Chapter 7.23 - Solubility Equilibria of Sparingly Soluble Salts

In the previous section, we completed a discussion on buffers. In this section we will see solubility equilibrium

• In this discussion, we will be considering ionic solids like NaCl (sodium chloride), CaCl₂ (calcium chloride), CaSO₄ (calcium sulphate), PbSO₄ (lead sulphate). LiF (lithium fluoride) etc.,
◼ Some of these solids like NaCl are readily soluble in water
    ♦ Readily soluble means:
    ♦ If sufficient quantity of water is available, all the added solid will dissolve in water
◼ Some others like LiF are sparingly soluble in water
    ♦ Sparingly soluble means:
    ♦ Even if sufficient quantity of water is available, only small quantities of those solids will dissolve in water

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• Let us see why some solids are readily soluble while some others are sparingly soluble. It can be written in 5 steps:
1. When a solid is added to water, two forces compete with each other:
    ♦ The force trying to keep the solid ions together
    ♦ The force trying to separate the solid ions from each other
2. Outcome of the competitoin:
    ♦ If the first force wins, the solid will be sparingly soluble
    ♦ If the second force wins, the solid will be readily soluble
3. The first force wins when the lattice enthalpy is large
• This is because, lattice enthalpy is the energy that binds the ions together
4. The second force wins when the solvation enthalpy is large
• Solvation enthalpy is the energy released when the solid ions are separated from each other and surrounded by the solvent molecules
• An example for solvation enthalpy can be written in 5 steps:
(i) When NaCl dissolves in water, Na⁺ and Cl^- ions are separated from each other
(ii) Each Na⁺ will be surrounded by water molecules
(iii) There will be an attractive force between each Na⁺ ions and it’s surrounding water molecules
• Thus new bonds are formed between Na⁺ and water molecules
(iv) Due to those new bond formations, energy will be released
• Similarly in the case of Cl^- ions also, energy will be released
(v) The energy thus released (per mole of the NaCl) is called solvation enthalpy of NaCl
5. Comparing the two enthalpies:
• If the lattice enthalpy of a solid is larger than it’s solvaion enthalpy, that solid will be sparingly soluble
• If the lattice enthalpy of a solid is smaller than it’s solvation enthalpy, that solid will be readily soluble

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• Let us see how solubility is expressed:
◼ The solubility is expressed in moles per liter
• This can be explained using an example. It can be written in 3 steps:
1. Suppose that, a maximum of x moles of a solid can be dissolved in 1 L of water at standard temperature (298 K)
2. That means, when x moles are dissolved, the solution becomes saturated
3. Then we write: solubility of that solid = x moles per liter = x M

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• Using this unit for solubility, we can classify the solids into three categories:
Category 1: Soluble solids
    ♦ Solubility of these solids is greater than 0.1 M
        ✰ That means, it is possible to dissolve more than 0.1 moles of these solids in 1 L water
Category 2: Slightly soluble solids
    ♦ Solubility of these solids is less than 0.1 M but greater than 0.01 M
        ✰ That means, it is possible to dissolve more than 0.01 moles of these solids in 1 L water
        ✰ That also means, it is impossible to dissolve more than 0.1 moles of these solids in 1 L water
Category 3: Sparingly soluble solids
    ♦ Solubility of these solids is less than 0.01 M
        ✰ That means, it is impossible to dissolve more than 0.01 moles of these solids in 1 L water

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Solubility product constant

• This can be explained using an example. It can be written in steps:
1. Consider the solution of BaSO₄ (Barium sulfate) in water
• The equation is: BaSO₄ (s) ⇌ Ba²⁺(aq) + SO₄^2-(aq)
• This is a reversible reaction
    ♦ The solid BaSO₄ molecules will be continuously dissociating into ions
         ✰ This is the forward reaction
    ♦ The ions will be continuously combining to give back the solid molecules
         ✰ This is the backward reaction
    ♦ At equilibrium, rates of both forward and backward reactions will be the same
2. The stoichiometric coefficients show that, equal quantities of Ba²⁺ and SO₄^2- will be present
3. We know that, every reversible reaction will have an equilibrium constant K
• Let us calculate the K for our present reaction which is:
BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄^2-  (aq)
4. We have: $\mathbf\small{\rm{K=\frac{[Ba^{2+}][SO_4^{2-}]}{[BaSO_4]}}}$
• The [BaSO₄] will be a constant because, it is a pure solid
• So the denominator is a constant
    ♦ That means, we have to multiply the constant K with another constant
    ♦ Multiplying a constant with another constant gives a third constant
• That means, K get modified
    ♦ We will denote the new K as K_sp
    ♦ K_sp is called the solubility product of BaSO₄
        ✰ The subscript 'sp' stands for 'Solubility product'
• So we can write: K_sp = [Ba²⁺][SO₄^2-]
5. Experiments show that:
• K_sp for BaSO₄ is 1.1 × 10^-10 at 298 K
• Since the stoichiometric coefficients are equal, [Ba²⁺] will be equal to [SO₄^2-]
• So we can write: [Ba²⁺] = [SO₄^2-] = √(K_sp) = √(1.1 × 10^-10) = 1.05 × 10^-5
6. That means, in 1 L, there will be 1.05 × 10^-5 moles each of [Ba²⁺] and [SO₄^2-]
• That means, 1.05 × 10^-5 moles of BaSO₄ has dissolved in 1 L water
7. But number of moles of BaSO₄ which dissolve in 1 L, is the solubility of BaSO₄
• Let us denote the solubility as S
• Then we can write S of BaSO₄ = 1.05 × 10^-5
    ♦ That means, 1.05 × 10-5 moles of BaSO₄ will dissolve in 1 L of water at 298 K

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• In the above discussion, we took BaSO₄ as an example
• One molecule of BaSO₄ gives only one anion (one SO₄^2-) and one cation (one Ba²⁺)
    ♦ So the stoichiometric coefficients were equal
• But in the case of solids like Zr₃(PO₄)₄ (Zirconium phosphate), one molecule will give more than one anion and cation
    ♦ The equation is: Zr₃(PO₄)₄ ⇌ 3Zr⁴⁺ + 4PO₄^3-
• So one molecule of Zr₃(PO₄)₄ will give four anions (four PO₄^3-) and three cations (three Zr⁴⁺)
• Let us find the S of Zr₃(PO₄)₄. It can be written in 4 steps:
1. We have: K_sp = [Zr⁴⁺]³[PO₄^3-]⁴
2. Let S be the solubility of Zr₃(PO₄)₄
    ♦ That means, S moles of Zr₃(PO₄)₄ dissolves in 1 liter of water
3. If S moles dissolve, 3S  moles of Zr⁴⁺ and 4S moles of PO₄^3- will be produced
• So from (1), we get: K_sp = (3S)³ (4s)⁴ = 6912 × S⁷
⇒ $\mathbf\small{\rm{S=\left(\frac{K_{sp}}{6912} \right)^{\frac{1}{7}}}}$
4. The value of K_sp of Zr₃(PO₄)₄ can be obtained from the data book
• Using that value, we can easily calculate the S of Zr₃(PO₄)₄

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• Let us try to find a general equation for calculating S. It can be written in steps:
1. We know the special method of writing the formula of ionic compounds. Let us see some examples:
    ♦ NaCl can be written as: Na⁺Cl^-
    ♦ MgCl₂ can be written as: Mg²⁺(Cl^-)₂
    ♦ Zr₃(PO₄)₄ can be written as: (Zr⁴⁺)₃(PO₄^3-)₄
• Based on the above examples, we can write a general form for such ionic compounds as: M_x^p+X_y^q- (Details here)
2. Once we write the above general form, it is easy to write the general dissociation equation. It will be:
M_x^p+X_y^q- ⇌ xM^p+ + yX^q-
3. Once we write the above general dissociation equation, it is easy to write the general form of solubility product. It will be:
K_sp = [M^p+]^x[X^q-]^y
4. Now, if S moles of (M_x^p+X_y^q-) dissolve in water,
    ♦ There will be (xS) moles of M^p+ in water
        ✰ Then [M^p+] will be xS
    ♦ There will be (yS) moles of X^q- in water
        ✰ Then [X^q-] will be yS
5. Then the result in (3) becomes: K_sp = [xS]^x[yS]^y
⇒ K_sp = (x^x × S^x × y^y × S^y) = x^xy^yS^(x+y)
⇒ $\mathbf\small{\rm{S^{(x+y)}=\frac{K_{sp}}{x^x y^y}}}$
◼ Thus we get Eq.7.12: $\mathbf\small{\rm{S=\left(\frac{K_{sp}}{x^x y^y}\right)^{\frac{1}{(x+y)}}}}$
• The K_sp values can be obtained from the data book
6. Note that, in the final Eq.7.12, the charges +p and -q do not appear. So the charges possessed by the ions do not have any effect on S
• The subscripts x and y in M_x^p+X_y^q- are important because, they determine the stoichiometric coefficients

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• Next we will see the situation when the reaction has not reached equilibrium
• It can be written in 5 steps:
1. Let us measure the concentrations when the reading in the stop watch is t
• Assume that, at that time t, equilibrium is not reached
2. Let the concentrations be [M^p+]_t and [X^q-]_t
• These concentrations are not equilibrium concentrations
• So the product [M^p+]_t^x × [X^q-]_t^y will not give K_sp
3. We will denote the product by another letter Q
• So we get Eq.7.13: Q = [M^p+]_t^x × [X^q-]_t^y
4. If Q is less than K_sp, the system will be trying to increase Q upto K_sp
    ♦ For that, [M^p+]_t and [X^q-]_t should increase
    ♦ For that, more and more solid will dissolve
        ✰ This is forward reaction
5. If Q is greater than K_sp, the system will be trying to lower Q upto K_sp
    ♦ For that, [M^p+]_t and [X^q-]_t should decrease
    ♦ For that, more and more ions will combine to form the solid
        ✰ This is backward reaction
    ♦ When more solids are formed in this way, precipitation will begin

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Now we will see a solved example
Solved example 7.86
Calculate the solubility of A₂X₃ in pure water, assuming that neither kind of ion
reacts with water. The solubility product of A₂X₃, K_sp = 1.1 × 10^-23
Solution:
1. The dissolution of A₂X₃ will be according to the equation:
A₂X₃ ⇌ 2A^p+ + 3X^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of A₂X₃ dissolve in water,
   ♦ 2 moles of A^p+ are produced
   ♦ 3 moles of X^q- are produced
3. So when S moles of A₂X₃ dissolve in water,
   ♦ 2S moles of A^p+ are produced
   ♦ 3S moles of X^q- are produced
4. Then we can write:
K_sp = [A^p+]²[X^q-]³ = (2S)²(3S)³ = 108 × S⁵
⇒ 1.1 × 10^-23 = 108 × S⁵
⇒ S = 1.0 × 10^-5

Solved example 7.87
The values of K_sp of two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 × 10^-15 and 6 × 10^-17 respectively. Which salt is more soluble? Explain.
Solution:
• We will solve the problem in two parts
Part (a): Analysis of Ni(OH)₂
1. The dissolution of Ni(OH)₂ will be according to the equation:
Ni(OH)₂ ⇌ Ni^p+ + 2OH^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of Ni(OH)₂ dissolve in water,
   ♦ 1 mole of Ni^p+ is produced
   ♦ 2 moles of OH^q- are produced
3. So when S moles of Ni(OH)₂ dissolve in water,
   ♦ S moles of Ni^p+ are produced
   ♦ 2S moles of OH^q- are produced
4. Then we can write:
K_sp = [Ni^p+]¹[OH^q-]² = (S)¹(2S)² = 4 × S³
⇒ 2.0 × 10^-15 = 4 × S³
⇒ S = 7.94 × 10^-6
Part (b): Analysis of AgCN
1. The dissolution of AgCN will be according to the equation:
AgCN ⇌ Ag^p+ + CN^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of AgCN dissolve in water,
   ♦ 1 mole of Ag^p+ is produced
   ♦ 1 mole of CN^q- is produced
3. So when S moles of AgCN dissolve in water,
   ♦ S moles of Ag^p+ are produced
   ♦ S moles of CN^q- are produced
4. Then we can write:
K_sp = [Ag^p+]¹[CN^q-]¹ = (S)¹(S)¹ = S²
⇒ 6 × 10^-17 = S²
⇒ S = 7.75 × 10^-9
◼  Comparing the two S values, we see that, Ni(OH)₂ has greater solubility

Solved example 7.88
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Solution:
1. The dissolution of Sr(OH)₂ will be according to the equation:
Sr(OH)₂ ⇌ Sr^p+ + 2OH^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of Sr(OH)₂ dissolve in water,
   ♦ 1 mole of Sr^p+ is produced
   ♦ 2 moles of OH^q- are produced
3. So when S moles of Sr(OH)₂ dissolve in water,
   ♦ S moles of Sr^p+ are produced
   ♦ 2S moles of OH^q- are produced
4. But we are given the value of S: 19.13 g/L
• The molar mass of Sr(OH)₂ is 121.63 grams
    ♦ So 19.13 grams will contain (^19.13⁄_121.63) = 0.1573 moles of Sr(OH)₂
    ♦ That means solubility (S) is 0.1573 moles per L
5. Thus from (3), we get:
    ♦ concentration of strontium ions = S = 0.1573 moles per liter
    ♦ concentration of hydroxyl ions = 2S = (2 × 0.1573) = 0.3146 moles per liter
6. We have: pOH = -1 × logarithm of hydroxyl ion concentration
= -log₁₀(0.30) = 0.5228
• So pH = (14 - pOH) = (14 - 0.5228) = 13.48

Solved example 7.89
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide,
lead chloride and mercurous iodide at 298K from their solubility product constants given in Standard Table. Determine also the molarities of individual ions.
Solution:
Part (a): Analysis of Ag₂CrO₄ (Silver chromate)
1. The dissolution of Ag₂CrO₄ will be according to the equation:
Ag₂CrO₄ ⇌ 2Ag^p+ + CrO₄^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of Ag₂CrO₄ dissolve in water,
   ♦ 2 moles of Ag^p+ are produced
   ♦ 1 mole of CrO₄^q- is produced
3. So when S moles of Ag₂CrO₄ dissolve in water,
   ♦ 2S moles of Ag^p+ are produced
   ♦ S moles of CrO₄^q- are produced
4. Then we can write:
K_sp = [Ag^p+]²[CrO₄^q-]¹ = (2S)²(S)¹ = 4 × S³
5. From the table, we have: K_sp of Ag₂CrO₄ = 1.1 × 10^-12
• So we can write: 1.1 × 10^-12 = 4 × S³
⇒ S = 6.5 × 10^-5
6. Molarities of individual ions:
    ♦ Molarity of Ag^p+ = 2S = 2 × 6.5 × 10^-5  = 13.0 × 10^-5 
    ♦ Molarity of CrO₄^q- = S = 6.5 × 10^-5

Part (b): Analysis of BaCrO₄ (Barium chromate)
1. The dissolution of BaCrO₄ will be according to the equation:
BaCrO₄ ⇌ Ba^p+ + CrO₄^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of BaCrO₄ dissolve in water,
   ♦ 1 moles of Ba^p+ are produced
   ♦ 1 mole of CrO₄^q- is produced
3. So when S moles of BaCrO₄ dissolve in water,
   ♦ 12S moles of Ba^p+ are produced
   ♦ S moles of CrO₄^q- are produced
4. Then we can write:
K_sp = [Ba^p+]¹[CrO₄^q-]¹ = (S)¹(S)¹ = S²
5. From the table, we have: K_sp of BaCrO₄ = 1.2 × 10^-10
• So we can write: 1.2 × 10^-10 = S²
⇒ S = 1.1 × 10^-5
6. Molarities of individual ions:
    ♦ Molarity of Ba^p+ = S = 1.1 × 10^-5
    ♦ Molarity of CrO₄^q- = S = 1.1 × 10^-5 xxx

Part (c): Analysis of Fe(OH)₃ (Ferric hydroxide)
1. The dissolution of Fe(OH)₃ will be according to the equation:
Fe(OH)₃ ⇌ Fe^p+ + 3OH^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of Fe(OH)₃ dissolve in water,
   ♦ 1 mole of Fe^p+ is produced
   ♦ 3 moles of OH^q- are produced
3. So when S moles of Fe(OH)₃ dissolve in water,
   ♦ S moles of Fe^p+ are produced
   ♦ 3S moles of OH^q- are produced
4. Then we can write:
K_sp = [Fe^p+]¹[OH^q-]³ = (S)¹(3S)³ = 27 × S⁴
5. From the table, we have: K_sp of Fe(OH)₃ = 1.0 × 10^-38
• So we can write: 1.0 × 10^-38 = 27 × S⁴
⇒ S = 1.38 × 10^-10
6. Molarities of individual ions:
    ♦ Molarity of Fe^p+ = S = 1.39 × 10^-10
    ♦ Molarity of OH^q- = 3S = 3 × 1.38 × 10^-10  = 4.17 × 10^-10

Part (d): Analysis of PbCl₂ (Ferric hydroxide)
1. The dissolution of PbCl₂ will be according to the equation:
PbCl₂ ⇌ Pb^p+ + 2Cl^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of PbCl₂ dissolve in water,
   ♦ 1 mole of Pb^p+ is produced
   ♦ 2 moles of Cl^q- are produced
3. So when S moles of PbCl₂ dissolve in water,
   ♦ S moles of Pb^p+ are produced
   ♦ 2S moles of Cl^q- are produced
4. Then we can write:
K_sp = [Pb^p+]¹[Cl^q-]² = (S)¹(2S)² = 4 × S³
5. From the table, we have: K_sp of PbCl₂ = 1.6 × 10^-5
• So we can write: 1.6 × 10^-5 = 4 × S³
⇒ S = 1.59 × 10^-2
6. Molarities of individual ions:
    ♦ Molarity of Pb^p+ = S = 1.59 × 10^-2 
    ♦ Molarity of Cl^q- = 2S = 2 × 1.59 × 10^-2  = 3.18 × 10^-2  xxx

Part (e): Analysis of Hg₂I₂ (Mercurous Iodide)
1. The dissolution of Hg₂I₂ will be according to the equation:
Hg₂I₂ ⇌ Hg₂^p+ + 2I^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of Hg₂I₂ dissolve in water,
   ♦ 2 mole of Hg₂^p+ is produced
   ♦ 2 moles of I^q- are produced
3. So when S moles of PbCl₂ dissolve in water,
   ♦ 2S moles of Hg₂^p+ are produced
   ♦ 2S moles of I^q- are produced
4. Then we can write:
K_sp = [Hg₂^p+]¹[I^q-]² = (S)¹(2S)² = 4 × S³
5. From the table, we have: K_sp of Hg₂I₂ = 4.5 × 10^-29
• So we can write: 4.5 × 10^-29 = 4 × S³
⇒ S = 2.24 × 10^-10
6. Molarities of individual ions:
    ♦ Molarity of Hg₂^p+ = S = 2.24 × 10^-10
    ♦ Molarity of I^q- = 2S = 2 × 2.24 × 10^-10  = 4.48 × 10^-10

Solved example 7.90
The solubility product constant of Ag₂CrO₄ and AgBr are 1.1 × 10^-12 and
5.0 × 10^-13 respectively. Calculate the ratio of the molarities of their saturated
solutions.
Solution:
• We will solve the problem in two parts
Part (a): Analysis of Ag₂CrO₄ (Silver chromate)
1. The dissolution of Ag₂CrO₄ will be according to the equation:
Ag₂CrO₄ ⇌ 2Ag^p+ + CrO₄^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of Ag₂CrO₄ dissolve in water,
   ♦ 2 moles of Ag^p+ are produced
   ♦ 1 mole of CrO₄^q- is produced
3. So when S moles of Ag₂CrO₄ dissolve in water,
   ♦ 2S moles of Ag^p+ are produced
   ♦ S moles of CrO₄^q- are produced
4. Then we can write:
K_sp = [Ag^p+]²[CrO₄^q-]¹ = (2S)²(S)¹ = 4 × S³
⇒ 1.1 × 10^-12 = 4 × S³
⇒ S = 6.5 × 10^-5

Part (b): Analysis of AgBr
1. The dissolution of AgBr will be according to the equation:
AgBr ⇌ Ag^p+ + Br^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of AgCN dissolve in water,
   ♦ 1 mole of Ag^p+ is produced
   ♦ 1 mole of Br^q- is produced
3. So when S moles of AgBr dissolve in water,
   ♦ S moles of Ag^p+ are produced
   ♦ S moles of Br^q- are produced
4. Then we can write:
K_sp = [Ag^p+]¹[Br^q-]¹ = (S)¹(S)¹ = S²
⇒ 5 × 10^-13 = S²
⇒ S = 7.07 × 10^-7
◼  In a saturated solution, the 'maximum possible quantity' of the solute will be present
• The maximum possible quantity is S
• So the required ratio is:
$\mathbf\small{\rm{\frac{\text{Solubility of}\,Ag_2CrO_4}{\text{Solubility of}\,AgBr}=\frac{6.5 \times 10^{-5}}{7.07 \times 10^{-7}}}}$ = 91.9

Solved example 7.91
What is the minimum volume of water required to dissolve 1 gram of calcium sulphate at 298 K (For calcium sulphate, K_sp is 9.1 × 10^-6)
Solution:
1. The dissolution of CaSO₄ will be according to the equation:
CaSO₄ ⇌ Ca^p+ + SO₄^q-
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of CaSO₄ dissolve in water,
   ♦ 1 mole of Ca^p+ is produced
   ♦ 1 mole of SO₄^q- is produced
3. So when S moles of AgBr dissolve in water,
   ♦ S moles of Ca^p+ are produced
   ♦ S moles of SO₄^q- are produced
4. Then we can write:
K_sp = [Ca^p+]¹[SO₄^q-]¹ = (S)¹(S)¹ = S²
⇒ 9.1 × 10^-6 = S²
⇒ S = 3.02 × 10^-3
5. So in 1 liter water, a maximum of 3.02 × 10^-3 moles of CaSO₄ will dissolve
• We have to convert this mole into grams
• Molar mass of CaSO₄ is 136.14 grams
   ♦ That means, one mole of CaSO₄ is 136.14 grams
• So 3.02 × 10^-3 moles is (3.02 × 10^-3 × 136.14) = 0.411 grams
   ♦ That means, 1 liter water is required to dissolve 0.411 grams of CaSO₄
• So the volume of water required to dissolve one gram of CaSO₄ = ¹⁄_0.411 = 2.43 liters

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• In the next section, we will see common ion effect on solubility


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