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Describe the solubility of \(\mathrm{CaCO}_{3}\) in each of the following solutions compared to its solubility in water: (a) in \(0.10 \mathrm{M} \mathrm{NaCl}\) solution; \((\mathrm{b})\) in \(0.10 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) solution; (c) \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\); (d) \(0.10 \mathrm{M}\) HCl solution. (Answer same, less soluble, or more soluble.)

Short Answer

Expert verified
(a) same; (b) less soluble; (c) less soluble; (d) more soluble

Step by step solution

01

In the 0.10 M NaCl solution, there will be an increase in electrolytes but no common ions with the solid CaCO3. Because of this, the solubility of calcium carbonate will not be significantly affected by the electrolyte (NaCl), and the solubility will be the same as in pure water. Thus, the answer is: same #Step 2: Solubility in 0.10 M Ca(NO3)2 solution#

In the 0.10 M Ca(NO3)2 solution, there is a common ion presence, Ca2+, which will affect the solubility of CaCO3. The common ion effect states that the presence of a common ion will supress the solubility of a poorly soluble salt, such as CaCO3 in this case. The Ca2+ ions from the Ca(NO3)2 will shift the equilibrium to the left, decreasing the solubility of CaCO3. Thus, the answer is: less soluble #Step 3: Solubility in 0.10 M Na2CO3 solution#
02

In the 0.10 M Na2CO3 solution, CO32- is the common ion present. Similar to step 2, the common ion effect comes into play. The presence of CO32- will shift the equilibrium to the left, reducing the solubility of CaCO3. Thus, the answer is: less soluble #Step 4: Solubility in 0.10 M HCl solution#

In the 0.10 M HCl solution, the H+ ions will react with the CO32- ions from the calcium carbonate, forming HCO3- ions. This reaction will essentially remove the CO32- ions from the solution and drive the equilibrium to the right, which increases the solubility of calcium carbonate. Thus, the answer is: more soluble

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Key Concepts

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

Solubility Equilibrium
Understanding how substances dissolve in various substances is paramount in fields such as chemistry and environmental science. Solubility equilibrium is a type of dynamic equilibrium that exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. The solubility product constant (\(K_{sp}\)) is a numerically equal value to the product of the ions' concentrations, each raised to the power of its stoichiometric coefficient.

When the system is at equilibrium, the solution is saturated and any additional substance will not dissolve. In fact, the undissolved substance and dissolved ions exist in a balance where their concentrations no longer change over time. For example, calcium carbonate, \(CaCO_3\), if not affected by other factors, will reach a point where no more of it can be dissolved into the water.
Common Ion Effect
The common ion effect refers to the decrease in solubility of an ionic compound when another compound that provides a common ion is added to the solution. This principle is a direct application of Le Chatelier's Principle, which predicts that the addition of an ion common to a dissolved substance will shift the equilibrium to favor the undissolved state, thereby reducing its solubility.

For instance, \(CaCO_{3}\) is less soluble in a \(Ca(NO_{3})_{2}\) solution because the common ion \(Ca^{2+}\) is already present, and the equilibrium will shift to reduce its concentration by forming more solid \(CaCO_{3}\). This effect has practical implications in everything from the handling of waste in the environment to the formulation of pharmaceuticals.
Solubility in Different Solutions
The solubility of a substance can vary widely depending on the other components in the solution. In some cases, a salt such as \(CaCO_{3}\) may be equally soluble in water and a solution that does not contain common ions, such as \(NaCl\) solution. However, introducing a common ion like \(Ca^{2+}\) or \(CO_{3}^{2-}\) in the solution will significantly alter solubility. For instance, solubility decreases in the presence of \(Ca(NO_{3})_{2}\) or \(Na_{2}CO_{3}\) due to the common ion effect discussed above. On the other hand, the addition of \(HCl\) to the solution leads to an increase in solubility, as additional reactions, such as the formation of \(HCO_{3}^{-}\), occur which remove \(CO_{3}^{2-}\) ions, pushing the dissolution equilibrium forward.
Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction and its reverse reaction occur at the same rate, leading to the maintenance of constant properties in the system. It is a state in which reactants and products are present in concentrations which have no further tendency to change with time. This concept is crucial as it applies to all chemical reactions and is foundational to understanding how reactions occur in nature.

Chemical equilibriums can be affected by changes in concentration, temperature, pressure, and volume. Applying this to solubility, when a compound reaches its solubility equilibrium in a solution, dissolving and precipitating occur at equal rates, with saturated concentrations staying constant. If additional common ions are introduced or removed, the equilibrium will shift according to Le Chatelier's Principle, thereby either favoring the formation of precipitate or increased solubility.

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Most popular questions from this chapter

The acid-base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the \(\mathrm{pH}\) is \(4.68\). What is the \(\mathrm{p} K_{a}\) for bromcresol green?

(a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubilityproduct constant? (b) Write the expression for the solubility-product constant for each of the following strong electrolytes: \(\mathrm{AgI}, \mathrm{SrSO}_{4}, \mathrm{Fe}(\mathrm{OH})_{2}\), and \(\mathrm{Hg}_{2} \mathrm{Br}_{2}\).

Suppose you want to do a physiological experiment that calls for a \(\mathrm{pH} 6.5\) buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 M\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH} 6.50 ?\) (lgnore any volume change.)

How many grams of sodium lactate \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COONa}\right.\) or \(\left.\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) should be added to \(1.00 \mathrm{~L}\) of \(0.150 \mathrm{M}\) lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to form a buffer solution with \(\mathrm{pH} 4.00\) ? Assume that no volume change occurs when the sodium lactate is added.

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) \(\mathrm{ZnCO}_{3}\), (b) \(\mathrm{ZnS},(\mathrm{c}) \mathrm{BiI}_{3}\) (d) \(\mathrm{AgCN}\), (e) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\)

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