbes, as well as slows down the oxidation process, ultimately reducing the rate of spoilage.

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Most popular questions from this chapter

Which of these hydrated cations are colorless: \(\mathrm{Fe}^{2+}(a q), \mathrm{Zn}^{2+}(a q), \mathrm{Cu}^{+}(a q), \mathrm{Cu}^{2+}(a q), \mathrm{V}^{5+}(a q)\) \(\mathrm{Ca}^{2+}(a q), \mathrm{Co}^{2+}(a q), \mathrm{Sc}^{3+}(a q), \mathrm{Pb}^{2+}(a q) ?\) Explain your choice.

When aqueous potassium cyanide is added to a solution of copper(II) sulfate, a white precipitate, soluble in an excess of potassium cyanide, is formed. No precipitate is formed when hydrogen sulfide is bubbled through the solution at this point. Explain.

Draw structures of all the geometric and optical isomers of each of these cobalt complexes: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}\) (d) \(\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\) (e) \(\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\)

The formation constant for the reaction \(\mathrm{Ag}^{+}+\) \(2 \mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) is \(1.5 \times 10^{7}\) and that for the reaction \(\mathrm{Ag}^{+}+2 \mathrm{CN}^{-} \rightleftharpoons\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) is 1.0 \(\times 10^{21}\) at \(25^{\circ} \mathrm{C}\) (see Table 17.4 ). Calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for the reaction $$\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}+2 \mathrm{CN}^{-} \rightleftharpoons\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}+2 \mathrm{NH}_{3}$$

As we read across the first-row transition metals from left to right, the +2 oxidation state becomes more stable in comparison with the +3 state. Why is this so?
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