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Calcium chloride is a strong electrolyte and is used to "salt" streets in the winter to melt ice and snow. Write a reaction to show how this substance breaks apart when it dissolves in Water.

Short Answer

Expert verified
When calcium chloride (CaCl2) dissolves in water, it dissociates completely into its ions, forming a calcium ion (\(Ca^{2+}\)) and two chloride ions (\(2Cl^-\)). The balanced chemical equation for this reaction is: \(CaCl_2(s) \rightarrow Ca^{2+}(aq) + 2Cl^-(aq)\)

Step by step solution

01

Step 1. Writing the formula of Calcium chloride (CaCl2)

As given in the problem, calcium chloride has the chemical formula CaCl2.
02

Step 2. Identify the ions formed when it dissolves in water

When calcium chloride dissolves in water, it dissociates into two types of ions: 1. Calcium ion: \(Ca^{2+}\) 2. Chloride ions: \(Cl^-\) Note that calcium ion has a charge of +2, and chloride ion has a charge of -1.
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Step 3. Write the balanced chemical equation for CaCl2 dissolving in water

Now we know the ions formed during the dissociation of calcium chloride, we can write the balanced chemical equation: \(CaCl_2(s) \rightarrow Ca^{2+}(aq) + 2Cl^-(aq)\) This equation shows that when solid calcium chloride (CaCl2) dissolves in water, it breaks apart into one calcium ion (\(Ca^{2+}\)) and two chloride ions (\(2Cl^-\)) in aqueous solution.

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Key Concepts

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

Strong Electrolytes
Understanding strong electrolytes is essential when studying substances like calcium chloride (CaCl2). A strong electrolyte is any compound that completely dissociates into its ions when dissolved in water. This means that when you add a substance like CaCl2 to water, it will separate into positively charged calcium ions (\(Ca^{2+}\)) and negatively charged chloride ions (\(Cl^-\)) without leaving any undissociated molecules behind.

Strong electrolytes tend to increase the electrical conductivity of water because they release a significant number of ions, which are charge carriers. Substances that are strong electrolytes include most salts, inorganic acids like hydrochloric acid (HCl), and bases such as sodium hydroxide (NaOH). In the context of this exercise, by knowing that calcium chloride is a strong electrolyte, we can predict that it will efficiently break apart into ions in a solution, which is why it's effective in melting ice and snow by disrupting the structure of ice through its ions.
Ionic Dissociation
Ionic dissociation is the process in which an ionic compound separates into its constituent ions when dissolved in a solvent, typically water. This concept is at the heart of understanding how calcium chloride behaves in aqueous solutions. As we look at CaCl2, which is composed of calcium ions and chloride ions held together by ionic bonds in a solid state, upon dissolving, these bonds are overcome by the interaction with water molecules.

The dissociation can be visualized as water molecules surrounding the CaCl2 unit and pulling the ions apart due to their attraction to the positive and negative charges of the water molecules (a polar solvent). This process is crucial for processes like de-icing, as the free ions in the solution work to disrupt the ice's crystalline structure. Remember, the extent of dissolution and dissociation is crucial in applications that require ions in free states, such as electrolyte solutions for batteries or physiological fluids.
Chemical Equations
) ions after dissolution, specifically one calcium ion and two chloride ions, which is reflected by the numbers in front of the chloride ions. The process described is a physical change rather than a chemical reaction since the ions are not chemically altered, only their physical state changes. Additionally, the equation shows the conservation of charge; the charge is balanced on both sides, which is an essential principle in chemistry. Being able to interpret and write chemical equations is a fundamental skill in chemistry that provides a clear and concise way of conveying the changes occurring during chemical processes.

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Most popular questions from this chapter

Hydrochloric acid (75.0 mL of 0.250 \(M\) ) is added to 225.0 mL of 0.0550 \(M\) Ba(OH) \(_{2}\) solution. What is the concentration of the excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in this solution?

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 \(65.0 \mathrm{ppm}\) HCl into the stream. (See Exercise 123 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of 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 \(\mathrm{CaO}\) 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.

A 500.0 -mL sample of \(0.200 \mathrm{M}\) sodium phosphate is mixed with \(400.0 \mathrm{mL}\) of \(0.289 M\) barium chloride. What is the mass of the solid produced?

What volume of each of the following bases will react completely with \(25.00 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HCl? a. \(0.100 M\) NaOH c. \(0.250 M\) KOH b. \(0.0500 M \operatorname{Sr}(O H)_{2}\)

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow\) b. \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow\) c. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow\)

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