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Many plants are poisonous because their stems and leaves contain oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) or sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) . When ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-},\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride, \(\mathrm{CaCl}_{2},\) in aqueous solution.

Short Answer

Expert verified
The net ionic equation for the reaction between sodium oxalate (Na2C2O4) and calcium chloride (CaCl2) in aqueous solution is: \(C2O4^{2-} (aq) + Ca^{2+} (aq) \rightarrow CaC2O4 (s)\)

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction between sodium oxalate and calcium chloride is: Na2C2O4 (aq) + CaCl2 (aq) → CaC2O4 (s) + 2 NaCl (aq)
02

Identify the spectator ions

Spectator ions are ions that remain unchanged during the reaction and do not participate in it. In this case, the spectator ions are sodium (Na⁺) and chloride (Cl⁻) ions.
03

Write the net ionic equation

To write the net ionic equation, we'll remove the spectator ions from the balanced chemical equation. The net ionic equation is: C2O4²⁻ (aq) + Ca²⁺ (aq) → CaC2O4 (s) This is the net ionic equation for the reaction between sodium oxalate and calcium chloride in aqueous solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxalate Ion
The oxalate ion, represented as \( \text{C}_2\text{O}_4^{2-} \), is derived from oxalic acid \( \text{H}_2\text{C}_2\text{O}_4 \). It functions primarily as an anion in chemical reactions, which means it carries a negative charge. When substances like oxalic acid or sodium oxalate dissolve in water, they release oxalate ions.

Oxalate ions can be harmful if ingested in large amounts, as they can combine with calcium in the body to form insoluble compounds, potentially leading to health issues such as the formation of kidney stones. This happens because calcium oxalate is insoluble, meaning it does not dissolve well in water, and can accumulate in organs.

In a laboratory analysis, the presence of oxalate ions is often detected by converting them into an insoluble form with a known reagent. For example, in the reaction with calcium chloride, the oxalate ions are precipitated out as calcium oxalate. This precipitation process is helpful in quantifying the amount of oxalate in a sample by isolating it as a solid compound.
Calcium Oxalate
Calcium oxalate \( \text{CaC}_2\text{O}_4 \) is the compound formed when calcium ions \( \text{Ca}^{2+} \) react with oxalate ions \( \text{C}_2\text{O}_4^{2-} \). It forms as an insoluble precipitate in aqueous solutions. This means that it does not readily dissolve.

This precipitation reaction is useful in a variety of analytical procedures. When analyzing samples for oxalate content, calcium chloride is often used because it reacts with oxalate ions to produce a visible solid, calcium oxalate. This allows researchers to filter and weigh the precipitate, thereby determining the amount of oxalate ions present in the original sample.
  • Typical reactions involve mixing solutions of a soluble calcium salt, like calcium chloride, with a solution containing oxalate ions.
  • The formation of calcium oxalate precipitate is indicative of the presence of oxalate ions and can be used for qualitative identification and quantitative analysis.
Moreover, due to its poor solubility, calcium oxalate also has implications outside of the laboratory. In biology and medicine, calcium oxalate is known for forming kidney stones, a painful condition that can occur when these compounds accumulate in the urinary tract.
Spectator Ions
Spectator ions are ions that exist in the same form on both the reactant and product sides of a chemical equation. In other words, they do not actually participate in the reaction itself. Instead, they remain dissolved in the solution.

In the given reaction between sodium oxalate and calcium chloride, the spectator ions are sodium \( \text{Na}^+ \) and chloride \( \text{Cl}^- \). These ions do not take part in the formation of calcium oxalate and remain in the aqueous solution throughout the reaction.
  • Writing net ionic equations involves removing these spectator ions to streamline and highlight the actual chemical change occurring.
  • In the mentioned reaction, the net ionic equation is written as \( \text{C}_2\text{O}_4^{2-} (aq) + \text{Ca}^{2+} (aq) \rightarrow \text{CaC}_2\text{O}_4 (s) \), focusing only on the ions that combine to form the precipitate.
Understanding spectator ions is crucial for simplifying chemical equations and concentrating on the core interactions during a chemical reaction. This practice assists in analyzing and predicting the outcomes of reactions, making them more comprehensible even to those new to chemistry.

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Most popular questions from this chapter

Chlorisondamine chloride \(\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{N}_{2}\right)\) is a drug used in the treatment of hypertension. A 1.28-g sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, 0.104 g silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second (L/s) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{L} / \mathrm{s}\) of water that contains 65.0 \(\mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 135 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$\mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(i) $$ What mass of CaO is consumed in an 8.00-h work day by this plant? d. The original stream water contained 10.2 \(\mathrm{ppm} \mathrm{Ca}^{2+}\) . Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If 90.0% of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

It took \(25.06 \pm 0.05 \mathrm{mL}\) of a sodium hydroxide solution to titrate a 0.4016-g sample of KHP (see Exercise 83). Calculate the concentration and uncertainty in the concentration of the sodium hydroxide solution. (See Appendix 1.5.) Neglect any uncertainty in the mass

Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is $$\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$$ Balance this equation.

In most of its ionic compounds, cobalt is either Co(II) or Co(III). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A 0.256-g sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. A second sample of 0.416 g of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was 0.145 g. a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.

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