Common Ion Effect On Solubility Pogil Answers
Common Ion Effect On Solubility Pogil
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Learning Objectives

Common Ion Effect on Solubility 5 Extension Questions 15. In a precipitation reaction (otherwise known as a double replacement or metathesis reaction) the extent to which a precipitate is formed is dependent on the solubility product constant of the precipitate. The role that the common ion effect plays in solutions is mostly visible in the decrease of solubility of solids. Through the addition of common ions, the solubility of a compound generally decreases due to a shift in equilibrium. The common ion effect also plays a.

Recognize common ions from various salts, acids, and bases.
Calculate concentrations involving common ions.
Calculate ion concentrations involving chemical equilibrium.

The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.

Introduction

The solubility products K_sp's are equilibrium constants in hetergeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:

[mathrm{[Na^+] + [K^+] = [Cl^-]} label{1}]

Consideration of charge balance or mass balance or both leads to the same conclusion. The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that K_sp is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. Adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion. The exceptions generally involve the formation of complex ions, which is discussed later.

Common Ions

When (ce{NaCl}) and (ce{KCl}) are dissolved in the same solution, the (mathrm{ {color{Green} Cl^-}}) ions are common to both salts. In a system containing (ce{NaCl}) and (ce{KCl}), the (mathrm{ {color{Green} Cl^-}}) ions are common ions.

(mathrm{NaCl rightleftharpoons Na^+ + {color{Green} Cl^-}})
(mathrm{KCl rightleftharpoons K^+ + {color{Green} Cl^-}})
(mathrm{CaCl_2 rightleftharpoons Ca^{2+} + {color{Green} 2 Cl^-}})
(mathrm{AlCl_3 rightleftharpoons Al^{3+} + {color{Green} 3 Cl^-}})
(mathrm{AgCl rightleftharpoons Ag^+ + {color{Green} Cl^-}})

For example, when (ce{AgCl}) is dissolved into a solution already containing (ce{NaCl}) (actually (ce{Na+}) and (ce{Cl-}) ions), the (ce{Cl-}) ions come from the ionization of both (ce{AgCl}) and (ce{NaCl}). Thus, (ce{[Cl- ]}) differs from (ce{[Ag+]}). The following examples show how the concentration of the common ion is calculated.

Example (PageIndex{1})

What are (ce{[Na+]}), (ce{[Cl- ]}), (ce{[Ca^2+]}), and (ce{[H+]}) in a solution containing 0.10 M each of (ce{NaCl}), (ce{CaCl2}), and (ce{HCl})?

Solution

Due to the conservation of ions, we have

[mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10: ce M}nonumber.]

but

[begin{alignat}{3}
&ce{[Cl- ]} &&= && && :textrm{0.10 (due to NaCl)}nonumber
& && && + &&mathrm{:0.20: (due: to: CaCl_2)}nonumber
& && && + &&mathrm{:0.10: (due: to: HCl)}nonumber
& &&= && &&mathrm{:0.40: M}nonumber
end{alignat}]

Exercise (PageIndex{1})

John poured 10.0 mL of 0.10 M (ce{NaCl}), 10.0 mL of 0.10 M (ce{KOH}), and 5.0 mL of 0.20 M (ce{HCl}) solutions together and then he made the total volume to be 100.0 mL. What is (ce{[Cl- ]}) in the final solution?

[mathrm{[Cl^-] = dfrac{0.1: Mtimes 10: mL+0.2: Mtimes 5.0: mL}{100.0: mL} = 0.020: M}nonumber]

Le Châtelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.

Example (PageIndex{2})

Consider the lead(II) ion concentration in this saturated solution of PbCl₂. The balanced reaction is

[ PbCl_{2 (s)} rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)}nonumber]

Defining (s) as the concentration of dissolved lead(II) chloride, then:

[[Pb^{2+}] = snonumber ]

[[Cl^- ] = 2snonumber]

These values can be substituted into the solubility product expression, which can be solved for (s):

[begin{align*} K_{sp} &= [Pb^{2+}] [Cl^-]^2 [4pt] &= s times (2s)^2 [4pt] 1.7 times 10^{-5} &= 4s^3 [4pt] s^3 &= frac{1.7 times 10^{-5}}{4} [4pt] &= 4.25 times 10^{-6} [4pt] s &= sqrt[3]{4.25 times 10^{-6}} [4pt] &= 1.62 times 10^{-2}, mol dm^{-3} end{align*}]

The concentration of lead(II) ions in the solution is 1.62 x 10^-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term 'common ion effect'.

Look at the original equilibrium expression again:

[ PbCl_2 ; (s) rightleftharpoons Pb^{2+} ; (aq) + 2Cl^- ; (aq)nonumber ]

What happens to that equilibrium if extra chloride ions are added? According to Le Châtelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.

Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.

A Simple Example

If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.

[[Pb^{2+}] = s label{2}]

The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution. In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.

So we assume:

[[Cl^- ] = 0.100; M label{3}] Emuladores xbox 360 rgh download.

Evernote ink. The rest of the mathematics looks like this:

begin{equation} begin{split} K_{sp}& = [Pb^{2+}][Cl^-]^2 & = s times (0.100)^2 1.7 times 10^{-5} & = s times 0.00100 end{split} end{equation}

therefore:

begin{equation} begin{split} s & = dfrac{1.7 times 10^{-5}}{0.0100} & = 1.7 times 10^{-3} , text{M} end{split} label{4} end{equation}

Common Ion Effect On Solubility Pogil