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The addition of \(\mathrm{NaCl}\) to \(\mathrm{AgCl}\) decreases the solubility of \(\mathrm{AgCl}\) because (1) solubility product decreases (2) due to the common ion effect of \(\mathrm{Cl}\) (3) solubility becomes unsaturated (4) solution becomes supersaturated

Short Answer

Expert verified
The solubility of \( \mathrm{AgCl} \) decreases due to the common ion effect of \( \mathrm{Cl}^- \) from \( \mathrm{NaCl} \).

Step by step solution

01

- Understand the problem

Identify the key substances in the problem: \(\mathrm{NaCl}\) and \(\mathrm{AgCl}\). Also focus on the term 'decreases the solubility' of \(\mathrm{AgCl}\).
02

- Solubility Product Concept

Recall that the solubility product constant (Ksp) of \( \mathrm{AgCl} \) is an equilibrium constant that relates to the concentrations of its ions in a saturated solution.
03

- Common Ion Effect

Adding \( \mathrm{NaCl} \) introduces more \( \mathrm{Cl^-} \) ions into the solution since \( \mathrm{NaCl} \) dissociates completely in water. According to the common ion effect, the addition of an ion common to the dissolved substance will decrease its solubility.
04

- Apply Common Ion Effect

Explain that the additional \( \mathrm{Cl^-} \) ions from \( \mathrm{NaCl} \) shift the equilibrium of \( \mathrm{AgCl} \) dissociation: \[ \mathrm{AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)} \] \ Hence, according to Le Chatelier's principle, the system will try to decrease the concentration of \( \mathrm{Cl^-} \) by forming more \( \mathrm{AgCl} \), thus decreasing its solubility.

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Key Concepts

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

Solubility Product Constant (Ksp)
The solubility product constant, represented as \( K_{sp} \), is a fundamental concept in chemistry that helps us understand the solubility of sparingly soluble salts. It expresses the product of the concentrations of the constituent ions each raised to the power of their stoichiometric coefficients in a saturated solution. For example, for silver chloride (\(AgCl\)), which dissociates in water as \[ \mathrm{AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)} \], the solubility product constant \( K_{sp} \) is given by: \[ K_{sp} = [ Ag^{+} ][ Cl^{-} ] \] Here, \( [Ag^{+}] \) and \( [Cl^{-}] \) are the molar concentrations of the silver and chloride ions in the solution. If the solution becomes saturated, the concentrations of these ions will reach a point where the product equals the \( K_{sp} \) value. When this value is exceeded, the salt will begin to precipitate out of the solution.
Understanding \( K_{sp} \) is crucial because it tells us how much of the salt can dissolve in a given amount of solvent at a specific temperature. This helps predict whether a precipitate will form under given conditions.
Common Ion Effect
The common ion effect is a phenomenon observed in the solubility of ionic compounds. It occurs when a solution already contains one of the ions from a dissolved salt. The introduction of additional common ions reduces the solubility of the salt. In our specific example, adding \( NaCl \) to a solution containing \( AgCl \) introduces more chloride ions (\( Cl^{-} \)), since NaCl dissociates completely in water:
\[ \mathrm{NaCl(s) \rightarrow Na^{+}(aq) + Cl^{-}(aq)} \] Therefore, the increase in \( Cl^{-} \) ion concentration from the \( NaCl \) shifts the equilibrium of \( AgCl \) dissociation:
\[ \mathrm{AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)} \] According to the common ion effect, because the solution now has an excess of \( Cl^{-} \) ions, the solubility of \( AgCl \) will decrease. There's less tendency for \( AgCl \) to dissociate because the product of its ions in the solution would exceed \( K_{sp} \). Thus, the introduction of a common ion effectively reduces the solubility of the ionic compound in the solution.
Le Chatelier's Principle
Le Chatelier's principle is a critical rule in chemistry that helps predict how a system at equilibrium will respond to changes in concentration, temperature, or pressure. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system will readjust itself to counteract the change and re-establish equilibrium.
When NaCl is added to an \( AgCl \) solution, the additional \( Cl^{-} \) ions create a disturbance in the existing equilibrium:
\[ \mathrm{AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)} \] With more \( Cl^{-} \) ions present, according to Le Chatelier's principle, the equilibrium will shift to the left to reduce the disturbance caused by the increased concentration of \( Cl^{-} \). This shift results in the formation of more solid \( AgCl \) and consequently decreases its solubility in the solution. The system does this to re-establish equilibrium, aligning with the principle that the reaction will move to counteract the addition of the common ion. Understanding Le Chatelier's principle allows us to predict and explain such changes in solubility and the formation of precipitates in various chemical contexts.

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Most popular questions from this chapter

The \(\mathrm{pH}\) of a solution produced when an aqueous solution of strong acid \(\mathrm{pH} 5\) mixed with equal volume of an aqueous solution of strong acid of \(\mathrm{pH} 3\) is (1) \(3.3\) (2) \(3.5\) (3) \(4.5\) (4) \(4.0\)

\(\Lambda\) chemist who is concerned with large-scale manufacture of useful compounds is primarily intcrested in (1) minimizing the cncrgy consumption (2) maximizing the backward reaction (3) minimizing the reverse reaction (4) decrcasing the acidity of the product

When equal volumes of the following solutions are mixed, precipitation of \(\mathrm{AgCl}\left(K_{\mathrm{pp}}=1.8 \times 10^{10}\right)\) will occur only with (1) \(10^{-4} \mathrm{M}\left(\mathrm{Ag}^{-}\right)\) and \(10^{-4} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (2) \(10^{-5} \mathrm{M}\left(\Lambda \mathrm{g}^{-}\right)\) and \(10^{-5} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (3) \(10^{-6} \mathrm{M}\left(\Lambda \mathrm{g}^{-}\right)\) and \(10^{-6} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (4) \(10^{-10} \mathrm{M}\left(\Lambda \mathrm{g}^{-}\right)\) and \(10^{-10} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\)

Which of the following statements is falsc? (1) Hydrolysis of the salt of strong acid and weak base is called cationic hydrolysis (2) Hydrolysis of the salt of weak acid and strong base is known as anionic hydrolysis (3) Aqueous solution of aluminium chloride is acidic due to hydrolysis of \(\mathrm{Al}^{3}\) ion (4) Aqueous solution of sodium carbonate is basic due to hydrolysis of \(\mathrm{Na}\) ion

\(\Lambda\) saturated solution of \(\mathrm{II}_{2} \mathrm{~S}\) in \(0.1 \mathrm{M} \mathrm{IICl}\) at \(25^{\circ} \mathrm{C}\) contains a \(\mathrm{S}^{2-}\) ion concentration of \(10^{-23} \mathrm{~mol} \mathrm{~L}^{-1}\). The solubility products of some sulphidcs are: \(\mathrm{CuS}=10^{-44}\), \(\mathrm{FeS}=10^{-14} ; \mathrm{MnS}=10^{-15}\) and \(\mathrm{CdS}=10^{-25} .\) If \(0.01 \mathrm{M}\) solutions of these salts in \(1 \mathrm{M}\) IICl are saturated with \(\mathrm{H}_{2} \mathrm{~S}\), which of these will be precipitated? (1) All (2) All except MnS (3) All except MnS and FeS (4) Only CuS

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