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Chapter 7: Problem 38

In 1 L saturated solution of \(\mathrm{AgCl}\left[\mathrm{Ksp}(\mathrm{AgCl})=1.6 \times 10^{-10}\right], 0.1 \mathrm{~mol}\) of \(\mathrm{CuCl}\left[K_{s p}(\mathrm{CuCl})=1.0 \times 10^{-6}\right]\) is added. The resultant concentration of \(\mathrm{Ag}^{+}\)in the solution is \(1.6 \times 10^{-x}\). The value of " \(x\) " is

Short Answer

Expert verified
x = 9

Step by step solution

01

Understanding Saturated Solution Equilibria

In a saturated solution of AgCl, \[AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq)\]The solubility product expression for AgCl is given by:\[K_{sp}(AgCl) = [Ag^+][Cl^-] = 1.6 \times 10^{-10}\]
02

Saturated Solution of CuCl

Similarly, for CuCl:\[CuCl (s) \rightleftharpoons Cu^+ (aq) + Cl^- (aq)\]The solubility product expression is:\[K_{sp}(CuCl) = [Cu^+][Cl^-] = 1.0 \times 10^{-6}\]By adding 0.1 mol of CuCl, Cu^+ ions and Cl^- ions are introduced into the solution.
03

Analyzing Chloride Ion Concentration

Adding 0.1 mol of CuCl means the concentration of Cl^- ions increases in the solution. Initially, before AgCl dissolution, we have a contribution from CuCl, so:\[[Cl^-] = [Cl^-]_{CuCl} = 0.1 \text{ mol/L}\]
04

Solving for Ag+ Concentration

In the presence of high [Cl^-] from CuCl, AgCl dissolution is affected. Using the expression for AgCl solubility, substitute the known Cl^- concentration:\[[Ag^+] = \frac{K_{sp}(AgCl)}{[Cl^-]} = \frac{1.6 \times 10^{-10}}{0.1} = 1.6 \times 10^{-9}\]Thus, the concentration of \([Ag^+]\) is calculated from the reduced solubility due to common ion effect.
05

Finding the Value of x

We know from the problem statement that \([Ag^+] = 1.6 \times 10^{-x}\).Comparing it with our calculated \([Ag^+] = 1.6 \times 10^{-9}\), \(x\) is clearly 9.

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Key Concepts

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

Common Ion Effect
The common ion effect occurs when a compound's solubility is affected by the presence of a common ion in the solution. In other words, when two different compounds that share a common ion are present, the solubility of the less soluble compound is reduced. This happens because the increased concentration of the common ion shifts the equilibrium position according to Le Chatelier's principle.
For example, in a saturated solution of AgCl, adding CuCl introduces additional chloride ions. The common ion here is the chloride ion \(Cl^-\).
As a result, the increased concentration of chloride ions from the \(CuCl\) decreases the solubility of \(AgCl\). This is because the system will try to counterbalance the increased chloride concentration by reducing the concentration of silver ions, \(Ag^+\), in the solution.
This effect is important in chemistry as it explains why some solutions have their solubilities reduced when other compounds are present.
Ionic Equilibrium
Ionic equilibrium refers to a state where the rate of formation of ions in the solution is equal to the rate of recombination back into the solid state, thus maintaining a constant concentration of ions. This equilibrium can exist in various types of reactions involving ionic compounds.
For a saturated solution, the equilibrium constant, or solubility product constant \(K_{sp}\), gives us insight into the ions' concentration at that specific state. For example, in the saturated solution of \(AgCl\), the equilibrium equation is described by:
\[AgCl \, (s) \rightleftharpoons Ag^+ \, (aq) + Cl^- \, (aq)\].
The equilibrium constant \(K_{sp}\) helps determine the solubility of \(AgCl\) in the presence of ions. If more of one ion is added, as in the case of adding CuCl which introduces more Cl鈦� ions, the equilibrium will shift, reducing the concentration of Ag鈦� ions accordingly. This understanding of ionic equilibrium is fundamental to predicting how concentrations will change under various conditions.
Saturated Solution
A saturated solution is a solution that contains the maximum amount of solute that can dissolve under the given temperature and pressure conditions. Beyond this point, any added solute will remain undissolved and will not increase the concentration of the solute in the solution.
In the case of AgCl in water, once the solution becomes saturated, the concentrations of Ag鈦� and Cl鈦� ions remain constant at equilibrium. If any other source of these ions is added to the solution, such as through the dissolution of another compound like CuCl, the solubility product equilibrium and thus, the concentrations of the ions may be affected.
This concept is significant as it demonstrates how a solution can reach a state of dynamic balance, and how additional solutes can influence this state by altering the concentration of ions in solution.

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Most popular questions from this chapter

The minimum volume of water required to dissolve \(0.1 \mathrm{~g}\) lead (II) chloride to get a saturated solution \(\left(K_{\mathrm{Sp}}\right.\) of \(\mathrm{PbCl}_{2}=3.2 \times 10^{-8}\); atomic mass of \(\mathrm{Pb}=207 \mathrm{u}\) ) is : (a) \(1.798 \mathrm{~L}\) (b) \(0.36 \mathrm{~L}\) (c) \(17.95 \mathrm{~L}\) (d) \(0.18 \mathrm{~L}\)

If the solubility product of \(\mathrm{AB}_{2}\) is \(3.20 \times 10^{-11} \mathrm{M}^{3}\), then the solubility of \(\mathrm{AB}_{2}\) in pure water is \(\times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\). [Assuming that neither kind of ion reacts with water]

The \(K_{s p}\) of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is \(1.1 \times 10^{-12}\) at \(298 \mathrm{~K}\). The solubility (in \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) in a \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\) solution is (a) \(1.1 \times 10^{-11}\) (b) \(1.1 \times 10^{-10}\) (c) \(1.1 \times 10^{-12}\) (d) \(1.1 \times 10^{-9}\)

For the reaction : $$ \mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ The forward reaction at constant temperature is favoured by (a) introducing an inert gas at constant volume (b) introducing chlorine gas at constant volume (c) introducing an inert gas at constant pressure (d) increasing the volume of the container (e) introducing \(\mathrm{PCl}_{5}\) at constant volume

An aqueous solution contains \(0.10 \mathrm{MH}_{2} \mathrm{~S}\) and \(0.20 \mathrm{M} \mathrm{HCl}\). If the equilibrium constants for the formation of \(\mathrm{HS}^{-}\)from \(\mathrm{H}_{2} \mathrm{~S}\) is \(1.0 \times 10^{-7}\) and that of \(\mathrm{S}^{2}\) from HS ions is \(1.2 \times 10^{-13}\) then the concentration of \(\mathrm{S}^{2-}\) ions in aqueous solution is : (a) \(5 \times 10^{-8}\) (b) \(3 \times 10^{-20}\) (c) \(6 \times 10^{-21}\) (d) \(5 \times 10^{-19}\)
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